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. Therefore, n=10 is a losing position for Alice. Correct.

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=2 mod5. For example, n=2,7,12, etc. Let me check n=12.

n=12 mod5=2, so it's a losing position. Alice has to move first. She can take 1 or 4.

If she takes 1, leaves 11 (W). Bob can take 1, leaving 10 (L). Then Alice is at 10 (L), so she can't win. Alternatively, Bob could take 4 from 11, leaving 7 (L). Either way, Bob can force Alice into a losing position.

If Alice takes 4 from 12, leaving 8 (W). Bob can take 1, leaving 7 (L). So same result. Therefore, n=12 is indeed a losing position. So the pattern holds.

Therefore, I think my answer is correct. The number is 809.

**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 congruent to 0 mod 5: These are 5, 10, 15, ..., 2020. The count is \( \left\lfloor \frac{2024}{5} \right\rfloor = 404 \).
- Numbers congruent to 2 mod 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 $CE$?<｜Assistant｜><think>

Okay, so I have this geometry problem here about two rectangles, ABCD and EFGH, with some points collinear and some lying on a circle. The goal is to find the length of CE. Let me try to parse all the given information step by step.

First, let me note down the given data:

- Rectangles ABCD and EFGH.
- Points D, E, C, F are collinear. So, they all lie on a straight line.
- Points A, D, H, G all lie on a circle. That means these four points are concyclic or lie on the same circle.
- The sides BC = 16, AB = 107, FG = 17, and EF = 184. Need to find CE.

Hmm, okay. Let me try to visualize this. Since ABCD is a rectangle, opposite sides are equal. So, AB = CD = 107, and BC = AD = 16. Similarly, for rectangle EFGH, EF = GH = 184, and FG = EH = 17.

Since D, E, C, F are collinear, that line must be the line containing points from both rectangles. Let me try to sketch a rough diagram in my mind. Let me consider rectangle ABCD with points labeled in order, so that AB is the top side, BC the right side, CD the bottom, and DA the left side. Similarly, EFGH would have EF as the top, FG the right, GH the bottom, and HE the left.

But since D, E, C, F are collinear, maybe the line DEF C is a straight line. Wait, the order is D, E, C, F. So starting from D, going through E, then C, then F? Hmm, so D is a corner of rectangle ABCD, and E is a corner of EFGH. Since they are collinear with C and F, perhaps the rectangles are placed such that this line is a common side or something?

Alternatively, maybe the rectangles are arranged so that side DC of rectangle ABCD is along the same line as EF of rectangle EFGH? Wait, but EF is a side of the other rectangle. Hmm, perhaps not. Let me think again.

Alternatively, since D, E, C, F are collinear, the line passes through D, then E, then C, then F. So, starting at D (a corner of ABCD), moving to E (a corner of EFGH), then to C (another corner of ABCD), then to F (another corner of EFGH). That suggests that the rectangles are placed such that this line is a transversal passing through both rectangles.

Also, points A, D, H, G lie on a circle. So, these four points are concyclic. That must impose some condition on the positions of the rectangles.

Hmm, perhaps coordinate geometry would help here. Let me try to assign coordinates to the points to model this situation.

Let me set up a coordinate system. Let me place point D at the origin (0,0) to simplify calculations. Since ABCD is a rectangle, if D is (0,0), then C would be along the x-axis, so C is (DC, 0). Since DC is equal to AB, which is 107, so DC = 107. Therefore, point C is at (107, 0). Then, since BC = 16, which is the vertical side, point B would be at (107, 16), and point A would be at (0, 16). Wait, but hold on, in a rectangle, the sides AB and DC are the lengths, and BC and AD are the widths. So, if AB = 107, then the length from A to B is 107, so if A is at (0,16), then B would be at (107,16), and D is at (0,0), so C is at (107,0). That makes sense.

So, coordinates:

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

Now, rectangle EFGH. We need to place this rectangle such that points E, F are on the line passing through D, E, C, F. Since D is (0,0) and C is (107,0), the line D-E-C-F must be the x-axis? Wait, because D and C are on the x-axis. So if D, E, C, F are collinear, then they all lie on the x-axis. Therefore, points E and F must also lie on the x-axis.

Therefore, E and F are on the x-axis. Since E is a corner of rectangle EFGH, and F is another corner. Since EFGH is a rectangle, the sides EF and FG are adjacent sides. Given that EF = 184 and FG =17, so EF is the length along the x-axis from E to F, which is 184 units. Wait, but EF is a side of the rectangle, so if E and F are on the x-axis, then the rectangle EFGH must have EF as its base on the x-axis, and then the other sides going up or down.

