Solution:
By properties of logarithms, we can simplify the given equation to $x\log_xy=4y\log_yx=10$. Let us break this into two separate equations:
\[x\log_xy=10\]
\[4y\log_yx=10.\]
We multiply the two equations to get:
\[4xy\left(\log_xy\log_yx\right)=100.\]
Also by properties of logarithms, we know that $\log_ab\cdot\log_ba=1$; thus, $\log_xy\cdot\log_yx=1$. Therefore, our equation simplifies to:
\[4xy=100\implies xy=\boxed{025}.\]
~Technodoggo
Convert the two equations into exponents:
\[x^{10}=y^x~(1)\]
\[y^{10}=x^{4y}~(2).\]
Take $(1)$ to the power of $\frac{1}{x}$:
\[x^{\frac{10}{x}}=y.\]
Plug this into $(2)$:
\[x^{(\frac{10}{x})(10)}=x^{4(x^{\frac{10}{x}})}\]
\[{\frac{100}{x}}={4x^{\frac{10}{x}}}\]
\[{\frac{25}{x}}={x^{\frac{10}{x}}}=y,\]
So $xy=\boxed{025}$
~alexanderruan
Similar to solution 2, we have:
$x^{10}=y^x$ and $y^{10}=x^{4y}$
Take the tenth root of the first equation to get
$x=y^{\frac{x}{10}}$
Substitute into the second equation to get
$y^{10}=y^{\frac{4xy}{10}}$
This means that $10=\frac{4xy}{10}$, or $100=4xy$, meaning that $xy=\boxed{25}$.
~MC413551

Answer: 025

<｜begin▁of▁sentence｜><｜User｜>There exist real numbers $x$ and $y$, both greater than 1, such that $\log_x\left(y^x\right)=\log_y\left(x^{4y}\right)=10$. Find $xy$.<｜Assistant｜><think>

Okay, so I have this problem here: There exist real numbers x and y, both greater than 1, such that log base x of y^x equals 10, and log base y of x^{4y} also equals 10. I need to find the product xy. Hmm, logarithms with variables as bases and exponents. Let me try to recall logarithm properties and see how to approach this.

First, let me write down the given equations. The first one is log_x(y^x) = 10. The second one is log_y(x^{4y}) = 10. Since both equations are equal to 10, maybe I can set them equal to each other or find a way to relate x and y through these equations.

Starting with the first equation: log_x(y^x) = 10. I remember that log_b(a^c) = c * log_b(a). So applying that property here, this becomes x * log_x(y) = 10. Hmm, so x times the logarithm of y with base x equals 10. Maybe I can rewrite the logarithm in terms of natural logs or something else?

Alternatively, maybe it's better to convert the logarithmic equations into exponential form. Let me try that. For the first equation, log_x(y^x) = 10 means that x^10 = y^x. Because log_b(a) = c is equivalent to b^c = a. So here, the base is x, the result is 10, so x^10 = y^x. Got that equation: x^{10} = y^{x}.

Similarly, the second equation: log_y(x^{4y}) = 10. Converting that, we get y^{10} = x^{4y}. So the second equation is y^{10} = x^{4y}.

Now, so we have two equations:

1) x^{10} = y^{x}

2) y^{10} = x^{4y}

Hmm, so we need to solve for x and y here. Since both equations involve exponents with x and y in both the base and exponent, this might be a bit tricky. Maybe taking logarithms on both sides would help? Let me try that.

Starting with the first equation: x^{10} = y^{x}. Taking natural logarithm of both sides: ln(x^{10}) = ln(y^x). Using the power rule again, that becomes 10 ln x = x ln y. Similarly, for the second equation: y^{10} = x^{4y}, take natural logs: 10 ln y = 4y ln x.

So now we have two equations:

1) 10 ln x = x ln y

2) 10 ln y = 4y ln x

Hmm, so maybe we can denote some variables here to simplify. Let me let a = ln x and b = ln y. Then, since x and y are greater than 1, a and b are positive real numbers.

Substituting, the first equation becomes 10a = x b. Wait, but x is e^{a}, since a = ln x. Similarly, y is e^{b}. So substituting x = e^{a} and y = e^{b}, then:

First equation: 10a = e^{a} * b

Second equation: 10b = 4 e^{b} * a

Hmm, so now we have:

1) 10a = e^{a} * b

2) 10b = 4 e^{b} * a

Hmm, this seems a bit complicated, but maybe we can solve for one variable in terms of the other and substitute.

From the first equation, let's solve for b: b = (10a)/e^{a}

Then plug this into the second equation:

10*(10a / e^{a}) = 4 e^{b} * a

Simplify left side: 100a / e^{a} = 4 a e^{b}

We can divide both sides by a, since a is positive (because x >1, so ln x ≠0), so:

100 / e^{a} = 4 e^{b}

Then, divide both sides by 4:

25 / e^{a} = e^{b}

So e^{b} = 25 e^{-a}

Take natural log of both sides: b = ln(25) - a

But from earlier, we had b = (10a)/e^{a}

Therefore, setting those equal:

(10a)/e^{a} = ln(25) - a

Hmm, so now we have an equation with just a. Let me write that:

(10a)/e^{a} + a = ln(25)

Factor out a:

a (10/e^{a} + 1) = ln(25)

Hmm, this is a transcendental equation, which might not have an algebraic solution. Maybe we can find a substitution or guess a solution?

Alternatively, perhaps there is a clever substitution or relationship between x and y. Let me think again about the original equations.

We have x^{10} = y^{x} and y^{10} = x^{4y}

Maybe we can express y in terms of x from the first equation and substitute into the second?

From the first equation: y^x = x^{10}, so y = (x^{10})^{1/x} = x^{10/x}

Similarly, from the second equation: x^{4y} = y^{10}, so substituting y from above:

x^{4y} = (x^{10/x})^{10} = x^{100/x}

Therefore, x^{4y} = x^{100/x}

Since the bases are the same and x >1, the exponents must be equal:

4y = 100/x

So 4y = 100/x => y = 25/x

Ah! That's a useful relationship: y = 25/x. So y is 25 over x. So now we can substitute this into one of the previous equations.

We had y = x^{10/x} from the first equation, and also y =25/x from the second equation. Therefore, setting them equal:

x^{10/x} = 25/x

Hmm, so x^{10/x} * x =25

Which is x^{10/x +1} =25

Hmm, that's still a bit complicated. Let me write that again:

x^{10/x} = 25/x

Multiply both sides by x:

x^{10/x} * x =25

Which is x^{10/x +1} =25

Alternatively, perhaps take natural logs of both sides of the equation x^{10/x} =25/x.

Taking ln of both sides:

(10/x) ln x = ln(25/x) = ln25 - ln x

So:

(10/x) ln x + ln x = ln25

Factor out ln x:

ln x (10/x +1) = ln25

Hmm, this is similar to the equation I had earlier with a substitution. Let me let a = ln x again, then:

a (10/e^{a} +1 ) = ln25

Wait, but maybe with the substitution y =25/x, perhaps we can find x and y more directly?

We have y =25/x, so xy =25. Wait, is that possible? If y =25/x, then xy=25. So is the answer 25? Wait, but let me check that.

Wait, hold on. If we have y =25/x, then indeed xy=25. But let me confirm if that's consistent with the equations.

Wait, but let me see. Let me suppose that xy=25, so y=25/x. Let me check if this holds with the original equations.

Wait, but let me see if that's the case. Let me suppose that xy=25, so y=25/x. Let me plug this into the first equation: x^{10}= y^x = (25/x)^x. So x^{10} = (25^x)/(x^x). Multiply both sides by x^x:

x^{10} * x^x =25^x

x^{x +10} =25^x

Take both sides to the power 1/x:

x^{(x +10)/x} =25

x^{1 +10/x} =25

Hmm, so x^{1 +10/x} =25. Let me see if this can be true for some x.

Alternatively, let me suppose that x is 5. Let me test x=5. Then y=25/5=5. So x=y=5. Let me check the original equations.

First equation: log_5(5^5)= log_5(3125)=5, but the problem says it should be 10. So that's not working. Hmm, so x=5 and y=5 is not a solution.

Hmm, so maybe my assumption that xy=25 is incorrect? Wait, but we derived y=25/x from the equations. Wait, let me check that step again.

From the second equation, after substituting y from the first equation:

We had from first equation: y =x^{10/x}

Then substituted into the second equation: x^{4y}= y^{10}

So x^{4y}= (x^{10/x})^{10}=x^{100/x}

Therefore, exponents must be equal: 4y =100/x, so y=25/x. That seems correct. So that relationship must hold. Therefore, y=25/x, so xy=25. But when I tested with x=5, it didn't work. So maybe there is another x and y where xy=25, but x and y are not 5?

Wait, perhaps x and y are different numbers such that their product is 25. Let me try x= something else. Let me suppose x=25, then y=1. But y must be greater than 1, so that's invalid. How about x= sqrt(25)=5, but that gives y=5 again. Hmm.

Alternatively, maybe x= something like 25^{1/ something}. Let me think. Let me try to proceed with the equation x^{1 +10/x}=25.

Let me set t = x. So the equation is t^{1 +10/t}=25. Let me take natural logs:

(1 +10/t) ln t = ln25.

Hmm, this equation might require some trial and error. Let me see if I can find a t that satisfies this.

Suppose t=5. Then left side: (1 +10/5) ln5 = (1+2) ln5 =3 ln5 ≈3*1.609≈4.828. But ln25≈3.218. Not equal.

Hmm, too big. Let me try t= something larger. Let me try t=10.

Left side: (1 +10/10) ln10 =2 * ln10≈2*2.302≈4.605. Still bigger than ln25≈3.218.

Hmm, need to make the left side smaller. Let me try t=20.

(1 +10/20) ln20 = (1.5)*ln20≈1.5*2.995≈4.493. Still too big.

Hmm, maybe t=25?

(1 +10/25) ln25 = (1.4)*3.218≈4.505. Still too big.

Hmm, going the other way. Maybe t= something less than 5?

Wait, but x has to be greater than 1. Let me try t=2.

(1 +10/2) ln2 =6 *0.693≈4.158. Still bigger than 3.218.

Hmm, t=1. Let me see, but x must be greater than1. Let me try t approaching 1 from the right. If t approaches 1, then (1 +10/t) approaches 11, and ln t approaches 0. So the product approaches 0. Not helpful.

Hmm, maybe I made a mistake in the derivation. Let me check again.

Starting over:

First equation: log_x(y^x)=10 ⇒ x^{10}=y^x.

Second equation: log_y(x^{4y})=10 ⇒ y^{10}=x^{4y}

From first equation: y = x^{10/x}

Substitute into second equation: (x^{10/x})^{10} =x^{4y}

So x^{100/x}=x^{4y} ⇒ 100/x =4y ⇒ y=25/x. So that's correct.

Therefore, y=25/x. So substituting back into first equation, x^{10}=(25/x)^x.

So x^{10} = (25^x)/(x^x). Multiply both sides by x^x:

x^{10} *x^x =25^x ⇒x^{x +10}=25^x.

Divide both sides by x^{x}:

x^{10}= (25^x)/(x^x)= (25/x)^x.

Wait, so x^{10}=(25/x)^x. Hmm, so x^{10}=(25/x)^x. Let me write this as (25/x)^x =x^{10}

Take both sides to the power 1/x:

25/x =x^{10/x}

Multiply both sides by x:

25 =x^{10/x +1}

Wait, so 25 =x^{1 +10/x}

Which is the same as before. Hmm. Let me set z = x. So 25 = z^{1 +10/z}

Take natural logs: ln25 = (1 +10/z) ln z

Let me denote f(z) = (1 +10/z) ln z. We need to find z>1 such that f(z)=ln25≈3.218.

Let me try z=5: f(5)= (1+2) ln5≈3*1.609≈4.828>3.218

z=10: (1+1) ln10≈2*2.302≈4.604>3.218

z=15: (1 +10/15) ln15≈(1.6667)*2.708≈4.413>3.218

z=20: (1+0.5) ln20≈1.5*2.995≈4.493>3.218

Hmm, still too big. Wait, maybe z=25?

(1 +10/25) ln25≈1.4*3.218≈4.505>3.218

Hmm, still higher. Wait, so as z increases, 10/z decreases, so 1 +10/z approaches 1, and ln z increases. So the product might have a minimum?

Wait, perhaps I need to try a larger z? Let me try z=50.

(1 +10/50) ln50 =1.2 * 3.912≈4.694. Still higher.

Hmm, z=100: (1 +0.1) ln100≈1.1*4.605≈5.065. Hmm, even higher. Wait, so as z increases, the function first increases? Wait, maybe I need to check derivative.

Alternatively, maybe trying smaller z?

Wait, z= sqrt(25)=5, but that was too big. Wait, let me try z= something like 3.

z=3: (1 +10/3) ln3≈(4.333)*1.098≈4.766>3.218

Hmm, z=4:

(1 +10/4) ln4= (1+2.5)*1.386≈3.5*1.386≈4.851>3.218

Hmm, z=1. Let me see z approaching 1 from the right. Let z=1.1:

(1 +10/1.1) ln1.1≈(1+9.0909)*0.0953≈10.0909*0.0953≈0.963. That's less than 3.218.

Ah! So at z=1.1, f(z)=0.963, which is less than 3.218, and at z=5, it's 4.828. So the function increases from z=1 to some point?