But since rectangles can be placed either above or below the x-axis. Let me think. Since points A, D, H, G are concyclic, and A is at (0,16), D is at (0,0), so the circle passes through (0,16) and (0,0). Points H and G are also on this circle. Since H and G are corners of rectangle EFGH, which is connected to the x-axis via E and F.

Let me try to assign coordinates to E and F first. Since they are on the x-axis, let me denote E as (e, 0) and F as (f, 0). Since EF = 184, the distance between E and F is |f - e| = 184. Also, since D, E, C, F are collinear on the x-axis, the order is D (0,0), then E (e,0), then C (107,0), then F (f,0). So the order along the x-axis must be D, E, C, F. Therefore, the coordinates must satisfy 0 < e < 107 < f, so that moving from D (0,0) to E (e,0), then to C (107,0), then to F (f,0). Therefore, the distance from E to C is 107 - e, and from C to F is f - 107. The total distance from E to F is (f - e) = 184. So, f = e + 184.

But also, since E is between D and C, e must be between 0 and 107, and F is beyond C, so f = e + 184 must be greater than 107. Since e < 107, then f = e + 184 would be at least 184, which is way beyond 107, so that's okay.

Now, rectangle EFGH has sides EF = 184 and FG =17. Since EF is along the x-axis from E(e,0) to F(f,0), then FG is a vertical side upwards or downwards from F. Since it's a rectangle, the next point G would be either (f, 17) or (f, -17). Similarly, H would be (e, 17) or (e, -17), depending on the direction. Let me assume for now that the rectangle is above the x-axis, so FG is upwards. So, point G would be (f,17), and H would be (e,17). Alternatively, if it's downward, G would be (f, -17) and H (e, -17). But since points A, D, H, G are on a circle, and A is at (0,16), D is (0,0), so if H and G are below the x-axis, their y-coordinates would be negative, but the circle through A(0,16), D(0,0), and two other points below might not pass through them easily. Maybe it's better to assume the rectangle is above. Let me check both possibilities.

First, let me suppose that the rectangle EFGH is above the x-axis. Then:

- E is (e,0)
- F is (f,0) = (e + 184, 0)
- G is (f,17) = (e + 184,17)
- H is (e,17)

Alternatively, if the rectangle is oriented the other way, but since FG is length 17, which is the vertical side, that should be correct.

Now, the points A(0,16), D(0,0), H(e,17), and G(e + 184,17) lie on a circle. So these four points must be concyclic. Let me write the equation for a circle passing through these four points.

First, let's recall that three non-collinear points determine a circle, so if four points lie on a circle, they must satisfy the same circle equation.

Let me first find the equation of the circle passing through A(0,16), D(0,0), and H(e,17). Then, check if G(e + 184,17) also lies on it.

Let me denote the general equation of a circle as:

x² + y² + 2gx + 2fy + c = 0

Plugging in point D(0,0):

0 + 0 + 0 + 0 + c = 0 ⇒ c = 0.

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

Now plug in point A(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,17):

(e)^2 + (17)^2 + 2g(e) - 16*(17) = 0

Compute:

e² + 289 + 2ge - 272 = 0 ⇒ e² + 2ge + 17 = 0. (Equation 1)

Now, we need point G(e + 184,17) to also lie on this circle. Let's plug G into the equation:

(e + 184)^2 + (17)^2 + 2g(e + 184) - 16*(17) = 0

Expand (e + 184)^2:

e² + 368e + (184)^2 + 289 + 2g(e + 184) - 272 = 0

Wait, let me compute step by step:

First, expand (e + 184)^2:

= e² + 2*184*e + 184²

Then, adding 17² (which is 289):

So total terms:

(e² + 368e + 184²) + 289 + 2g(e + 184) - 16*17 = 0

Compute 16*17 = 272, so:

e² + 368e + 184² + 289 + 2ge + 368g - 272 = 0

Combine constants:

184² is 33856, so 33856 + 289 = 34145; 34145 - 272 = 33873.

So:

e² + 368e + 2ge + 368g + 33873 = 0

Factor terms with e and g:

e² + (368 + 2g)e + 368g + 33873 = 0 (Equation 2)

Now, from Equation 1, we have:

e² + 2ge + 17 = 0 ⇒ e² = -2ge -17

We can substitute e² from Equation 1 into Equation 2.

Substituting into Equation 2:

(-2ge -17) + (368 + 2g)e + 368g + 33873 = 0

Let me expand this:

-2ge -17 + 368e + 2ge + 368g + 33873 = 0

Notice that -2ge and +2ge cancel out.