Wait, so between z=1.1 and z=5, the function increases from ~1 to ~4.8, so it must cross 3.218 somewhere between z=1.1 and z=5. Let me try z=2.

Wait, z=2 gives f(z)= (1+5)*ln2≈6*0.693≈4.158>3.218

Hmm, so between z=1.1 and z=2, the function goes from 0.96 to 4.15. So crosses 3.218 somewhere between z=1.5?

Wait, let me try z=1.5:

(1 +10/1.5) ln1.5≈(1+6.6667)*0.4055≈7.6667*0.4055≈3.111. Close to 3.218.

Hmm, 3.111 vs 3.218. So a bit lower. Let me try z=1.6:

(1 +10/1.6) ln1.6≈(1 +6.25)*0.4700≈7.25*0.4700≈3.4025. That's over 3.218.

So between z=1.5 and z=1.6, f(z) crosses 3.218.

Let me try z=1.55:

10/1.55≈6.4516, so 1+6.4516≈7.4516

ln(1.55)≈0.4383

Multiply: 7.4516 *0.4383≈3.266. Hmm, that's close to 3.218.

Hmm, z=1.55 gives ~3.266, which is a bit higher than 3.218.

Try z=1.53:

10/1.53≈6.5359, so 1+6.5359≈7.5359

ln(1.53)≈0.4245

Multiply: 7.5359 *0.4245≈3.200. That's very close to 3.218.

Hmm, 3.200 vs 3.218. Close. Let me try z=1.535:

10/1.535≈6.517, so 1+6.517≈7.517

ln(1.535)≈0.4273

Multiply: 7.517 *0.4273≈3.208. Hmm, getting closer.

z=1.54:

10/1.54≈6.4935, so total 7.4935

ln(1.54)= approx 0.4333

Multiply: 7.4935 *0.4333≈3.246. Hmm, over again.

Hmm, so between 1.53 and 1.54. Let me try z=1.535:

Wait, already did that. Maybe z=1.532:

10/1.532≈6.529, so total 7.529

ln(1.532)= approx 0.426

7.529 *0.426≈3.206. Hmm.

Hmm, this is getting tedious. Maybe there's an exact solution?

Wait, perhaps the equation can be manipulated to find an exact value?

We have x^{1 +10/x}=25. Let me suppose that 1 +10/x is an exponent that can be expressed as a logarithm. Alternatively, maybe x is 25^{something}?

Alternatively, let me suppose that x=5^k for some k. Let me try substituting x=5^k. Then y=25/x=25/(5^k)=5^{2 -k}

Let me see if this substitution helps.

First equation: x^{10}=y^x ⇒ (5^k)^{10}= (5^{2 -k})^{5^k}

Simplify left side:5^{10k}

Right side:5^{(2 -k)*5^k}

Therefore, exponents must be equal:

10k = (2 -k)*5^k

Hmm, so 10k = (2 -k)5^k. Let me see if this equation can be solved for k.

Trying k=1: 10*1= (2-1)*5^1 ⇒10=5. Not equal.

k=2: 10*2= (2-2)*5^2 ⇒20=0. Nope.

k=0.5: 10*(0.5)= (2 -0.5)*5^{0.5} ⇒5=1.5*sqrt(5)≈1.5*2.236≈3.354. Not equal.

k= something else. Let me try k= ?

Hmm, maybe k= ?

Alternatively, let me try k= log something. Not sure.

Alternatively, perhaps k=2?

Wait, tried that. Hmm.

Alternatively, maybe k= - something? But x must be greater than1, so 5^k>1 ⇒k>0.

Hmm, maybe k= 1/2?

Wait tried that. Hmm.

Alternatively, let me try k= log_5 (something). Hmm, not sure.

Alternatively, perhaps this approach isn't helpful. Maybe there's another way.

Wait, going back to the equations:

We have two equations:

1) x^{10} = y^x

2) y^{10}=x^{4y}

And we have y=25/x. Let me substitute y=25/x into the first equation:

x^{10} = (25/x)^x ⇒x^{10} =25^x /x^x ⇒x^{10}*x^x =25^x ⇒x^{x +10}=25^x

Let me write this as (x^{(x +10)/x}) =25 ⇒x^{1 +10/x}=25

Let me set t =10/x. Then, 10/x =t ⇒x=10/t. Substitute into the equation:

(10/t)^{1 + t} =25

Hmm, so (10/t)^{1 + t}=25. Let me see if this can be solved for t.

Take natural logs: (1 + t) ln(10/t)=ln25

Hmm, still complicated. Maybe guess a value for t?

Suppose t=2, so x=5. Then left side: (1+2) ln(10/2)=3 ln5≈3*1.609≈4.828 vs ln25≈3.218. Not equal.

t=4, so x=2.5. Then (1+4) ln(10/4)=5 ln(2.5)≈5*0.916≈4.58. Still too big.

t=5, x=2. (1+5) ln(10/5)=6 ln2≈6*0.693≈4.158. Still too big.

t=6, x≈1.6667. (1+6) ln(10/6)=7 ln(1.6667)≈7*0.5108≈3.576. Closer to 3.218.

t=7, x≈1.4286. (8) ln(10/7)=8*ln(1.4286)≈8*0.3567≈2.853. Now it's less than 3.218.

Hmm, so between t=6 and t=7. Let me try t=6.5, x≈10/6.5≈1.538.

Then (1+6.5)=7.5, ln(10/6.5)=ln(1.5385)≈0.432.

Multiply:7.5*0.432≈3.24. Close to 3.218.

Hmm, so t≈6.5 gives 3.24, which is close. Let me try t=6.6:

x=10/6.6≈1.5151

Then (1+6.6)=7.6, ln(10/6.6)=ln(1.5151)≈0.4155

Multiply:7.6*0.4155≈3.16. Hmm, that's under.

Hmm, so between t=6.5 and t=6.6. Let me try t=6.55:

x≈10/6.55≈1.5267

(1+6.55)=7.55, ln(10/6.55)=ln(1.5267)≈0.422

Multiply:7.55*0.422≈3.20. Close to 3.218.

Hmm, t=6.53:

x≈10/6.53≈1.5314

(1+6.53)=7.53, ln(10/6.53)=ln(1.5314)≈0.426

Multiply:7.53*0.426≈3.206. Hmm, very close.

Hmm, so t≈6.53, so x≈1.5314, then y=25/x≈25/1.5314≈16.33.

Wait, but this is getting into approximate values. The problem says "real numbers", so maybe the answer is 25? But when I tested x=5 and y=5, it didn't work, but maybe there's another solution where xy=25?

Wait, but according to the equations, we derived that y=25/x, so their product must be 25. Therefore, even if the individual values of x and y are not integers, their product is 25. Therefore, the answer is 25?

Wait, but in the problem statement, it says "there exist real numbers x and y, both greater than 1", so even if we can't find exact values for x and y, their product is fixed at 25. Therefore, maybe the answer is 25?

Wait, but let me confirm with the equations. Since we derived that y=25/x from the two equations, so regardless of the values of x and y, their product must be 25. Therefore, the answer is 25. So maybe I overcomplicated it by trying to find x and y?

Wait, let me see again:

We had the two equations:

1) x^{10}=y^x

2) y^{10}=x^{4y}

From equation 1, express y in terms of x: y =x^{10/x}

Substitute into equation 2:

(y)^10 = (x^{10/x})^{10}=x^{100/x}

And equation 2 says that equals x^{4y}, so:

x^{100/x}=x^{4y} ⇒100/x=4y ⇒ y=25/x.

Therefore, y=25/x, so xy=25. Therefore, regardless of the values of x and y, their product must be 25. Therefore, the answer is 25. So even if x and y are not integers, their product is 25. Therefore, the problem is designed so that you don't need to find x and y individually, just their product, which is 25.

Therefore, the answer is 25. I think I overcomplicated it by trying to solve for x and y. The key was to realize that substituting one equation into the other gives y=25/x, so product is 25.

Wait, but let me check with the approximate values I had before. If x≈1.53 and y≈16.33, then xy≈1.53*16.33≈25, which checks out. So even though x and y are not nice numbers, their product is exactly 25. Therefore, the problem is designed so that you can find xy without solving for x and y explicitly. So the answer is 25.

**Final Answer**
\boxed{25}
</think>

Given the equations \(\log_x(y^x) = 10\) and \(\log_y(x^{4y}) = 10\), we start by converting these logarithmic equations into their exponential forms.

For the first equation:
\[
\log_x(y^x) = 10 \implies x^{10} = y^x
\]

For the second equation:
\[
\log_y(x^{4y}) = 10 \implies y^{10} = x^{4y}
\]

We now have two equations:
1. \(x^{10} = y^x\)
2. \(y^{10} = x^{4y}\)

From the first equation, we express \(y\) in terms of \(x\):
\[
y = x^{10/x}
\]

Substituting this expression for \(y\) into the second equation:
\[
(x^{10/x})^{10} = x^{4y}
\]
\[
x^{100/x} = x^{4y}
\]

Since the bases are the same, we equate the exponents:
\[
\frac{100}{x} = 4y \implies y = \frac{25}{x}
\]

Thus, we find that \(y = \frac{25}{x}\). Therefore, the product \(xy\) is:
\[
xy = x \cdot \frac{25}{x} = 25
\]

Hence, the final answer is:
\[
\boxed{25}
\]


Solution:
Let's first try some experimentation. Alice obviously wins if there is one coin. She will just take it and win. If there are 2 remaining, then Alice will take one and then Bob will take one, so Bob wins. If there are $3$, Alice will take $1$, Bob will take one, and Alice will take the final one. If there are $4$, Alice will just remove all $4$ at once. If there are $5$, no matter what Alice does, Bob can take the final coins in one try. Notice that Alice wins if there are $1$, $3$, or $4$ coins left. Bob wins if there are $2$ or $5$ coins left.
After some thought, you may realize that there is a strategy for Bob. If there is n is a multiple of $5$, then Bob will win. The reason for this is the following: Let's say there are a multiple of $5$ coins remaining in the stack. If Alice takes $1$, Bob will take $4$, and there will still be a multiple of $5$. If Alice takes $4$, Bob will take $1$, and there will still be a multiple of $5$. This process will continue until you get $0$ coins left. For example, let's say there are $205$ coins. No matter what Alice does, Bob can simply just do the complement. After each of them make a turn, there will always be a multiple of $5$ left. This will continue until there are $5$ coins left, and Bob will end up winning.
After some more experimentation, you'll realize that any number that is congruent to $2$ mod $5$ will also work. This is because Bob can do the same strategy, and when there are $2$ coins left, Alice is forced to take $1$ and Bob takes the final coin. For example, let's say there are $72$ coins. If Alice takes $1$, Bob will take $4$. If Alice takes $4$, Bob will take $1$. So after they each make a turn, the number will always be equal to $2$ mod $5$. Eventually, there will be only $2$ coins remaining, and we've established that Alice will simply take $1$ and Bob will take the final coin.
So we have to find the number of numbers less than or equal to $2024$ that are either congruent to $0$ mod $5$ or $2$ mod $5$. There are $404$ numbers in the first category: $5, 10, 15, \dots, 2020$. For the second category, there are $405$ numbers. $2, 7, 12, 17, \dots, 2022$. So the answer is $404 + 405 = \boxed{809}$
~lprado
We will use winning and losing positions, where a $W$ marks when Alice wins and an $L$ marks when Bob wins.
$1$ coin: $W$
$2$ coins: $L$
$3$ coins: $W$
$4$ coins: $W$
$5$ coins: $L$
$6$ coin: $W$
$7$ coins: $L$
$8$ coins: $W$
$9$ coins: $W$
$10$ coins: $L$
$11$ coin: $W$
$12$ coins: $L$
$13$ coins: $W$
$14$ coins: $W$
$15$ coins: $L$
We can see that losing positions occur when $n$ is congruent to $0, 2 \mod{5}$ and winning positions occur otherwise. In other words, there will be $2$ losing positions out of every $5$ consecutive values of n. As $n$ ranges from $1$ to $2020$, $\frac{2}{5}$ of these values are losing positions where Bob will win. As $n$ ranges from $2021$ to $2024$, $2022$ is the only value where Bob will win. Thus, the answer is $2020\times\frac{2}{5}+1=\boxed{809}$
~alexanderruan
Denote by $A_i$ and $B_i$ Alice's or Bob's $i$th moves, respectively.
Case 1: $n \equiv 0 \pmod{5}$.
Bob can always take the strategy that $B_i = 5 - A_i$.
This guarantees him to win.
In this case, the number of $n$ is $\left\lfloor \frac{2024}{5} \right\rfloor = 404$.
Case 2: $n \equiv 1 \pmod{5}$.
In this case, consider Alice's following strategy: $A_1 = 1$ and $A_i = 5 - B_{i-1}$ for $i \geq 2$.
Thus, under Alice's this strategy, Bob has no way to win.
Case 3: $n \equiv 4 \pmod{5}$.
In this case, consider Alice's following strategy: $A_1 = 4$ and $A_i = 5 - B_{i-1}$ for $i \geq 2$.
Thus, under Alice's this strategy, Bob has no way to win.
Case 4: $n \equiv 2 \pmod{5}$.
Bob can always take the strategy that $B_i = 5 - A_i$.
Therefore, after the $\left\lfloor \frac{n}{5} \right\rfloor$th turn, there are two tokens leftover.
Therefore, Alice must take 1 in the next turn that leaves the last token on the table.
Therefore, Bob can take the last token to win the game.
This guarantees him to win.
In this case, the number of $n$ is $\left\lfloor \frac{2024 - 2}{5} \right\rfloor +1 = 405$.
Case 5: $n \equiv 3 \pmod{5}$.
Consider Alice's following strategy: $A_1 = 1$ and $A_i = 5 - B_{i-1}$ for $i \geq 2$.
By doing so, there will finally be 2 tokens on the table and Bob moves first. Because Bob has the only choice of taking 1 token, Alice can take the last token and win the game.
Therefore, in this case, under Alice's this strategy, Bob has no way to win.
Putting all cases together, the answer is $404 + 405 = \boxed{\textbf{(809) }}$.
Since the game Alice and Bob play is impartial (the only difference between player 1 and player 2 is that player 1 goes first (note that games like chess are not impartial because each player can only move their own pieces)), we can use the Sprague-Grundy Theorem to solve this problem. We will use induction to calculate the Grundy Values for this game.
We claim that heaps of size congruent to $0,2 \bmod{5}$ will be in outcome class $\mathcal{P}$ (win for player 2 = Bob), and heaps of size equivalent to $1,3,4 \bmod{5}$ will be in outcome class $\mathcal{N}$ (win for player 1 = Alice). Note that the mex (minimal excludant) of a set of nonnegative integers is the least nonnegative integer not in the set. e.g. mex$(1, 2, 3) = 0$ and mex$(0, 1, 2, 4) = 3$.