So remaining terms:

-17 + 368e + 368g + 33873 = 0

Combine constants:

-17 + 33873 = 33856

Thus:

368e + 368g + 33856 = 0

Factor out 368:

368(e + g) + 33856 = 0

Divide both sides by 8 to simplify:

46(e + g) + 4232 = 0

Wait, 368 divided by 8 is 46, and 33856 divided by 8 is 4232.

So:

46(e + g) = -4232 ⇒ e + g = -4232 / 46

Calculate that:

4232 ÷ 46: 46*90 = 4140, so 4232 - 4140 = 92, so 90 + 92/46 = 90 + 2 = 92. Wait, actually:

Wait 46*92 = 46*(90 +2) = 4140 + 92 = 4232. So 4232 /46 = 92. Therefore,

e + g = -92 ⇒ g = -92 - e

So, we have g expressed in terms of e. Now, let's go back to Equation 1:

e² + 2ge +17 =0

Substitute g = -92 - e into this equation:

e² + 2*(-92 - e)*e +17 =0

Expand:

e² - 184e - 2e² +17 =0

Combine like terms:

e² -2e² = -e²

So:

- e² -184e +17 =0 ⇒ Multiply both sides by -1:

e² + 184e -17 =0

So quadratic equation: e² + 184e -17 =0

Hmm, solving for e:

Using quadratic formula:

e = [-184 ± sqrt(184² + 68)] / 2

Wait discriminant is 184² + 4*1*17 = 184² + 68

Wait, discriminant D = (184)^2 + 4*1*17 = 33856 + 68 = 33924

Hmm, sqrt(33924). Let me see if this is a perfect square.

Let me check 184^2 is 33856, so sqrt(33924) is sqrt(33856 + 68) = sqrt(184² + 68). Hmm, not obvious. Let me compute sqrt(33924):

Let me see 184^2 = 33856, so 184.5^2 = (184 + 0.5)^2 = 184² + 2*184*0.5 + 0.25 = 33856 + 184 + 0.25 = 34040.25, which is higher than 33924. Hmm, so between 184 and 184.5.

Wait, maybe I made a mistake in calculation. Let me check discriminant again.

Wait, original equation after substitution was:

e² + 184e -17 =0

Thus, discriminant D = (184)^2 -4*1*(-17) = 33856 + 136 = 33992. Wait, hold on, I think I miscalculated earlier.

Wait, quadratic equation ax² +bx +c, discriminant is b² -4ac.

Here, a=1, b=184, c= -17.

Thus D = 184² -4*1*(-17) = 33856 + 68 = 33924. Wait, yes, that's correct. So sqrt(33924). Let me see:

Let me try 184.2 squared:

Wait, perhaps 184.2^2 = ?

Alternatively, maybe 184.2^2 is not helpful. Alternatively, perhaps factor 33924.

Divide by 4: 33924 /4 = 8481. Hmm, 8481. Let me see sqrt(8481). 92^2=8464, so 92^2=8464, so 92.1^2= 92^2 + 2*92*0.1 +0.1^2=8464+18.4+0.01=8482.41, which is close to 8481. Hmm, so sqrt(8481) is approx 92.09. So sqrt(33924)=sqrt(4*8481)=2*92.09≈184.18. Hmm, so irrational. That seems messy. Maybe I made a mistake in the setup?

Alternatively, perhaps the rectangle is below the x-axis? Let me check that possibility.

If the rectangle EFGH is below the x-axis, then points G and H would have y-coordinate -17. Let me try that.

So, if E is (e,0), F is (f,0) = (e + 184, 0). Then, rectangle EFGH would go downward, so G is (f, -17), H is (e, -17).

Then, points A(0,16), D(0,0), H(e, -17), G(f, -17) = (e + 184, -17) must lie on a circle.

Let me redo the circle equation with these points.

Again, general equation x² + y² + 2gx + 2fy + c =0.

Plug in D(0,0): 0 +0 +0 +0 +c=0 ⇒ c=0.

Plug in A(0,16): 0 + 256 +0 +32f =0 ⇒ 256 +32f=0 ⇒ f= -8, same as before.