$\text{heap}(0) = \{\} = *\text{mex}(\emptyset) = 0$

$\text{heap}(1) = \{0\} = *\text{mex}(0) = *$

$\text{heap}(2) = \{*\} = *\text{mex}(1) = 0$

$\text{heap}(3) = \{0\} = *\text{mex}(0) = *$

$\text{heap}(4) = \{0, *\} = *\text{mex}(0, 1) = *2$

$\text{heap}(5) = \{*, *2\} = *\text{mex}(1, 2) = 0$

$\text{heap}(6) = \{0, 0\} = *\text{mex}(0, 0) = *$

$\text{heap}(7) = \{*, *\} = *\text{mex}(1, 1) = 0$

$\text{heap}(8) = \{*2, 0\} = *\text{mex}(0, 2) = *$

$\text{heap}(9) = \{0, *\} = *\text{mex}(0, 1) = *2$

$\text{heap}(10) = \{*, *2\} = *\text{mex}(1, 2) = 0$

We have proven the base case. We will now prove the inductive hypothesis: If $n \equiv 0 \bmod{5}$, $\text{heap}(n) = 0$, $\text{heap}(n+1) = *$, $\text{heap}(n+2) = 0$, $\text{heap}(n+3) = *$, and $\text{heap}(n+4) = *2$, then $\text{heap}(n+5) = 0$, $\text{heap}(n+6) = *$, $\text{heap}(n+7) = 0$, $\text{heap}(n+8) = *$, and $\text{heap}(n+9) = *2$.

$\text{heap}(n+5) = \{\text{heap}(n+1), \text{heap}(n+4)\} = \{*, *2\} = *\text{mex}(1, 2) = 0$

$\text{heap}(n+6) = \{\text{heap}(n+2), \text{heap}(n+5)\} = \{0, 0\} = *\text{mex}(0, 0) = *$

$\text{heap}(n+7) = \{\text{heap}(n+3), \text{heap}(n+6)\} = \{*, *\} = *\text{mex}(1, 1) = 0$

$\text{heap}(n+8) = \{\text{heap}(n+4), \text{heap}(n+7)\} = \{*2, 0\} = *\text{mex}(2, 1) = *$

$\text{heap}(n+9) = \{\text{heap}(n+5), \text{heap}(n+8)\} = \{0, *\} = *\text{mex}(0, 1) = *2$

We have proven the inductive hypothesis. QED.
There are $2020*\frac{2}{5}=808$ positive integers congruent to $0,2 \bmod{5}$ between 1 and 2020, and 1 such integer between 2021 and 2024. $808 + 1 = \boxed{809}$.

~numerophile
We start with $n$ as some of the smaller values. After seeing the first 4 where Bob wins automatically, with trial and error we see that $2, 5, 7,$ and $10$ are spaced alternating in between 2 and 3 apart. This can also be proven with modular arithmetic, but  this is an easier solution for some people. We split them into 2 different sets with common difference 5: {2,7,12 ...} and {5,10,15...}. Counting up all the numbers in each set can be done as follows:
Set 1 ${2,7,12...}$
$2024-2=2022$ (because the first term is two)
$\lfloor \frac{2024}{5} \rfloor = 404$
Set 2 ${5,10,15}$
$\lfloor \frac{2024}{5} \rfloor = 404$

And because we forgot 2022 we add 1 more.
$404+404+1=809$
-Multpi12
(Edits would be appreciated)
LaTexed by BossLu99

Answer: 809

<｜begin▁of▁sentence｜><｜User｜>Alice and Bob play the following game. A stack of $n$ tokens lies before them. The players take turns with Alice going first. On each turn, the player removes either $1$ token or $4$ tokens from the stack. Whoever removes the last token wins. Find the number of positive integers $n$ less than or equal to $2024$ for which there exists a strategy for Bob that guarantees that Bob will win the game regardless of Alice's play.<｜Assistant｜><think>

Okay, so Alice and Bob are playing this game with a stack of n tokens. They take turns removing either 1 or 4 tokens, and the person who takes the last token wins. I need to figure out how many numbers n ≤ 2024 give Bob a winning strategy, meaning no matter how Alice plays, Bob can force a win. Hmm, let's start by understanding the game mechanics better.

First, since Alice goes first, Bob can only win if the starting position n is a losing position for Alice. So, I need to figure out for each n whether it's a winning or losing position. A position is a winning position if the current player can force a win no matter what the opponent does. Conversely, a losing position is one where no matter what the current player does, the opponent can force a win.

Let me try to model this with some examples. Let's start with small n and see if we can find a pattern.

Let's denote the positions as either W (winning) or L (losing). The base case is when n=0, which would mean the previous player has taken the last token and thus wins. But since the game ends when the last token is taken, n=0 isn't part of the game. The first position is n=1.

For n=1: The current player can take 1 token and win. So that's a W.

n=2: The current player can take 1 token, leaving 1 for the opponent, who then takes it and wins. Wait, but if the current player takes 1, opponent gets to take the last one. Alternatively, can the current player take 4? No, because there are only 2 tokens. So only option is to take 1. Therefore, the current player cannot win? Wait, but if you take 1, opponent takes the remaining 1 and wins. So actually, n=2 is a losing position? Wait, but hold on, maybe I made a mistake here.

Wait, let me think again. If n=2, the current player can only take 1 token, leaving 1 for the opponent. Then the opponent takes that 1 and wins. Therefore, the current player cannot force a win, so n=2 is a losing position (L). Hmm, so n=2 is L.

n=3: The current player can take 1 token, leaving 2 (which is L). Since leaving the opponent at an L position means they can't win, so if current player takes 1, opponent is at L, so current player can win. Therefore, n=3 is W.

Alternatively, could they take 4? No, so only option is 1. So yes, n=3 is W.

n=4: The current player can take 4 tokens and win immediately. So that's a W.

n=5: Let's see. The current player can take 1 or 4.

If they take 1, leaving 4. Then opponent can take 4 and win. If they take 4, leaving 1, opponent takes 1 and wins. So no matter what the current player does, opponent can win. Therefore, n=5 is L.

Wait, so n=5 is a losing position. Hmm.

n=6: Let's see. The current player can take 1 or 4.

If they take 1, leaving 5 (which is L). Since leaving opponent at L is good, so if taking 1 leads to L, then current player can win by taking 1. Therefore, n=6 is W.

Alternatively, taking 4 would leave 2 (which is L). Wait, so taking 4 would also leave opponent at L. So actually, either move works. So n=6 is definitely W.

Wait, so n=6 is W.

n=7: Let's see. Taking 1 leaves 6 (W), taking 4 leaves 3 (W). So opponent can win from either position. Therefore, if current player takes 1, opponent is at W, so opponent can win. Similarly, taking 4 also leaves opponent at W. Therefore, n=7 is L? Wait, so both options lead to opponent being able to win. Therefore, n=7 is L.

Wait, so n=7 is L.

n=8: Let's see. Taking 1 leaves 7 (L). So if current player takes 1, opponent is at L, so current player can win. Therefore, n=8 is W.

Alternatively, taking 4 leaves 4 (W). So taking 4 would leave opponent at W, but since taking 1 is better, the current player can choose to take 1 and win. So n=8 is W.

n=9: Taking 1 leaves 8 (W), taking 4 leaves 5 (L). So if current player takes 4, leaving opponent at L, so that's a win. Therefore, n=9 is W.

n=10: Taking 1 leaves 9 (W), taking 4 leaves 6 (W). So both options leave opponent at W. Therefore, n=10 is L?

Wait, so n=10 is L.

Hmm, let's tabulate this so far:

n : 0 1 2 3 4 5 6 7 8 9 10

Result: - W L W W L W L W W L

Wait, so starting from n=1:

1: W

2: L

3: W

4: W

5: L

6: W

7: L

8: W

9: W

10: L

Hmm, trying to see a pattern here. Let's see:

Looking at the sequence of results from n=1 to n=10:

W, L, W, W, L, W, L, W, W, L

Hmm, seems like every 5 numbers, there's a pattern? Let's see:

Looking at positions 1-5: W, L, W, W, L

Positions 6-10: W, L, W, W, L

Wait, so the pattern repeats every 5? Let me check n=11.

n=11: Taking 1 leaves 10 (L). So taking 1 would leave opponent at L, so current player can win. Therefore, n=11 is W.

Similarly, n=12: Taking 1 leaves 11 (W), taking 4 leaves 8 (W). So both options leave opponent at W, so n=12 is L?

Wait, let's check:

n=11: W

n=12: Let's see. If current player takes 1, leaves 11 (W). Opponent can win from there. If takes 4, leaves 8 (W). Opponent can win. Therefore, n=12 is L.

n=13: Taking 1 leaves 12 (L). So taking 1 leads to opponent at L, so n=13 is W.

n=14: Taking 1 leaves 13 (W), taking 4 leaves 10 (L). So taking 4 would leave opponent at L, so n=14 is W.

n=15: Taking 1 leaves 14 (W), taking 4 leaves 11 (W). Both options leave opponent at W, so n=15 is L.

Hmm, so n=11 to 15:

11: W, 12: L, 13: W, 14: W, 15: L

Same as the first five. So the pattern is indeed repeating every 5. Let me confirm n=16:

n=16: Taking 1 leaves 15 (L). So taking 1 leads to opponent at L, so n=16 is W.

Similarly, n=17: Taking 1 leaves 16 (W), taking 4 leaves 13 (W). So both options leave opponent at W, so n=17 is L?

Wait, n=17:

Wait, taking 1 leaves 16 (W). So opponent can win from there. Taking 4 leaves 13 (W), opponent can win. Therefore, n=17 is L.

Wait, but according to the previous pattern, n=17 would be position 17 mod 5? Let's see:

Wait, 1 mod 5 is 1, 2 mod5=2, etc. Let's see:

Looking at n mod 5:

n=1: 1 mod5=1 → W

n=2: 2 mod5=2 → L

n=3: 3 mod5=3 → W

n=4: 4 mod5=4 → W

n=5: 0 mod5=0 → L

n=6: 1 mod5=1 → W

n=7: 2 mod5=2 → L

n=8: 3 mod5=3 → W

n=9: 4 mod5=4 → W

n=10:0 mod5=0 → L

So the pattern is that for n mod5:

If n mod5 is 1,3,4 → W

If n mod5 is 2 or 0 → L

Wait, except n=5 mod5=0 is L, n=2 mod5=2 is L, and the rest (1,3,4) are W. So the losing positions are n ≡0 or 2 mod5.

Wait, so the losing positions are numbers congruent to 0 or 2 modulo 5. Let me check n=12: 12 mod5=2 → L, yes. n=15 mod5=0 → L, yes. n=17 mod5=2 → L, yes. n=7 mod5=2 → L, yes. So that seems consistent.

Therefore, the losing positions are numbers where n mod5 is 0 or 2. Therefore, for Bob to have a winning strategy, the starting position n must be a losing position, because then no matter what Alice does, she has to move to a winning position, and Bob can then always move back to a losing position, eventually forcing Alice into a losing position again until the end.

Therefore, the numbers n ≤2024 where n ≡0 or 2 mod5 are the losing positions, so Bob can win from those. Therefore, the count is the number of integers ≤2024 congruent to 0 or 2 mod5.

Wait, but let me confirm once again with n=5, which is 0 mod5, that's a losing position. So if n=5, Alice has to take 1 or 4. If she takes 1, leaves 4, which is a winning position for Bob (he can take 4 and win). If she takes 4, leaves 1, which is a winning position for Bob (he takes 1 and wins). So yes, n=5 is a losing position for Alice, so Bob can win. Similarly, n=2: Alice can only take 1, leaving 1 for Bob, who takes it and wins. So yes, n=2 is a losing position for Alice. So that's correct.

Therefore, the total numbers are those n ≤2024 where n mod5 is 0 or 2. Let's compute how many numbers ≤2024 are ≡0 mod5 or ≡2 mod5.

First, numbers ≡0 mod5 up to 2024: floor(2024/5) = 404.8, so 404 numbers (since 5×404=2020). Wait, but 2020 is included, so 404 numbers.

Numbers ≡2 mod5 up to 2024: Let's see, the numbers are 2,7,12,..., up to the largest number ≤2024.

The sequence is 2 +5k ≤2024. So 5k ≤2022 → k ≤2022/5=404.4, so k=0 to 404, which is 405 numbers. Wait, let's check:

First term: 2 =2+5×0, last term: let's see 2 +5×404=2+2020=2022. 2022 is ≤2024, so that's okay. So the number of terms is 405.

Therefore total numbers are 404 +405=809.

Wait, but let me confirm:

For numbers ≡0 mod5: from 5×1=5 up to 5×404=2020. So the count is 404 numbers (since 2020/5=404).

Numbers ≡2 mod5: starting at 2, then 7,...,2022. The formula for the nth term is 2 +5(n-1). Set that ≤2024:

2 +5(n-1) ≤2024 →5(n-1) ≤2022 →n-1 ≤404.4 →n-1=404 →n=405. So 405 numbers.

Total is 404+405=809. Hmm, but let me check with another approach.

Total numbers from 1 to 2024:2024 numbers.

Each residue mod5 occurs either floor(2024/5) or floor(2024/5)+1 times.

2024 divided by5 is 404.8, so floor is 404. So residues 1,2,3,4,0.

The residues 1 to 5:

Numbers congruent to r mod5, for r=0: numbers are 5,10,...,2020: total 404.

For r=1: 1,6,...,2021 (since 2021=1+5×404). So 405 numbers.

Similarly, r=2: 2,7,...,2022 (2022=2+5×404). 405 numbers.

r=3: 3,8,...,2023 (2023=3+5×404). 405 numbers.

r=4:4,9,...,2024 (2024=4+5×404). 405 numbers.

Wait, hold on, 2024 mod5 is 2024-5×404=2024-2020=4. So residue 4. So the counts are:

r=0: 404

r=1: 405 (since 1+5×404=2021 ≤2024)

r=2:405 (2+5×404=2022 ≤2024)

r=3:405 (3+5×404=2023 ≤2024)

r=4:405 (4+5×404=2024)

Therefore, residues 0 has 404, others have 405.

Therefore, the numbers with residue 0 or 2 mod5 are 404 (for 0) +405 (for 2)= 809. So that's correct.

Therefore, the answer should be 809. Wait, but let me check with n=2024. Since 2024 mod5 is 4, so it's a winning position (since 4 mod5 is W). So 2024 is not counted. The last number counted for residue 2 is 2022, which is 2 mod5. So that's okay.

Wait, but let me check with n=2020, which is 0 mod5, so included. So total numbers are 404 +405=809. Hmm, so that's the answer?

Wait, but let me check with smaller numbers to see if my pattern holds.

For example, up to n=5:

Numbers where Bob can win are n=2 and 5. So 2 numbers. Let's see with the formula:

For n=5, numbers ≤5:

residues 0 mod5: 5 →1 number

residues 2 mod5:2 →1 number

Total 2, which matches. So that works.

Similarly, up to n=10:

Numbers where n≡0 or2 mod5: 0mod5:5,10 →2 numbers; 2mod5:2,7 →2 numbers. Total 4. Let's see actual losing positions up to 10: n=2,5,7,10. That's 4 numbers. Correct.

So the formula works here. So seems correct.

Therefore, the answer is 809. Wait, but let me check once again the count for residue 2 mod5.

The first term is 2, last term is 2022. The number of terms is ((2022 -2)/5)+1 = (2020/5)+1=404+1=405. Correct.

Similarly for residue 0 mod5: first term 5, last term 2020: ((2020-5)/5)+1=(2015/5)+1=403+1=404. Correct.

So total is 405+404=809. So I think that's the answer.

But just to be thorough, let me check n=0 mod5 and n=2 mod5.

Wait, but the problem states "positive integers n less than or equal to 2024". So n starts at 1. So residue 0 mod5 starts at 5, which is okay.

Therefore, the answer is 809. Hmm, but let me see if there's any mistake in the initial pattern assumption.

Wait, let me check n=7. n=7 mod5=2, so it's a losing position. Let's see:

If n=7, Alice has to move first. She can take 1 or 4.

If she takes 1, leaves 6. 6 mod5=1, which is a winning position for Bob. So Bob can take 1, leaving 5 (losing for Alice). Wait, but let's see:

Wait, if Alice takes 1 from 7, leaving 6. Bob can choose to take 1, leaving 5 (L). Then Alice is forced to move from 5 (L), so whatever she does, Bob can win. So yes, Bob can force a win.

Alternatively, if Alice takes 4 from 7, leaving 3. 3 mod5=3 (W). Bob can take 1, leaving 2 (L). Then Alice is at 2 (L), so she has to leave 1, and Bob takes it. So yes, Bob can win. Therefore, n=7 is indeed a losing position for Alice, so Bob can win. So that's correct.

Another test: n=10 (0 mod5). Alice can take 1 or 4.

If she takes 1, leaves 9 (W). Bob can take 1, leaving 8 (W). Wait, but maybe Bob can take 4? Let's see:

Wait, from 9, Bob can choose to take 4, leaving 5 (L). Then Alice is at 5 (L), so whatever she does, Bob can win. So yes, Bob can force a win.

Alternatively, if Alice takes 4 from 10, leaving 6 (W). Bob can take 1, leaving 5 (L). So same result. So yes, n=10 is a losing position for Alice.

Therefore, the pattern holds. So I think the answer is 809. Wait, but let me check once more the arithmetic.

Total numbers ≡0 mod5 up to 2024: 5,10,...,2020. The number of terms is 2020/5 = 404.

Numbers ≡2 mod5 up to 2024: 2,7,...,2022. The number of terms: Let's compute (2022 -2)/5 +1 = 2020/5 +1= 404 +1=405. So total 404+405=809. Yes.

Therefore, the answer is 809. Hmm, but the problem says "positive integers n less than or equal to 2024", so I think that's correct. Wait, but let me check if 0 is included in any way, but n is positive, so starting at 1. So yes, 809 is the answer.

Wait, but hold on a second. Let me check n=1. If n=1, Alice takes 1 and wins. So Bob can't win here. So n=1 is a winning position for Alice, so not counted. Similarly, n=2 is losing for Alice, so Bob can win. So our count includes n=2, which is correct. So all the numbers counted are indeed the ones where Bob can win, so the total is 809. I think that's correct.

**Final Answer**
\boxed{809}
</think>

To determine the number of positive integers \( n \leq 2024 \) for which Bob has a winning strategy, we need to identify the losing positions in the game. A losing position is one where the current player cannot force a win regardless of the opponent's moves.

We analyze the game by examining small values of \( n \) and identifying a pattern:

- \( n = 1 \): Winning (W)
- \( n = 2 \): Losing (L)
- \( n = 3 \): Winning (W)
- \( n = 4 \): Winning (W)
- \( n = 5 \): Losing (L)
- \( n = 6 \): Winning (W)
- \( n = 7 \): Losing (L)
- \( n = 8 \): Winning (W)
- \( n = 9 \): Winning (W)
- \( n = 10 \): Losing (L)

We observe a repeating pattern every 5 numbers. The losing positions are those where \( n \equiv 0 \pmod{5} \) or \( n \equiv 2 \pmod{5} \).

To find the count of such numbers up to 2024:
- Numbers \( \equiv 0 \pmod{5} \): These are 5, 10, 15, ..., 2020. The count is \( \left\lfloor \frac{2024}{5} \right\rfloor = 404 \).
- Numbers \( \equiv 2 \pmod{5} \): These are 2, 7, 12, ..., 2022. The count is \( \left\lfloor \frac{2022 - 2}{5} \right\rfloor + 1 = 405 \).

Summing these counts gives the total number of losing positions:
\[ 404 + 405 = 809 \]

Thus, the number of positive integers \( n \leq 2024 \) for which Bob has a winning strategy is \(\boxed{809}\).


Solution:
This is a conditional probability problem. Bayes' Theorem states that
\[P(A|B)=\dfrac{P(B|A)\cdot P(A)}{P(B)}\]

in other words, the probability of $A$ given $B$ is equal to the probability of $B$ given $A$ times the probability of $A$ divided by the probability of $B$. In our case, $A$ represents the probability of winning the grand prize, and $B$ represents the probability of winning a prize. Clearly, $P(B|A)=1$, since by winning the grand prize you automatically win a prize. Thus, we want to find $\dfrac{P(A)}{P(B)}$.
Let us calculate the probability of winning a prize. We do this through casework: how many of Jen's drawn numbers match the lottery's drawn numbers?
To win a prize, Jen must draw at least $2$ numbers identical to the lottery. Thus, our cases are drawing $2$, $3$, or $4$ numbers identical.
Let us first calculate the number of ways to draw exactly $2$ identical numbers to the lottery. Let Jen choose the numbers $a$, $b$, $c$, and $d$; we have $\dbinom42$ ways to choose which $2$ of these $4$ numbers are identical to the lottery. We have now determined $2$ of the $4$ numbers drawn in the lottery; since the other $2$ numbers Jen chose can not be chosen by the lottery, the lottery now has $10-2-2=6$ numbers to choose the last $2$ numbers from. Thus, this case is $\dbinom62$, so this case yields $\dbinom42\dbinom62=6\cdot15=90$ possibilities.
Next, let us calculate the number of ways to draw exactly $3$ identical numbers to the lottery. Again, let Jen choose $a$, $b$, $c$, and $d$. This time, we have $\dbinom43$ ways to choose the identical numbers and again $6$ numbers left for the lottery to choose from; however, since $3$ of the lottery's numbers have already been determined, the lottery only needs to choose $1$ more number, so this is $\dbinom61$. This case yields $\dbinom43\dbinom61=4\cdot6=24$.
Finally, let us calculate the number of ways to all $4$ numbers matching. There is actually just one way for this to happen.
In total, we have $90+24+1=115$ ways to win a prize. The lottery has $\dbinom{10}4=210$ possible combinations to draw, so the probability of winning a prize is $\dfrac{115}{210}$. There is actually no need to simplify it or even evaluate $\dbinom{10}4$ or actually even know that it has to be $\dbinom{10}4$; it suffices to call it $a$ or some other variable, as it will cancel out later. However, let us just go through with this. The probability of winning a prize is $\dfrac{115}{210}$. Note that the probability of winning a grand prize is just matching all $4$ numbers, which we already calculated to have $1$ possibility and thus have probability $\dfrac1{210}$. Thus, our answer is $\dfrac{\frac1{210}}{\frac{115}{210}}=\dfrac1{115}$. Therefore, our answer is $1+115=\boxed{116}$.
~Technodoggo
For getting all $4$ right, there is only $1$ way.
For getting $3$ right, there is $\dbinom43$ multiplied by $\dbinom61$ = $24$ ways.
For getting $2$ right, there is $\dbinom42$ multiplied by $\dbinom62$ = $90$ ways.
$\frac{1}{1+24+90}$  = $\frac{1}{115}$
Therefore, the answer is $1+115 = \boxed{116}$
~e___

Answer: 116

<｜begin▁of▁sentence｜><｜User｜>Jen enters a lottery by picking $4$ distinct numbers from $S=\{1,2,3,\cdots,9,10\}.$ $4$ numbers are randomly chosen from $S.$ She wins a prize if at least two of her numbers were $2$ of the randomly chosen numbers, and wins the grand prize if all four of her numbers were the randomly chosen numbers. The probability of her winning the grand prize given that she won a prize is $\tfrac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$.<｜Assistant｜><think>

Okay, so Jen is in a lottery where she picks 4 distinct numbers from the set S = {1,2,3,...,9,10}. Then, 4 numbers are randomly chosen from S. She wins a prize if at least two of her numbers are among the randomly chosen numbers. She wins the grand prize if all four of her numbers are the chosen ones. We need to find the probability that she wins the grand prize given that she won a prize. The answer should be in the form m/n where m and n are coprime, and then find m + n.

Hmm, let me start by recalling that conditional probability is involved here. The probability of winning the grand prize given that she won a prize is equal to the probability of winning the grand prize divided by the probability of winning a prize. So, mathematically, that would be P(Grand Prize) / P(At least two numbers match).

First, let me compute P(Grand Prize). Since the grand prize is when all four of her numbers are chosen, that's the probability that the four numbers drawn exactly match her four numbers. Since the lottery is choosing 4 numbers out of 10, the total number of possible combinations is C(10,4). The number of favorable outcomes for the grand prize is just 1 (since there's only one combination that matches her numbers exactly). Therefore, P(Grand Prize) = 1 / C(10,4).

Now, P(At least two numbers match) is the probability that at least two of her numbers are among the four chosen. To compute this, it might be easier to calculate the complementary probability and subtract it from 1. The complementary events would be: winning nothing (zero numbers match) or winning a smaller prize (exactly one number matches). Wait, but actually, the problem states she wins a prize if at least two numbers are chosen. So the complementary event is that she doesn't win a prize, which is when fewer than two numbers match, i.e., 0 or 1 numbers match. Therefore, P(At least two numbers match) = 1 - [P(0 matches) + P(1 match)].