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

Now plug in H(e, -17):

e² + (-17)^2 + 2g*e -16*(-17) =0 ⇒ e² +289 +2ge +272 =0 ⇒ e² + 2ge + 561 =0 (Equation 1a)

Now plug in G(e +184, -17):

(e +184)^2 + (-17)^2 +2g(e +184) -16*(-17)=0

Expand (e +184)^2 = e² + 368e + 184²

Add (-17)^2 =289, so total:

e² +368e + 33856 +289 + 2ge + 368g + 272 =0

Wait, let me do step by step:

First, (e +184)^2 + (-17)^2 = e² + 368e + 184² + 289

Then, 2g(e +184) = 2ge + 368g

Then, -16*(-17) = +272, so adding all terms:

e² + 368e + 33856 + 289 + 2ge + 368g + 272 =0

Combine constants:

33856 + 289 = 34145; 34145 +272 = 34417

Thus equation becomes:

e² + 368e + 2ge + 368g + 34417 =0 (Equation 2a)

Again, from Equation 1a: e² +2ge = -561

Substitute into Equation 2a:

(-561) + 368g + 34417 + (368e)?

Wait, wait, let me see:

Wait, Equation 1a: e² +2ge = -561

So in Equation 2a:

e² + (368e + 2ge) + 368g + 34417 =0

Wait, perhaps better to substitute e² = -2ge -561 from Equation 1a into Equation 2a.

So Equation 2a:

(e²) + 368e + 2ge + 368g + 34417 =0

Substitute e² = -2ge -561:

(-2ge -561) + 368e + 2ge + 368g + 34417 =0

Simplify terms:

-2ge +2ge cancels out.

Left with:

-561 + 368e + 368g + 34417 =0

Combine constants:

-561 +34417 = 33856

Thus:

368e + 368g + 33856 =0

Same as before! So 368(e + g) +33856 =0 ⇒ same as previous case.

Thus, e + g = -33856 /368 = let's compute that:

33856 ÷ 368. Let's divide numerator and denominator by 8: 33856/8=4232; 368/8=46. So 4232/46=92. So e +g = -92, same as before.

Therefore, g = -92 -e.

Then, plugging back into Equation 1a:

e² +2ge +561 =0

Substitute g = -92 -e:

e² +2*(-92 -e)*e +561 =0

Expand:

e² -184e -2e² +561 =0

Combine terms:

- e² -184e +561 =0 ⇒ Multiply by -1:

e² +184e -561 =0

Ah, here the equation is e² +184e -561 =0, which is different from the previous case. Let's compute discriminant here:

D = 184² +4*561 = 33856 + 2244 = 36100

Ah! 36100 is a perfect square, since 190² = 36100. So sqrt(36100)=190.

Therefore, solutions:

e = [-184 ±190]/2

So two solutions:

First solution: (-184 +190)/2 =6/2=3

Second solution: (-184 -190)/2= (-374)/2= -187

Now, since in this scenario, the rectangle is below the x-axis, so points E and F are on the x-axis with E between D(0,0) and C(107,0). So e must be between 0 and 107. The solutions e=3 and e=-187. e=-187 would place E far to the left of D(0,0), which would not be between D and C. So e=3 is valid.

Therefore, e=3. Then, g = -92 -e = -92 -3 = -95.

Therefore, the coordinates are:

E is (3,0)

F is (e +184,0)= 3+184=187, so F is (187,0)

G is (187, -17)

H is (3, -17)

Now, let's check if points A(0,16), D(0,0), H(3,-17), G(187,-17) lie on a circle.

We can verify with the circle equation.

Earlier, we had the equation x² + y² +2gx -16y =0, with g = -95, so:

Equation becomes:

x² + y² + 2*(-95)x -16y =0 ⇒ x² + y² -190x -16y =0

Let me check point G(187, -17):

187² + (-17)^2 -190*187 -16*(-17)

Compute each term:

187²: 187*187. Let's compute 180²=32400, 2*180*7=2520, 7²=49. So (180+7)^2= 32400 +2520 +49= 32400+2520=34920 +49=34969

(-17)^2=289

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

-16*(-17)= +272

Adding all together:

34969 +289 -35530 +272

Compute step by step:

34969 +289 = 35258

35258 -35530 = -272

-272 +272 =0. Perfect, so G lies on the circle.

Similarly, check H(3,-17):

3² + (-17)^2 -190*3 -16*(-17)

9 +289 -570 +272

9+289=298; 298 -570= -272; -272 +272=0. Good.

Point A(0,16):

0 + 256 -0 -256=0. Correct.

Point D(0,0): 0+0 -0 -0=0. Correct.

So that works. Therefore, the correct configuration is with the rectangle EFGH below the x-axis, with E at (3,0), F at (187,0), G at (187,-17), H at (3,-17).

Therefore, now, the problem asks for the length of CE.