So I need to compute P(0 matches) and P(1 match). Let's see.

First, the total number of possible lottery draws is C(10,4). Let me compute that first. C(10,4) is 10! / (4! * 6!) = (10*9*8*7)/(4*3*2*1) = 210. So total possible combinations are 210.

Now, the number of ways to have 0 matches: that means all four numbers chosen are from the 6 numbers she didn't pick. Since she picked 4 numbers, there are 10 - 4 = 6 numbers not chosen by her. So the number of ways to choose 4 numbers from those 6 is C(6,4). Similarly, for exactly 1 match: that would be choosing 1 number from her 4 and 3 numbers from the remaining 6. So that's C(4,1)*C(6,3).

Let me compute these:

Number of 0 matches: C(6,4) = 15.

Number of 1 match: C(4,1)*C(6,3) = 4 * 20 = 80. Wait, C(6,3) is 20? Let me check: 6*5*4 / 3*2*1 = 20. Yes, that's right. So 4*20 = 80.

Therefore, the number of unfavorable outcomes (0 or 1 matches) is 15 + 80 = 95. Therefore, the number of favorable outcomes for at least two matches is total outcomes minus 95, so 210 - 95 = 115. Alternatively, we can compute it directly as the sum of 2, 3, and 4 matches. Let me verify that way too to check.

Number of exactly 2 matches: C(4,2)*C(6,2). C(4,2) is 6, C(6,2) is 15, so 6*15 = 90.

Exactly 3 matches: C(4,3)*C(6,1) = 4*6 = 24.

Exactly 4 matches: C(4,4)*C(6,0) = 1*1 = 1.

Adding those up: 90 +24 +1 = 115. That matches the previous result. Good.

Therefore, P(At least two matches) = 115 / 210.

Therefore, the conditional probability P(Grand Prize | At least two matches) is (1/210) divided by (115/210) = 1/115.

Wait, hold on. Wait, the grand prize probability is 1/210, and the denominator is 115/210, so when you divide them, the 210 cancels, giving 1/115. So the probability is 1/115. Therefore, m is 1 and n is 115. But 1 and 115 are coprime, so m + n is 116?

Wait, but that seems too straightforward. Let me check again.

Wait, hold on. Wait, actually, the problem says "the probability of her winning the grand prize given that she won a prize". So the conditional probability is P(Grand Prize) / P(Winning a prize). Since winning the grand prize is a subset of winning a prize, that's correct.

But let me confirm the numbers again. The total number of possible draws is 210. The number of grand prize outcomes is 1. The number of prize-winning outcomes (at least two matches) is 115. Therefore, the conditional probability is indeed 1/115. So m/n is 1/115, so m + n is 116. Hmm, but I feel like maybe I made a mistake here because 115 seems a bit high, but let me think again.

Alternatively, perhaps I made a mistake in calculating the number of prize-winning outcomes. Let me check again:

Total combinations: 210.

Number of 0 matches: C(6,4)=15.

Number of 1 match: C(4,1)*C(6,3)=4*20=80. So total non-prize is 15+80=95. Therefore prize is 210-95=115. That's correct.

Therefore, the conditional probability is 1/115. Hmm, so 1/115 reduces to 1/115, since 115 factors into 5*23, and 1 is coprime with everything. Therefore, m + n is 1 + 115 = 116.

Wait, but I have a feeling that maybe the problem is interpreted differently? Let me check the problem again.

"Jen enters a lottery by picking 4 distinct numbers from S={1,2,3,⋯,9,10}. 4 numbers are randomly chosen from S. She wins a prize if at least two of her numbers were 2 of the randomly chosen numbers, and wins the grand prize if all four of her numbers were the randomly chosen numbers. The probability of her winning the grand prize given that she won a prize is m/n where m and n are relatively prime positive integers. Find m+n."

Hmm, so the problem is exactly as I thought. So maybe 116 is the answer? But let me see if there's another way to approach this.

Alternatively, maybe the problem is considering the number of possible overlapping numbers? Wait, but no, the calculations seem straightforward. Alternatively, perhaps I made a mistake in the combinations.

Wait, let me recalculate C(6,4). 6 choose 4 is 15, yes. 6*5*4*3/(4*3*2*1) = 15. Correct.

C(6,3) is 20, yes. 6*5*4/(3*2*1)=20. Correct. So 4*20 is 80. 15+80=95. 210-95=115. So that's correct.

Alternatively, maybe the problem is in the conditional probability? Let me think again. The conditional probability is P(Grand Prize) / P(Prize). Since Grand Prize is a subset of Prize, that's correct. So yes, 1/115.

Alternatively, maybe the question is phrased as "the probability of her winning the grand prize given that she won a prize", so perhaps the numerator is the number of grand prize outcomes over the number of prize outcomes. Since all outcomes are equally likely, the probability is just (number of grand prize)/ (number of prize outcomes). Since there is 1 grand prize outcome and 115 prize outcomes, so 1/115. So that's the same as before.

Therefore, the answer is 1 + 115 = 116. Hmm, but I need to check if 1 and 115 are coprime. The prime factors of 115 are 5 and 23. Since 1 shares no common factors with 115, yes, so the fraction is reduced. Therefore, 116 is the answer.

Wait, but I have a doubt because in some lotteries, sometimes the "winning a prize" might have different categories, but the problem says "wins a prize if at least two of her numbers were 2 of the randomly chosen numbers". So exactly, so at least two matching numbers. So the calculation is correct. Hmm.

Alternatively, maybe I made a mistake in the total number of combinations? Let me check again. 10 choose 4 is 210. Yes, that's correct. 10C4 = 210. So that's right.

Alternatively, maybe the problem is in the interpretation of "at least two of her numbers were 2 of the randomly chosen numbers". Wait, maybe the wording is tricky? "at least two of her numbers were 2 of the randomly chosen numbers". Hmm, maybe that's ambiguous? Wait, perhaps it's supposed to be "at least two of her numbers are among the four randomly chosen numbers". Which is how I interpreted it. So that's correct.

Alternatively, maybe "two of the randomly chosen numbers" refers to exactly two? But the wording says "at least two of her numbers were 2 of the randomly chosen numbers". Hmm, maybe the wording is confusing. Wait, perhaps it's a translation issue? Let me parse the sentence again.

"She wins a prize if at least two of her numbers were 2 of the randomly chosen numbers".

Hmm, maybe it's supposed to be "if at least two of her numbers are among the two randomly chosen numbers". Wait, but that can't be, because the lottery chooses 4 numbers. Wait, the problem says "4 numbers are randomly chosen from S". So the winning numbers are 4. So "at least two of her numbers were 2 of the randomly chosen numbers" – perhaps the wording is trying to say that at least two of her numbers are in the four chosen numbers. So that is, the intersection between her numbers and the winning numbers has size at least two. So that's how I interpreted it, so that's correct.

Therefore, I think my calculation is correct, so the answer is 116. Hmm, but maybe I made a mistake in the problem's grand prize condition? Let me check again. "wins the grand prize if all four of her numbers were the randomly chosen numbers". So that's exactly the 4 numbers, so that's 1 case. So yes, that's correct.

Alternatively, maybe the problem is from an official competition, like AIME, and the answer is different? Wait, but since I can't check external sources, I have to go with my calculation. Alternatively, maybe I made a mistake in the combinations for the number of prize outcomes?

Wait, let me recalculate the number of ways to have exactly 2 matches: C(4,2)*C(6,2). So C(4,2)=6, C(6,2)=15, so 6*15=90. Exactly 3 matches: C(4,3)*C(6,1)=4*6=24. Exactly 4 matches:1. So total is 90+24+1=115. That's correct. So 115/210 is the probability of a prize, so the conditional probability is 1/115. Hmm.

Alternatively, maybe the problem wants the probability that given she has at least two matches, what's the probability she has exactly four matches. So that is 1 divided by 115. So yes, that's the same as 1/115. So m + n is 116. Hmm.

Alternatively, perhaps I made a mistake in the problem's initial step. Let me think of another approach. Let me think of it as hypergeometric distribution. The number of matches is hypergeometric with parameters N=10, K=4 (the number of success states in the population), n=4 (number of draws), and k is the number of observed successes.

The probability of exactly k matches is C(4, k)*C(6, 4 -k)/C(10,4). So the probability of at least two is sum from k=2 to 4 of [C(4,k)*C(6,4−k)] / C(10,4). Which is exactly what I computed. So the numerator for the conditional probability is the probability of k=4, which is [C(4,4)*C(6,0)] / C(10,4) = 1/210. The denominator is the sum from k=2 to 4 of those terms, which is (C(4,2)C(6,2) + C(4,3)C(6,1) + C(4,4)C(6,0))/C(10,4) = (90 +24 +1)/210 = 115/210. So the ratio is 1/115. So that's correct.

Therefore, I think the answer is 116. Hmm, but maybe I made a mistake in the problem's phrasing? Let me check once again.

Wait, the problem says "the probability of her winning the grand prize given that she won a prize". So that is, given that she has at least two matching numbers, what's the probability she has all four. So yes, that is exactly the conditional probability. So 1/115. So m + n is 1 + 115 = 116. Therefore, I think that is the answer.

Alternatively, maybe I made a mistake in the calculation of C(6,4). Wait, 6 choose 4 is 15. 6*5*4*3/(4*3*2*1) = (6*5)/ (2*1) = 15. Correct. So that's okay.

Alternatively, maybe the problem counts the numbers differently? Wait, the problem says "4 distinct numbers from S = {1,2,...,10}". So 10 numbers. Choosing 4. So that's correct. So I think my answer is correct. So I think the answer is 116.

Wait, but let me see, maybe the problem is from an AIME problem, and the answer is different? Wait, but since I can't look it up, I have to go with my calculations. Alternatively, maybe I made a mistake in the conditional probability formula?

Wait, conditional probability formula is P(A|B) = P(A ∩ B)/P(B). But since A is a subset of B (winning grand prize implies winning a prize), so P(A ∩ B) is just P(A). Therefore, it's P(A)/P(B). So that's correct. So 1/210 divided by 115/210 is 1/115. So that's correct.

Hmm, okay, I think I've checked it thoroughly. So I think the answer is 116.

**Final Answer**
\boxed{116}
</think>

Jen enters a lottery by picking 4 distinct numbers from the set \( S = \{1, 2, 3, \ldots, 9, 10\} \). Four numbers are randomly chosen from \( S \). She wins a prize if at least two of her numbers are among the chosen numbers, and wins the grand prize if all four of her numbers are chosen. We need to find the probability of her winning the grand prize given that she won a prize.

First, we calculate the total number of ways to choose 4 numbers from 10, which is \( \binom{10}{4} = 210 \).

The probability of winning the grand prize is the probability that all four of her numbers are chosen. There is only 1 favorable outcome for this, so:
\[
P(\text{Grand Prize}) = \frac{1}{210}
\]

Next, we calculate the probability of winning at least two numbers. We use the complementary probability of winning 0 or 1 numbers:
- The number of ways to choose 0 matching numbers (all from the 6 numbers not picked by Jen) is \( \binom{6}{4} = 15 \).
- The number of ways to choose 1 matching number (1 from Jen's 4 and 3 from the other 6) is \( \binom{4}{1} \binom{6}{3} = 4 \times 20 = 80 \).

Thus, the number of unfavorable outcomes (0 or 1 matches) is \( 15 + 80 = 95 \). Therefore, the number of favorable outcomes for at least two matches is:
\[
210 - 95 = 115
\]

The probability of winning at least two numbers is:
\[
P(\text{At least two matches}) = \frac{115}{210}
\]

The conditional probability of winning the grand prize given that she won a prize is:
\[
\frac{P(\text{Grand Prize})}{P(\text{At least two matches})} = \frac{\frac{1}{210}}{\frac{115}{210}} = \frac{1}{115}
\]

Since 1 and 115 are coprime, the fraction is in simplest form. Therefore, \( m = 1 \) and \( n = 115 \), giving \( m + n = 1 + 115 = 116 \).

\[
\boxed{116}
\]


Solution:
We use simple geometry to solve this problem.

We are given that $A$, $D$, $H$, and $G$ are concyclic; call the circle that they all pass through circle $\omega$ with center $O$. We know that, given any chord on a circle, the perpendicular bisector to the chord passes through the center; thus, given two chords, taking the intersection of their perpendicular bisectors gives the center. We therefore consider chords $HG$ and $AD$ and take the midpoints of $HG$ and $AD$ to be $P$ and $Q$, respectively.

We could draw the circumcircle, but actually it does not matter for our solution; all that matters is that $OA=OH=r$, where $r$ is the circumradius.
By the Pythagorean Theorem, $OQ^2+QA^2=OA^2$. Also, $OP^2+PH^2=OH^2$. We know that $OQ=DE+HP$, and $HP=\dfrac{184}2=92$; $QA=\dfrac{16}2=8$; $OP=DQ+HE=8+17=25$; and finally, $PH=92$. Let $DE=x$. We now know that $OA^2=(x+92)^2+8^2$ and $OH^2=25^2+92^2$. Recall that $OA=OH$; thus, $OA^2=OH^2$. We solve for $x$:
\begin{align*}
(x+92)^2+8^2&=25^2+92^2 \\
(x+92)^2&=625+(100-8)^2-8^2 \\
&=625+10000-1600+64-64 \\
&=9025 \\
x+92&=95 \\
x&=3. \\
\end{align*}
The question asks for $CE$, which is $CD-x=107-3=\boxed{104}$.
~Technodoggo
Suppose $DE=x$. Extend $AD$ and $GH$ until they meet at $P$. From the [Power of a Point Theorem](https://artofproblemsolving.com/wiki/index.php/Power_of_a_Point_Theorem), we have $(PH)(PG)=(PD)(PA)$. Substituting in these values, we get $(x)(x+184)=(17)(33)=561$. We can use guess and check to find that $x=3$, so $EC=\boxed{104}$.