Point C is (107,0), and point E is (3,0). Wait, but hold on, in the collinear points D,E,C,F, the order is D(0,0), E(3,0), C(107,0), F(187,0). So CE is the distance from C(107,0) to E(3,0). Wait, but that would be 107 -3 =104? But that seems straightforward, but let me confirm.

Wait, CE is from C(107,0) to E(3,0). Since they are on the x-axis, the distance is |107 -3| =104. But the problem might be expecting a different answer? Wait, but let me check again.

Wait, but in the problem statement, it says "Rectangles ABCD and EFGH are drawn such that D,E,C,F are collinear." So the order is D, E, C, F. So from D to E to C to F. So E is between D and C, and C is between E and F. Therefore, the distance from C to E is indeed CE = DC - DE. Since DC is 107 (as ABCD is a rectangle with AB=107), and DE is the distance from D(0,0) to E(3,0) which is 3. Therefore, CE = 107 -3 =104. But let me check coordinates again.

Wait, in coordinates, C is at (107,0), E is at (3,0). The distance between them is |107 -3| =104. So that's straightforward. But the problem gives FG=17 and EF=184, which we used in the calculation, so perhaps 104 is the answer. But let me see if I made any mistake here.

Wait, but in the problem statement, it says "what is the length of CE". If it's 104, that's straightforward, but maybe I made an error in assuming the rectangle is below the x-axis? Let me check again.

Wait, in the problem, the four points A,D,H,G are concyclic. In the case where the rectangle is above the x-axis, we ended up with a quadratic equation with an irrational solution, which is possible, but when we took the rectangle below, we got an integer solution, which is nicer, so that's likely the correct configuration.

Alternatively, maybe the problem expects CE to be CF - CE? Wait no, CE is just the distance between C and E. Wait, but in the problem statement, the order is D,E,C,F, so moving from D to E to C to F, so E is between D and C, so CE is from C to E going backwards, so the length is indeed DC - DE. Since DC is 107, DE is 3, so 104. Hmm.

Wait, but let me check if the problem perhaps refers to CE as a different segment? Maybe in another configuration?

Alternatively, perhaps I made a mistake in the coordinate assignments. Let me confirm again.

Wait, in the problem, rectangles ABCD and EFGH. The order of the letters in the rectangles matters. For rectangle ABCD, the order is such that AB is a side, BC is next, etc. So if it's labeled in order, then ABCD is arranged so that moving from A to B to C to D and back to A makes the rectangle. So if I placed A at (0,16), B at (107,16), C at (107,0), D at (0,0), that is correct.

Similarly, rectangle EFGH: E to F to G to H. If E is (3,0), F is (187,0), then G would be (187,-17), H would be (3,-17), so that's a rectangle. Yes, that works.

Therefore, CE is from (107,0) to (3,0), so distance 104. But let me check the problem again to see if I misread something.

Wait, the problem says "what is the length of CE". Hmm, but maybe CE is not along the x-axis? Wait, but D,E,C,F are collinear on the x-axis, so CE is along the x-axis. So the length should be 104. But let me check the problem's given numbers again.

Wait, BC=16, which is the vertical side, so that's correct (from (107,16) to (107,0)). AB=107, from (0,16) to (107,16). FG=17, which is the vertical side of the other rectangle, so from (187,0) to (187,-17), which is 17 units. EF=184, from (3,0) to (187,0), which is 184 units. So that's all correct.

Therefore, CE is indeed 107 -3 =104. Hmm, but maybe I made a mistake in the calculation of e?

Wait, in the quadratic equation when the rectangle is below, we had e² +184e -561 =0, which had solutions e=3 and e=-187. So e=3 is correct. So CE is 107 -3 =104. Hmm, but maybe the problem expects CE to be CF - CE? Wait, no, CE is just the distance between C and E. Wait, but in the problem statement, the points are D,E,C,F collinear in that order, so the distance from C to E would be CE = |C - E| = 107 -3 =104. So that's the answer.

Wait, but let me check if the problem might have meant CE as in the segment from C to E going the other way via the circle? But no, CE is just the straight line between them, which is along the x-axis. Hmm.

Alternatively, perhaps I made a mistake in the circle equation? Let me confirm with coordinates.

The circle passes through A(0,16), D(0,0), H(3,-17), G(187,-17). Let me compute the circle's equation again.

We had the equation x² + y² -190x -16y =0.

Let me write it in standard form by completing the squares.

x² -190x + y² -16y =0

x² -190x + (190/2)^2 - (190/2)^2 + y² -16y + (16/2)^2 - (16/2)^2 =0

So,

(x - 95)^2 - 95² + (y -8)^2 - 64 =0

Thus,

(x -95)^2 + (y -8)^2 = 95² +64

Compute 95²: 9025, so 9025 +64=9089. So radius squared is 9089, center at (95,8). That seems okay.

Check point A(0,16):

(0-95)^2 + (16-8)^2 = 9025 +64=9089. Correct.

Point H(3,-17):

(3-95)^2 + (-17 -8)^2 = (-92)^2 + (-25)^2= 8464 +625=9089. Correct.

Point G(187,-17):

(187-95)^2 + (-17 -8)^2 = (92)^2 + (-25)^2= same as above, 8464+625=9089. Correct.

So all points lie on the circle. Therefore, the configuration is correct, so CE is indeed 104. Hmm, but the problem is from an Olympiad perhaps? Maybe the answer is different. Wait, let me check the problem again.

Wait, the problem says "Rectangles ABCD and EFGH are drawn such that D,E,C,F are collinear." So the order is D-E-C-F, so E is between D and C, and F is beyond C. So CE is the distance from C to E, which is indeed 107 -3 =104. Hmm, but maybe I misread the problem? Let me check once more.

Wait, the problem says "what is the length of CE". If CE is from point C to point E, which are both on the x-axis, then yes, 104. Alternatively, maybe the problem uses a different notation where CE is a different segment? For example, in rectangle EFGH, is CE a diagonal or something? Wait, but CE would be between point C (of ABCD) and E (of EFGH). So yes, the straight line between them, which is along the x-axis. So 104.

Alternatively, maybe I made a mistake in the calculation of e? Let me check the quadratic equation again.

We had for the lower rectangle:

Equation 1a: e² +2ge +561 =0

Equation 2a led us to e +g = -92, so g = -92 -e

Substituting into Equation 1a:

e² +2*(-92 -e)*e +561 = e² -184e -2e² +561 = -e² -184e +561 =0 ⇒ e² +184e -561 =0

Solutions:

e = [-184 ± sqrt(184² +4*561)]/2

Compute discriminant:

184² = 33856

4*561 = 2244

Total discriminant: 33856 +2244 =36100, sqrt is 190.

Thus, e=( -184 ±190)/2

So ( -184 +190)/2 =6/2=3, and the other is negative. So e=3 is correct.

Therefore, CE is 107 -3 =104. Hmm, but maybe the problem expects CE to be CF minus something? Wait, CF is from C(107,0) to F(187,0), which is 80 units. But CE is 104, which is longer than CF? Wait no, CF is 187-107=80, so CE is from C back to E, so 107-3=104. That's correct.

Alternatively, maybe the problem is in the USA and uses different notation for rectangles? Like the order of the letters? Let me check the rectangle EFGH. If EFGH is labeled such that E-F-G-H-E, then the coordinates are as we have. So E(3,0), F(187,0), G(187,-17), H(3,-17). So that's a rectangle.

Alternatively, maybe the rectangle is labeled differently? For example, maybe EFGH has EF as vertical side? But the problem states FG=17 and EF=184, so EF is the horizontal side of length 184, FG is vertical 17. So that's correct.

Hmm, I think my answer is correct. Maybe the problem is expecting 104, but let me see if there's an alternative approach.

Alternatively, since A,D,H,G are concyclic, perhaps using power of a point or cyclic quadrilateral properties.

Let me think. Since A,D,H,G are on a circle, then the power of point C with respect to the circle could be considered? Wait, but maybe not necessary.

Alternatively, in the coordinate system, since we found CE =104, maybe that's the answer. Alternatively, perhaps I made a mistake in assuming the rectangle is below the x-axis? Let me see if there is another possible configuration.

Wait, if the rectangle is above the x-axis, then the quadratic equation was e² +184e -17 =0, which had solutions with sqrt(33924). Let me see what that would give.

sqrt(33924). Let me see 184^2 =33856, so sqrt(33924)=sqrt(33856 +68)= approx 184 + 68/(2*184) approximately, so about 184 + 0.185≈184.185. So e = [ -184 ±184.185 ] /2.

Taking the positive root: (-184 +184.185)/2≈0.185/2≈0.0925. So e≈0.0925, which is between 0 and 107. Then CE would be 107 -0.0925≈106.9075, which is not an integer, but the problem gives all sides as integers, so probably the intended answer is 104, so the configuration with the rectangle below the x-axis is correct.

Therefore, I think the answer is 104. Wait, but let me check once again.

Wait, in the problem statement, it says "A,D,H,G all lie on a circle". In the configuration with the rectangle below the x-axis, H is (3,-17) and G is (187,-17). So those are the two lower corners of the rectangle EFGH. So yes, those points along with A and D lie on the circle we found. So that's correct.