~alexanderruan
~diagram by Technodoggo
We find that \[\angle GAB = 90-\angle DAG = 90 - (180 - \angle GHD) = \angle DHE.\]
Let $x = DE$ and $T = FG \cap AB$. By similar triangles $\triangle DHE \sim \triangle GAB$ we have $\frac{DE}{EH} = \frac{GT}{AT}$. Substituting lengths we have $\frac{x}{17} = \frac{16 + 17}{184 + x}.$ Solving, we find $x = 3$ and thus $CE = 107 - 3 = \boxed{104}$.
~AtharvNaphade ~coolruler ~eevee9406
One liner: $107-\sqrt{92^2+25^2-8^2}+92=\boxed{104}$
~Bluesoul

Explanation
Let $OP$ intersect $DF$ at $T$ (using the same diagram as Solution 2).
The formula calculates the distance from $O$ to $H$ (or $G$), $\sqrt{92^2+25^2}$, then shifts it to $OD$ and the finds the distance from $O$ to $Q$, $\sqrt{92^2+25^2-8^2}$. $107$ minus that gives $CT$, and when added to $92$, half of $FE=TE$, gives $CT+TE=CE$
Let $\angle{DHE} = \theta.$ This means that $DE = 17\tan{\theta}.$ Since quadrilateral $ADHG$ is cyclic, $\angle{DAG} = 180 - \angle{DHG} = 90 - \theta$.
Let $X = AG \cap DF.$ Then, $\Delta DXA \sim \Delta FXG,$ with side ratio $16:17.$ Also, since $\angle{DAG} = 90 - \theta, \angle{DXA} = \angle{FXG} = \theta.$ Using the similar triangles, we have $\tan{\theta} = \frac{16}{DX} = \frac{17}{FX}$ and $DX + FX = DE + EF = 17\tan{\theta} + 184$.
Since we want $CE = CD - DE = 107 - 17\tan{\theta},$ we only need to solve for $\tan{\theta}$ in this system of equations. Solving yields $\tan{\theta} = \frac{3}{17},$ so $CE = \boxed{104.}$
~PureSwag
Using a ruler (also acting as a straight edge), draw the figure to scale with one unit = 1mm. With a compass, draw circles until you get one such that $A,D,H,G$ are on the edge of the drawn circle. From here, measuring with your ruler should give $CE = \boxed{104.}$
Note: 1 mm is probably the best unit to use here just for convenience (drawing all required parts of the figure fits into a normal-sized scrap paper 8.5 x 11); also all lines can be drawn with a standard 12-inch ruler
~kipper

Answer: 104

<｜begin▁of▁sentence｜><｜User｜>Rectangles $ABCD$ and $EFGH$ are drawn such that $D,E,C,F$ are collinear. Also, $A,D,H,G$ all lie on a circle. If $BC=16$,$AB=107$,$FG=17$, and $EF=184$, what is the length of $CE$?
[asy] import graph; unitsize(0.1cm);  pair A = (0,0);pair B = (70,0);pair C = (70,16);pair D = (0,16);pair E = (3,16);pair F = (90,16);pair G = (90,33);pair H = (3,33); dot(A^^B^^C^^D^^E^^F^^G^^H); label("$A$", A, S);label("$B$", B, S);label("$C$", C, N);label("$D$", D, N);label("$E$", E, S);label("$F$", F, S);label("$G$", G, N);label("$H$", H, N); draw(E--D--A--B--C--E--H--G--F--C); [/asy]<｜Assistant｜><think>

Okay, so I need to find the length of CE in this problem with the two rectangles ABCD and EFGH. Let me start by understanding the given information and the diagram.

First, rectangles ABCD and EFGH are drawn such that points D, E, C, F are collinear. That means these four points lie on a straight line. Also, points A, D, H, G all lie on a circle. The given side lengths are BC = 16, AB = 107, FG = 17, and EF = 184. I need to find CE.

Let me sketch a rough diagram in my mind based on the Asymptote code provided. Rectangle ABCD has AB = 107 as the base, BC = 16 as the height. So, if A is at (0,0), then B would be at (107,0), C at (107,16), and D at (0,16). Wait, but in the Asymptote code, the coordinates are given as A(0,0), B(70,0), but the problem states AB=107. Hmm, maybe the Asymptote code is just an example with different coordinates? The problem's actual coordinates might be different. So I should ignore the specific coordinates in the Asymptote code and go by the problem's given lengths.

Let me try to visualize it again. Rectangle ABCD: AB is the bottom side, BC is the right side, CD is the top, and DA is the left side. Since it's a rectangle, all angles are right angles. Similarly, rectangle EFGH has sides EF and FG. Since D, E, C, F are collinear, that line must be the line containing DC and EF? Wait, not sure yet.

Wait, the problem says D, E, C, F are collinear. So points D, E, C, F all lie on a straight line. Since D and C are vertices of rectangle ABCD, DC is one of its sides. Wait, in rectangle ABCD, DC is the top side, right? Because ABCD is a rectangle, so if AB is the bottom, BC is the right side, CD is the top, and DA is the left. So DC is the top side from D to C. But if D, E, C, F are collinear, then E and F must be on the line DC extended? Hmm, but E and F are vertices of rectangle EFGH. So maybe rectangle EFGH is placed such that its side EF is along the line DC, but extended?

Also, the points A, D, H, G lie on a circle. So these four points are concyclic. That must be important for applying some circle theorems or power of a point.

Given the sides: BC = 16, so the height of rectangle ABCD is 16. AB = 107, so the base is 107. For rectangle EFGH, FG = 17 and EF = 184. Since EFGH is a rectangle, EF and FG are adjacent sides, so EF is the base (184) and FG is the height (17). So the rectangle EFGH is longer in the EF direction.

Let me try to set up a coordinate system to model this. Let me place point A at the origin (0,0). Then since AB is 107, point B would be at (107, 0). Since BC is 16, point C is at (107,16), and point D is at (0,16). That makes rectangle ABCD with length 107 and height 16.

Now, points D, E, C, F are collinear. Since D is (0,16) and C is (107,16), the line DC is the top side of rectangle ABCD, which is horizontal at y=16. So the line through D, E, C, F must be this horizontal line. Therefore, points E and F must also lie on the line y=16. Since E and F are part of rectangle EFGH, which has sides EF = 184 and FG =17. Since EF is part of the line y=16, then EF is a horizontal side of length 184. So E and F are on y=16, with EF = 184. But since E and F are also connected to form the rectangle EFGH, then FG is vertical? Wait, since EFGH is a rectangle, if EF is horizontal, then FG would be vertical. Since FG is 17, then from point F, moving up 17 units to G, and similarly from E, moving up 17 units to H? Wait, but the direction (up or down) depends on the orientation.

Wait, but in the Asymptote code, point E is at (3,16), F is at (90,16), G is at (90,33), H is at (3,33). So in that example, the rectangle EFGH is above the line DC (since y=16 to y=33). So maybe in the actual problem, the rectangle EFGH is above the line DC. So points G and H would be above the line y=16.

But in the problem statement, the coordinates might be different, but the configuration is similar. So let me proceed with coordinates.

Let me denote coordinates:

Let me set coordinate system with point A at (0,0). Then:

- A: (0,0)
- B: (107, 0)
- C: (107,16)
- D: (0,16)

Now, points D, E, C, F are collinear on the line y=16. So E and F are on this line. Let me denote E as (e,16) and F as (f,16). Since EF = 184, the distance between E and F is 184. Since they are on a horizontal line, |f - e| = 184. So either f = e + 184 or e = f + 184. But since E and F are part of the rectangle EFGH, and the rectangle has sides EF and FG. Since FG is 17, which is the vertical side, then from F, moving upwards (assuming the rectangle is above the line DC) 17 units to G, so G would be (f,16 +17) = (f,33). Similarly, H would be (e,33). So the rectangle EFGH has coordinates E(e,16), F(f,16), G(f,33), H(e,33). That makes sense.

Now, the other condition is that points A, D, H, G lie on a circle. So these four points must be concyclic. Let me note their coordinates:

- A: (0,0)
- D: (0,16)
- H: (e,33)
- G: (f,33)

So these four points lie on a circle. I need to use this condition to find e and f, and then compute CE. Since CE is the distance from point C (107,16) to point E (e,16). Since both are on y=16, the distance CE is |107 - e|. So once I find e, I can compute CE.

So the key is to find e and f such that the four points A, D, H, G lie on a circle, with the constraints that |f - e| = 184 (since EF = 184). Let me write equations for the circle passing through A, D, H, G.

First, let's denote coordinates:

A(0,0), D(0,16), H(e,33), G(f,33).

We need these four points to lie on a circle. Let's find the equation of the circle passing through these points.

The general equation of a circle is x² + y² + 2gx + 2fy + c = 0. Wait, but maybe it's easier to use the fact that four points lie on a circle if the determinant of the following matrix is zero:

|x   y   x²+y²  1|
|0   0    0     1|
|0  16  256   1|
|e  33  e²+33² 1|
|f  33  f²+33² 1|

Wait, actually, the determinant for concyclic points can be set up, but it might be a bit involved. Alternatively, since points A, D, H, G are on a circle, we can use the power of a point or cyclic quadrilateral properties.

Alternatively, since points A, D, H, G are concyclic, the power of point A with respect to the circle defined by D, H, G should satisfy the condition. Wait, but maybe using coordinates is better.

Let me first write the equation of the circle passing through A(0,0), D(0,16), H(e,33), G(f,33).

Since A and D are on the circle, let's plug them into the general circle equation.

Let the circle equation be x² + y² + 2gx + 2fy + c = 0.

For point A(0,0):

0 + 0 + 0 + 0 + c = 0 ⇒ c = 0.

So the equation simplifies to x² + y² + 2gx + 2fy = 0.

Now plug in point D(0,16):

0 + (16)^2 + 0 + 2f*(16) = 0 ⇒ 256 + 32f = 0 ⇒ 32f = -256 ⇒ f = -8.

So now the equation becomes x² + y² + 2gx - 16y = 0.

Now plug in point H(e,33):

e² + (33)^2 + 2g*e - 16*(33) = 0.

Similarly, plug in point G(f,33):

f² + (33)^2 + 2g*f - 16*(33) = 0.

So both H and G must satisfy the equation. Let me write equations for H and G:

For H(e,33):

e² + 1089 + 2g e - 528 = 0 ⇒ e² + 2g e + (1089 - 528) = 0 ⇒ e² + 2g e + 561 = 0. (Equation 1)

For G(f,33):

f² + 1089 + 2g f - 528 = 0 ⇒ f² + 2g f + 561 = 0. (Equation 2)

So equations 1 and 2 are:

e² + 2g e + 561 = 0,

f² + 2g f + 561 = 0.

Therefore, both e and f satisfy the same quadratic equation in terms of x: x² + 2g x + 561 = 0.

Therefore, e and f are the roots of the quadratic equation x² + 2g x + 561 = 0.

Therefore, by Vieta's formula, e + f = -2g,

and e*f = 561.

Also, we know from the rectangle EFGH that the distance between E and F is 184. Since E is (e,16) and F is (f,16), the distance EF is |f - e| = 184.

So, (f - e)^2 = (184)^2.

But since e and f are roots of the quadratic equation, we can relate (f - e)^2 to (e + f)^2 - 4ef.

Indeed,

(f - e)^2 = (f + e)^2 - 4ef.

We have e + f = -2g,

and ef = 561.

Therefore,

(f - e)^2 = ( (-2g)^2 ) - 4*561 = 4g² - 2244.

But (f - e)^2 = (184)^2 = 33856.

Therefore,

4g² - 2244 = 33856,

so 4g² = 33856 + 2244 = 36100,

so g² = 36100 /4 = 9025,

so g = sqrt(9025) = 95 or g = -95.

Therefore, g = ±95.

Therefore, e + f = -2g,

so if g = 95, then e + f = -190,

if g = -95, then e + f = 190.

But we also have that e*f = 561.

So we have two cases:

Case 1: g = 95,

then e + f = -190,

and e*f = 561.

Case 2: g = -95,

then e + f = 190,

and e*f = 561.

We need to see which case is possible.

Let me check both cases.

First, Case 1: e + f = -190, e*f = 561.

We can try to solve for e and f.

The quadratic equation would be x² + 2*95 x + 561 = x² + 190x + 561 = 0.

Wait, but discriminant D = 190² - 4*1*561 = 36100 - 2244 = 33856,

sqrt(33856) = 184. So roots are [-190 ± 184]/2.

Thus,

x = [ -190 + 184 ] / 2 = (-6)/2 = -3,

or x = [ -190 - 184 ] /2 = (-374)/2 = -187.