Therefore, I think the answer is 104. But since the problem is likely to have an integer answer, and 104 is an integer, that's probably correct. Wait, but let me check if the problem might have CE as CF - EF or something? Wait, CF is 187-107=80, but CE is 104, which is separate.

Alternatively, maybe I messed up the direction. Wait, in the problem, the order is D,E,C,F collinear. So starting at D, going through E, then C, then F. So the distance from D to E is e, E to C is 107 -e, and C to F is f -107 = (e +184) -107 = e +77. So CE is E to C, which is 107 -e. Since e=3, that's 104. So that's correct.

Therefore, I think the answer is 104. Wait, but let me check the problem again for any possible misread.

Wait, the problem says "Rectangles ABCD and EFGH". So ABCD has sides AB=107, BC=16, so that's correct. EFGH has EF=184, FG=17, so that's correct. The points D,E,C,F collinear, and A,D,H,G concyclic. All conditions are satisfied in this configuration. So I think 104 is correct.

Hmm, but maybe I made a mistake in the problem's rectangle labeling? For example, maybe ABCD is labeled differently? Let me confirm.

In rectangle ABCD, the order of the letters matters. If it's ABCD, then the sides are AB, BC, CD, DA. So if A is connected to B and D, then the rectangle is arranged so that moving around the rectangle, the points are in order A-B-C-D-A. So if A is at (0,16), B would be at (107,16), C at (107,0), D at (0,0). That is the standard labeling. So that's correct.

Similarly, EFGH: E-F-G-H-E. So E to F is the base, F to G is the side, etc. So coordinates as assigned are correct.

Therefore, I think the answer is 104. Wait, but let me check with another method to confirm.

Alternative approach:

Since A,D,H,G are concyclic, the power of point C with respect to the circle might relate some distances. But since C is on the line DEF C F, which is the x-axis, and the circle passes through D and A, perhaps some power of point applies.

Power of point C with respect to the circle is equal to CA * CD or something? Wait, power of a point formula: for a point P outside a circle, power is PA * PB where PA and PB are lengths of intersections from P to the circle. Alternatively, if the point is on the circle, power is zero.

Wait, point C is not necessarily on the circle. Let me compute the power of point C with respect to the circle ADHG.

The power should be equal to CA * CD? Wait, not sure. Let me recall the power of a point formula.

Power of point C with respect to the circle is equal to the product of the lengths from C to the points of intersection along any line through C. Since line CD passes through D and C, and D is on the circle, so the power of C is CD * CE, where E is the other intersection point. Wait, but in our case, the line D-E-C-F passes through D (on the circle), E, C, F. Since D is on the circle, the power of C would be CD * CE = (distance from C to D) * (distance from C to E). Wait, but CD is 107, CE is what we need to find. But also, the power of point C can be calculated as CO² - r², where O is the center of the circle.

Alternatively, since points A, D, H, G are on the circle, perhaps we can use some cyclic quadrilateral properties.

In cyclic quadrilateral ADHG, the product of the slopes of its diagonals is -1 if it's a rectangle, but not necessarily here. Alternatively, opposite angles sum to 180 degrees.

Alternatively, using coordinates, since we already solved it via coordinates and got 104, which is consistent, perhaps that's the answer.

Alternatively, let me see if 104 is among standard answers or if I miscalculated something.

Wait, in the problem statement, BC=16, AB=107, so the rectangle ABCD has length 107 and width 16. The other rectangle has sides 184 and17. The key was that the four points lie on a circle, which constrained the position of E to be at 3 units from D, leading to CE=104.

Alternatively, maybe the problem wants the length of CE as in the other direction? Like from C to E via the circle? But that would be longer, but the problem just asks for CE, which is the straight line.

Alternatively, maybe I made a mistake in the sign when solving the quadratic equation?

Wait, in the case of the lower rectangle, equation was e² +184e -561 =0. Solution e = [-184 ± sqrt(184² +4*561)]/2. Which is [-184 ± sqrt(36100)]/2 = [-184 ±190]/2. So the positive solution is (6)/2=3. Correct.

Therefore, I think the answer is 104. Maybe the problem is designed to get this answer. Alternatively, perhaps I made a mistake in the problem's given numbers?

Wait, the problem says FG=17 and EF=184. In rectangle EFGH, FG is adjacent to EF, so yes, those are the sides. So that's correct.

Hmm, I think I've checked it thoroughly. The answer should be 104. Wait, but let me see if 104 is the correct answer in the context of the problem's numbers.

Alternatively, maybe I messed up the direction of the rectangle. Suppose the rectangle EFGH is placed such that E is to the left of D? But then E would be at negative coordinate, but then H and G would be to the left, but the circle through A,D,H,G would still need to pass through those points. But in that case, the quadratic would have given e=-187, which is invalid as per the collinear order D,E,C,F. So no.

Alternatively, maybe the rectangle is placed such that F is between C and E? But the problem states D,E,C,F are collinear in that order, so F is after C.

Therefore, I think the answer is 104. Wait, but let me check with another approach.

Suppose we let CE = x, so since DC =107, then DE = DC - CE =107 -x. Wait, no, actually, if E is between D and C, then DE + EC = DC. So DE + CE =107. So CE =107 - DE. If we let DE = y, then CE =107 - y.

We need to find CE, so need to find y.

Now, in the rectangle EFGH, EF =184, so since E to F is along the line DC, which is the x-axis, then F is located at distance EF from E, so DF = DE + EF = y +184. But since F is beyond C, DF = DE + EC + CF? Wait, no, since D-E-C-F are colinear in order, so DF is DE + EC + CF? Wait no, D to E is y, E to C is CE =107 - y, and C to F is CF. Since EF =184, which is the distance from E to F, so EF = EC + CF ⇒ 184 = (107 - y) + CF ⇒ CF =184 - (107 - y)=77 + y.

Therefore, DF = DE + EF = y +184.

Alternatively, DF can also be expressed as DC + CF =107 + CF =107 +77 + y=184 + y. So that's consistent.

Now, the coordinates would be:

Let me set D at (0,0), so E is at (y,0), C is at (107,0), F is at (y +184,0).

The rectangle EFGH has sides EF=184 (along x-axis) and FG=17 (vertical). Assuming it's below the x-axis, then G is at (y+184, -17), H is at (y, -17).

Points A(0,16), D(0,0), H(y, -17), G(y+184, -17) must lie on a circle.

Using the same circle equation approach as before, we can derive the equation and find y.

Following the previous steps, we would end up with the same quadratic equation leading to y=3, so CE=107 -3=104. So this alternative approach also leads to the same result.

Therefore, I'm confident the answer is 104. Wait, but let me check once more with the circle's properties.

The four points A(0,16), D(0,0), H(3,-17), G(187,-17). The circle through these points has its center at (95,8) as found earlier. The distance from center (95,8) to each point:

To A(0,16): sqrt((95)^2 + ( -8)^2)=sqrt(9025 +64)=sqrt(9089)

To D(0,0): sqrt(95² +8²)= same.

To H(3,-17): sqrt((95-3)^2 + (8 +17)^2)=sqrt(92² +25²)=sqrt(8464+625)=sqrt(9089)

To G(187,-17): sqrt((187-95)^2 + (-17-8)^2)=sqrt(92² + (-25)^2)= same.

So all radii equal, correct.

Therefore, everything checks out. The length of CE is 104.

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

Given rectangles \(ABCD\) and \(EFGH\) with the points \(D, E, C, F\) collinear and \(A, D, H, G\) concyclic, we need to find the length of \(CE\).

1. **Coordinates Assignment**:
   - Place \(D\) at \((0, 0)\), \(C\) at \((107, 0)\), \(B\) at \((107, 16)\), and \(A\) at \((0, 16)\).
   - Points \(E\) and \(F\) lie on the x-axis between \(D\) and \(C\) (for \(E\)) and beyond \(C\) (for \(F\)).

2. **Rectangle EFGH Dimensions**:
   - \(EF = 184\) and \(FG = 17\). Assuming \(EFGH\) is below the x-axis, coordinates are:
     - \(E = (e, 0)\)
     - \(F = (e + 184, 0)\)
     - \(G = (e + 184, -17)\)
     - \(H = (e, -17)\)

3. **Circle through \(A, D, H, G\)**:
   - The circle equation is derived as \(x^2 + y^2 + 2gx - 16y = 0\).
   - Substituting points \(A(0, 16)\), \(D(0, 0)\), \(H(e, -17)\), and \(G(e + 184, -17)\) into the circle equation:
     - Solving the equations leads to \(e = 3\) (valid solution between 0 and 107).

4. **Calculating \(CE\)**:
   - With \(E\) at \((3, 0)\) and \(C\) at \((107, 0)\), the distance \(CE\) is:
     \[
     CE = 107 - 3 = 104
     \]

Thus, the length of \(CE\) is \(\boxed{104}\).