Therefore, e and f would be -3 and -187. Then |f - e| = |-187 - (-3)| = |-184| = 184, which matches.

Alternatively, in Case 2: e + f = 190, e*f = 561.

Quadratic equation: x² - 190x + 561 = 0 (since 2g = -190? Wait, no, wait:

Wait, in case 2, g = -95, so the quadratic equation is x² + 2*(-95)x + 561 = x² - 190x + 561 = 0.

Discriminant D = (190)^2 - 4*1*561 = same as before, 36100 - 2244 = 33856,

sqrt(33856)=184,

so roots are [190 ± 184]/2,

so x = (190 + 184)/2 = 374/2 = 187,

or x = (190 - 184)/2 = 6/2 = 3.

Thus, e and f would be 3 and 187, so |187 - 3| = 184, which also works.

Therefore, there are two possibilities for e and f:

Either e = -3 and f = -187,

or e = 3 and f = 187,

or vice versa. Wait, but since E and F are points on the line DC which is from D(0,16) to C(107,16). Wait, but in the problem statement, the points D, E, C, F are collinear in that order? The problem says "D, E, C, F are collinear", but the order isn't specified. So they could be arranged in different orders along the line.

Wait, but in the Asymptote code example, E is at (3,16) and F is at (90,16), so between D(0,16) and C(70,16) in that example. But in the actual problem, C is at (107,16). So perhaps in the problem, E is between D and C, and F is beyond C? Or maybe E and F are on either side?

Wait, but in the problem, since EFGH is a rectangle, and H and G are above, then E and F must be placed such that the rectangle can be formed. Let me think.

If in case 1, e = -3 and f = -187, then both E and F are to the left of D (which is at x=0). Then the rectangle EFGH would be to the left of D, but then points H and G would be at (-3,33) and (-187,33). Then the circle through A(0,0), D(0,16), H(-3,33), G(-187,33). Hmm, possible, but let's see if that works.

Alternatively, in case 2, e = 3 and f = 187. Then E is at (3,16), F is at (187,16). Since C is at (107,16), so F is to the right of C. Then the rectangle EFGH would extend from E(3,16) to F(187,16), then up to G(187,33) and H(3,33). Then the circle through A(0,0), D(0,16), H(3,33), G(187,33). That also seems possible.

But we need to check if both cases satisfy all conditions, or if one of them is invalid for some reason.

Wait, the problem didn't specify the positions of E and F relative to D and C, except that D, E, C, F are collinear. So they could be arranged in any order along the line. Let me see.

In case 1: If E is at x = -3 and F at x = -187, then the order along the line y=16 would be F (-187), E (-3), D (0), C (107). So D, E, C, F are not in a straight line in the order given, since F is to the left of E and D. The problem states D, E, C, F are collinear, but doesn't specify the order. So maybe that's okay, but perhaps the rectangle EFGH would have to be placed such that EF is between E and F, but if F is further left than E, then the rectangle would be to the left. However, the circle through A, D, H, G: Let's see if that works.

Alternatively, in case 2, E is at 3 and F at 187, so the order along the line is D (0), E (3), C (107), F (187). So D, E, C, F are collinear in that order, which seems to fit the problem's wording better, perhaps. Since the problem lists them as D, E, C, F, maybe they are in that order along the line. So that would suggest case 2 is the correct one.

Alternatively, maybe both are possible, but the problem might have a unique solution, so perhaps only one case gives a valid circle passing through all four points. Wait, but we derived the cases based on the circle condition, so both cases satisfy the circle condition. Hmm.

Wait, but let's check the coordinates in case 1:

If e = -3, f = -187,

then H is (-3, 33),

G is (-187, 33).

Then the four points on the circle are A(0,0), D(0,16), H(-3,33), G(-187,33).

Let me check if these four points are concyclic.

We can compute the circle equation for these points.

Using the general equation again.

We already did this earlier, but let me verify.

Earlier, with e = -3 and f = -187,

then in case 1, g = 95,

so the circle equation is x² + y² + 2*95x + 2*(-8)y = 0,

Wait, no, earlier steps:

Wait, when we set up the equation, with c = 0 and f (the coefficient) was found to be -8. Wait, maybe I confused variables here.

Wait, in the general equation, we had:

After plugging in A and D, we found that c = 0 and the coefficient of y term was 2f (from the general equation), but in the problem, f is also used as a coordinate. Maybe that's confusing. Let me retrace.

Wait, in the general circle equation, I used x² + y² + 2gx + 2fy + c = 0. Then after plugging in A(0,0), we found c = 0. Then plugging in D(0,16), we found 256 + 32f = 0 ⇒ f = -8. So the coefficient of y is 2f = -16, so the equation becomes x² + y² + 2gx -16y = 0.

Then, when plugging in H(e,33) and G(f,33), we derived equations leading to e and f being roots of x² + 2gx + 561 = 0.

Therefore, in case 1 where g = 95,

the quadratic equation is x² + 190x + 561 = 0,

which has roots e = -3 and f = -187,

so the circle equation is x² + y² + 190x -16y = 0.

Let me check if point A(0,0) is on this circle: 0 + 0 + 0 - 0 = 0, yes.

Point D(0,16): 0 + 256 + 0 - 256 = 0, yes.

Point H(-3,33):

(-3)^2 + 33^2 + 190*(-3) -16*33

= 9 + 1089 - 570 - 528

= (9 + 1089) - (570 + 528)

= 1098 - 1098 = 0. Good.

Point G(-187,33):

(-187)^2 + 33^2 + 190*(-187) -16*33

Calculate each term:

(-187)^2 = 187² = let's compute 180² + 2*180*7 + 7² = 32400 + 2520 + 49 = 32400+2520=34920 +49=34969

33² = 1089,

190*(-187) = -190*187. Let's compute 200*187 = 37400, minus 10*187=1870, so 37400 - 1870 = 35530, so -35530.

-16*33 = -528.

So total:

34969 + 1089 -35530 -528

= (34969 + 1089) - (35530 + 528)

= 36058 - 36058 = 0. Perfect.

Similarly, in case 2, with e = 3 and f = 187,

then the quadratic equation is x² - 190x + 561 = 0 (since g = -95, so 2g = -190),

so the circle equation would be x² + y² -190x -16y =0.

Check point H(3,33):

3² + 33² -190*3 -16*33

9 + 1089 - 570 - 528

Same as before: 1098 - 1098 =0.

Point G(187,33):

187² +33² -190*187 -16*33

Same as previous case but with positive 187:

187² = 34969,

33² =1089,

-190*187 = -35530,

-16*33 = -528,

Total: 34969 + 1089 -35530 -528 = same 0.

So both cases are valid. Therefore, there are two possible solutions? But the problem is asking for CE. Let's see what CE would be in each case.

In case 1: E is at x = -3, so CE is the distance from C(107,16) to E(-3,16): |107 - (-3)| = 110.

In case 2: E is at x = 3, so CE is |107 -3| = 104.

Hmm, so two possible answers? But the problem states "what is the length of CE", implying a unique answer. Therefore, perhaps one of the cases is invalid for another reason.

Wait, let's check the rectangle EFGH in each case.

In case 1: E is at (-3,16), F is at (-187,16). Then the rectangle EFGH would have sides EF = distance between E and F is 184 (which is correct), and FG = 17. Since F is at (-187,16), moving up 17 units would put G at (-187, 33). Similarly, H would be at (-3,33). So the rectangle is from E(-3,16) to F(-187,16) to G(-187,33) to H(-3,33). That is a valid rectangle.

However, in this case, the rectangle is to the left of point D(0,16). The problem didn't specify any restrictions on the position, so this is possible.

In case 2: E is at (3,16), F is at (187,16). Then the rectangle goes to G(187,33) and H(3,33). This is also a valid rectangle, to the right of D.

But the problem mentions "rectangles ABCD and EFGH are drawn such that D, E, C, F are collinear". There's no restriction on which side, so both could be possible. But the answer is different in each case. Hmm.

Wait, but maybe the circle through A, D, H, G must also pass through those points in a certain way. Wait, but both cases satisfy that. Alternatively, perhaps the Asymptote code's diagram is an example, but in the problem's coordinates, the positions might be fixed.

Wait, in the Asymptote code example, E is at (3,16), F is at (90,16), so in their example, E is between D(0,16) and C(70,16). But in the problem, C is at (107,16). So perhaps in the problem, E is between D and C, so E's x-coordinate is between 0 and 107, so case 2 where E is at 3 is valid, but case 1 E is at -3 would be to the left of D, which might not be intended? The problem didn't specify, but maybe the figure in the Asymptote suggests that E is between D and C. Since in the Asymptote code, E is at (3,16), which is between D(0,16) and C(70,16). So perhaps in the problem, E is between D and C, so case 2 is the correct one, giving CE = 104. But let me check.

Wait, but in case 2, F is at 187, which is beyond point C (107,16). So F is to the right of C. So the points D(0,16), E(3,16), C(107,16), F(187,16) are collinear in that order. So that's acceptable.

Alternatively, in case 1, the points would be F(-187,16), E(-3,16), D(0,16), C(107,16). So D, E, C, F are in the order E between F and D, but the problem lists them as D, E, C, F, so maybe the intended order is D, E, C, F from left to right, which would correspond to case 2. Hence, the answer is 104.

Wait, but let me check the problem statement again: "Rectangles ABCD and EFGH are drawn such that D,E,C,F are collinear." It doesn't specify the order, but the way it's written, D first, then E, then C, then F, so probably in that order along the line. So that would mean E is between D and C, and F is beyond C. So in that case, E is between D(0,16) and C(107,16), so E's x-coordinate is between 0 and 107, so case 2 where e=3 is valid, giving CE = 107 -3 = 104. The other case would have E at -3, which is to the left of D, so the order would be F, E, D, C, which doesn't match the problem's listing of D,E,C,F. Therefore, case 2 is the correct one, so CE is 104.

Wait, but let me confirm once more. If E is between D and C, then CE is 107 - e. If e=3, then 107-3=104. If in case 1, E is at -3, then CE would be 107 - (-3)=110, but in that case, E is not between D and C. Since the problem says D,E,C,F are collinear, but doesn't specify the order, but the way it's written suggests the order D-E-C-F, so E between D and C, F beyond C. Therefore, the answer should be 104.

Wait, but let me check if there is another constraint I missed. The problem says "A, D, H, G all lie on a circle". In case 2, points H is (3,33) and G is (187,33). So the circle passes through A(0,0), D(0,16), H(3,33), G(187,33). Let me see if this circle makes sense.

Alternatively, maybe there's a calculation mistake. Let me recheck the equations.

Wait, in case 2, with e = 3 and f = 187,

then the quadratic equation for x is x² -190x +561 =0,

which factors? Let me see: 190 is the sum, 561 is the product. 561 factors into 3*187, and 3 + 187 = 190. So yes, (x - 3)(x - 187) = x² -190x + 561. Correct.

Therefore, that's correct.

Alternatively, perhaps the problem requires CE to be 104, but let me see if the problem's Asymptote code has CE as 70 -3 =67? Wait, in the Asymptote code, C is at (70,16), E is at (3,16), so CE is 67, but in the problem, the dimensions are different. So the Asymptote code is just an example.

Alternatively, maybe I made a mistake in assuming the coordinate system. Let me check again.

Wait, in my coordinate system, I placed A at (0,0), so AB is along the x-axis, AD along the y-axis. That's standard. Then with AB =107, so B is (107,0), etc. That seems correct.

Alternatively, maybe the line D,E,C,F is not the top side of rectangle ABCD, but another line? Wait, but D and C are vertices of rectangle ABCD, so DC is a side, which is horizontal in my coordinate system. Since D,E,C,F are collinear, they must lie on the line DC extended. So that's correct.

Alternatively, maybe the rectangle EFGH is placed below the line DC? But then H and G would be below, but then the circle through A, D, H, G might not close. Let me see.

Wait, if rectangle EFGH is below, then FG would be downward. Let me see. Suppose instead that the rectangle is below the line DC. Then points H and G would be at (e, -17) and (f, -17). But then the circle through A(0,0), D(0,16), H(e, -17), G(f, -17). Maybe that's possible, but in that case, the calculation would be different. However, the problem didn't specify the position, but the Asymptote code has it above, so probably it's above.

Alternatively, maybe my mistake is in assuming that FG is vertical. Since EFGH is a rectangle, EF is horizontal (since EF is along DC's line, which is horizontal). Therefore, FG must be vertical, either up or down. Since in the Asymptote code it's up, let's stick with that.

Therefore, I think the correct answer is 104. But let me check once again.

Wait, in case 2, CE is 107 - 3 = 104. But let me see if there is another way to approach the problem that might confirm this.

Alternatively, since A, D, H, G are concyclic, we can use the power of point C with respect to the circle. Wait, but C is not necessarily on the circle. Alternatively, since A, D, H, G are on a circle, we can use the cyclic quadrilateral properties. For quadrilateral ADHG, the product of the slopes of its diagonals should be -1 if it's cyclic? Wait, no, cyclic quadrilaterals have opposite angles summing to 180, or the power of a point.

Alternatively, using coordinates, perhaps compute the circle's equation and see.

Alternatively, let me think of inversion or other methods, but maybe it's more straightforward.

Alternatively, since we have two possible answers, 104 and 110, but based on the problem's likely intended configuration, it's 104. However, let me check if in case 1, the circle through A, D, H, G would have CE = 110, but maybe that's also valid. Since the problem didn't specify, but the Asymptote code's example has CE as 70 -3 =67, which is similar to the case where E is between D and C. So probably the answer is 104.

Wait, but let me check with another method to confirm.

Let me consider the circle through A(0,0), D(0,16), H(e,33), G(f,33). Since these four points are on a circle, the perpendicular bisectors of AD, DH, HG, etc., must meet at the center.

The perpendicular bisector of AD: since A is (0,0) and D is (0,16), the midpoint is (0,8), and the line AD is vertical, so the perpendicular bisector is horizontal, y=8.

The perpendicular bisector of DH: D(0,16) and H(e,33). The midpoint is (e/2, (16+33)/2) = (e/2, 24.5). The slope of DH is (33-16)/(e -0) = 17/e. Therefore, the perpendicular bisector slope is -e/17.

Equation: y -24.5 = (-e/17)(x - e/2).

Similarly, the perpendicular bisector of AG: A(0,0) and G(f,33). Midpoint is (f/2, 16.5). Slope of AG is (33 -0)/(f -0) = 33/f. Perpendicular bisector slope is -f/33.

Equation: y -16.5 = (-f/33)(x - f/2).

The center of the circle must lie at the intersection of these bisectors, as well as the bisector of AD (y=8). Wait, but the perpendicular bisector of AD is y=8, so the center lies on y=8. Let me see if that helps.

Wait, the perpendicular bisector of AD is y=8. So the center (h,8) must also lie on the perpendicular bisectors of DH and AG.

Let me use the perpendicular bisector of DH:

The center (h,8) must satisfy:

8 -24.5 = (-e/17)(h - e/2)

=> -16.5 = (-e/17)(h - e/2)

Multiply both sides by 17:

-16.5 *17 = -e(h - e/2)

Compute 16.5 *17: 16*17=272, 0.5*17=8.5 ⇒ total 280.5

Thus,

-280.5 = -e(h - e/2)

Multiply both sides by -1:

280.5 = e(h - e/2)

Similarly, using the perpendicular bisector of AG:

The center (h,8) must satisfy:

8 -16.5 = (-f/33)(h - f/2)

=> -8.5 = (-f/33)(h - f/2)

Multiply both sides by 33:

-8.5 *33 = -f(h - f/2)

Compute 8.5 *33: 8*33=264, 0.5*33=16.5 ⇒ total 280.5

Thus,

-280.5 = -f(h - f/2)

Multiply by -1:

280.5 = f(h - f/2)

Therefore, we have two equations:

1) 280.5 = e(h - e/2)

2) 280.5 = f(h - f/2)

Therefore, e(h - e/2) = f(h - f/2)

Let me denote this as:

e(h - e/2) = f(h - f/2)

Let me rearrange:

eh - (e²)/2 = fh - (f²)/2

Bring all terms to left:

eh - fh - (e² - f²)/2 =0

Factor:

h(e - f) - ( (e - f)(e + f) ) /2 =0

Factor out (e - f):

(e - f)[ h - (e + f)/2 ] =0

Therefore, either e = f, which is impossible since EF=184, or:

h - (e + f)/2 =0 ⇒ h = (e + f)/2

Therefore, the x-coordinate of the center is the average of e and f.

So h = (e + f)/2.

Now, from before, in our earlier analysis using the quadratic equation, we had e + f = ±190. So h would be ±95.

Therefore, the center is at (h,8) where h = (e + f)/2.

Now, let's substitute h into equation 1:

280.5 = e( h - e/2 ) = e( (e + f)/2 - e/2 ) = e( f /2 ) = (e f)/2

Thus,

(e f)/2 = 280.5 ⇒ e f = 561, which matches our previous result (since e*f =561). So that's consistent.

Therefore, all equations are consistent.

Therefore, the two cases are valid, leading to CE being either 104 or 110. Since the problem likely expects the configuration where E is between D and C, giving CE =104, I think that is the answer.

Alternatively, perhaps there is another constraint I haven't considered. Let me think about the rectangle EFGH. Since it's a rectangle, the sides EF and FG are perpendicular. Since EF is horizontal, FG is vertical, so that's okay.

Wait, in case 1, where E is at (-3,16) and F is at (-187,16), then FG would go up to (-187, 33). The rectangle would be from E(-3,16) to F(-187,16) to G(-187,33) to H(-3,33). The length FG is 17, which is correct (from y=16 to 33). The length GH would be the same as EF, which is 184? Wait no, in a rectangle, opposite sides are equal. So EF = GH =184, and FG = EH =17. So yes, that's correct.

Similarly in case 2, everything is okay.

Hmm. Since both are valid, but the problem is from a competition, likely has a unique solution, so perhaps I made an error in assuming the coordinate system.

Wait, let me check the problem again: "Rectangles ABCD and EFGH are drawn such that D,E,C,F are collinear." So the order D-E-C-F is along the line, so E is between D and C, and F is beyond C. Therefore, in that case, E is between D(0,16) and C(107,16), so E's x-coordinate is between 0 and 107, so e is between 0 and 107. Therefore, in case 2, e=3 is valid, but in case 1, e=-3 is not between D and C. Therefore, case 2 is the correct one, so CE is 107 -3 =104.

Therefore, the answer should be 104. Wait, but let me confirm with another approach.

Alternatively, using coordinate geometry, let me suppose that CE = x, so since C is at (107,16), and E is at (e,16), then CE = 107 - e = x ⇒ e =107 -x.

Similarly, since EF =184, then F is at (e +184,16), so f = e +184 =107 -x +184 = 291 -x.

Alternatively, if E is to the left of D, then F could be at e -184, but in the case where E is between D and C, F is to the right of C, so f = e +184.

Therefore, in this scenario, points are D(0,16), E(e,16), C(107,16), F(e +184,16).

Then, H is (e, 33), G is (e +184, 33).

Now, the four points A(0,0), D(0,16), H(e,33), G(e +184,33) must lie on a circle.

Let me write the equation for these four points.

Let me use the general circle equation again.

Let the circle equation be x² + y² + 2gx + 2fy + c =0.

Plugging in A(0,0):

0 +0 +0 +0 +c =0 ⇒ c=0.

Plugging in D(0,16):

0 + 256 +0 +32f +0 =0 ⇒ 256 +32f =0 ⇒ f= -8.

Thus, the equation becomes x² + y² +2gx -16y =0.

Now plug in H(e,33):

e² + 33² + 2g e -16*33 =0 ⇒ e² + 1089 + 2ge -528 =0 ⇒ e² + 2ge +561 =0. (1)

Plug in G(e+184, 33):

(e+184)^2 +33² +2g(e+184) -16*33 =0

Expand (e+184)^2 = e² + 368e + 184²,

So:

e² + 368e + 184² + 1089 + 2ge + 368g -528 =0

Wait, let me compute step by step:

G's x-coordinate is e +184, y=33.

So plug into equation:

(e +184)^2 + 33^2 + 2g(e +184) -16*33 =0

Expand (e +184)^2 = e² + 2*184 e + 184²,

33² is 1089,

2g(e +184) = 2ge + 2g*184,

-16*33 = -528.

So altogether:

e² + 368e + 184² + 1089 + 2ge + 368g + 1089 -528 =0 ?

Wait, no, let me re-express:

Wait, the equation is:

(e +184)^2 + 33² + 2g(e +184) -16*33 =0

= [e² + 368e + 184²] + [1089] + [2ge + 368g] + [-528] =0

Combine like terms:

e² + 368e + 184² + 1089 + 2ge + 368g -528 =0

Simplify constants:

184² + (1089 -528) = 184² + 561.

Terms with e:

(368e + 2ge) = e(368 + 2g)

Terms with g:

368g

So overall:

e² + e(368 + 2g) + 368g + 184² + 561 =0.

But from equation (1):

e² + 2ge = -561.

So substitute e² +2ge = -561 into the above equation:

(-561) + 368e + 368g + 184² +561 =0

Simplify:

-561 +561 cancels,

368e + 368g + 184² =0

Factor 368:

368(e + g) + 184² =0

Divide both sides by 184:

2(e + g) + 184 =0 ⇒ 2(e + g) = -184 ⇒ e + g = -92 ⇒ g = -92 -e.

Now, from equation (1):

e² + 2g e +561 =0,

substitute g = -92 -e,

then,

e² + 2*(-92 -e)*e +561 =0,

Expand:

e² -184e -2e² +561 =0,

Combine like terms:

-e² -184e +561 =0,

Multiply by -1:

e² +184e -561 =0,

Solve for e:

e = [-184 ± sqrt(184² +4*561)] /2

Compute discriminant:

184² + 4*561.

184² = (180 +4)^2 = 180² + 2*180*4 +4² = 32400 + 1440 +16 = 32400+1440=33840 +16=33856.

4*561 = 2244.

Thus discriminant is 33856 +2244 = 36100.

sqrt(36100)=190.

Thus,

e = [ -184 ±190 ] /2.

So two solutions:

e = (-184 +190)/2 =6/2=3,

or e=(-184 -190)/2= (-374)/2= -187.

Therefore, e=3 or e=-187.

Therefore, corresponding to the two cases we had before.

Therefore, if e=3, then CE =107 -3=104,

if e=-187, CE=107 - (-187)= 107+187=294? Wait, no, wait CE is the distance between C(107,16) and E(e,16). If e=-187, then CE is |107 - (-187)|= 107+187=294. Wait, but earlier I thought it was 110. Wait, that's a discrepancy. Wait, what's wrong here?

Wait, in the first approach, when I considered case 1 as e=-3 and f=-187, but in this current approach, when e=-187, then f = e +184= -187 +184= -3. So in that case, E is at (-187,16), F is at (-3,16). So the distance CE would be |107 - (-187)|= 294, but in my first analysis I had case 1 as e=-3 and f=-187, leading to CE=110. So there's inconsistency here. Wait, perhaps I mixed up e and f in the first analysis.

Wait, in the first analysis, I considered E as (e,16) and F as (f,16), with |f -e|=184. So if e=-187, then f = e +184= -3. So E is at (-187,16), F at (-3,16). Then CE is distance from (107,16) to (-187,16) is 107 - (-187)= 294. But in my first analysis, I had case 1 as e=-3 and f=-187, which would give CE=107 - (-3)=110. So that was a different assignment.

Ah! Here's the mistake. In the first approach, I considered e and f as roots of the quadratic, so e and f can be in either order. So when I considered case 1 as e=-3 and f=-187, that corresponds to E being at (-3,16) and F at (-187,16), but then the distance between them is | -187 - (-3)|=184, which is correct. But in that case, the order along the line would be F (-187), E (-3), D (0), C (107). So D,E,C,F are collinear in the order E between F and D, but the problem lists them as D,E,C,F, so that would not be in order. Whereas if we take e=-187 and f=-3, then E is at (-187,16), F at (-3,16), so the order is E, F, D, C, which also doesn't match.

Wait, but in the second approach, I assumed that F is at e +184, so if E is at e, then F is at e +184. So in that case, if e is -187, then F is at -3, so the order is E (-187), F (-3), D (0), C (107). So D,E,C,F are not in order. Therefore, the only case where D,E,C,F are in order is when E is between D and C, so E is at 3, F is at 3+184=187, so F is at 187, which is beyond C (107). Thus, the order is D (0), E (3), C (107), F (187), which matches the problem's listing. Therefore, in this case, CE is 107 -3=104. The other solution gives CE=294, but in that case, the points are ordered E,F,D,C,F which doesn't fit the problem's collinear order. Therefore, the correct answer is 104.

Therefore, the length of CE is 104.

**Final Answer**
\boxed{104}
</think>

Given rectangles \(ABCD\) and \(EFGH\) with \(D, E, C, F\) collinear and \(A, D, H, G\) concyclic, we need to find the length of \(CE\). The given dimensions are \(BC = 16\), \(AB = 107\), \(FG = 17\), and \(EF = 184\).

1. **Coordinate Setup**:
   - Place \(A\) at \((0, 0)\), \(B\) at \((107, 0)\), \(C\) at \((107, 16)\), and \(D\) at \((0, 16)\).
   - Points \(D, E, C, F\) are collinear on the line \(y = 16\). Let \(E = (e, 16)\) and \(F = (f, 16)\) with \(|f - e| = 184\).

2. **Circle Condition**:
   - Points \(A(0,0)\), \(D(0,16)\), \(H(e, 33)\), and \(G(f, 33)\) lie on a circle.
   - The circle equation derived from these points is \(x^2 + y^2 + 2gx - 16y = 0\).

3. **Solving for \(e\) and \(f\)**:
   - Using the circle equation and the distance \(EF = 184\), we find:
     - \(e\) and \(f\) are roots of \(x^2 + 2gx + 561 = 0\).
     - Solving, we get two cases: \(e = 3, f = 187\) or \(e = -187, f = -3\).

4. **Determining Valid Configuration**:
   - The problem specifies \(D, E, C, F\) collinear in order, implying \(E\) is between \(D\) and \(C\), and \(F\) is beyond \(C\).
   - Thus, \(E = (3, 16)\) and \(F = (187, 16)\) is valid, giving \(CE = 107 - 3 = 104\).

The length of \(CE\) is \(\boxed{104}\).