Solution:
We write the base-$b$ two-digit integer as $\left( xy \right)_b$.
Thus, this number satisfies
\[ \left( x + y \right)^2 = b x + y \]
with $x \in \left\{ 1, 2, \cdots , b-1 \right\}$ and $y \in \left\{ 0, 1, \cdots , b - 1 \right\}$.
The above conditions imply $\left( x + y \right)^2 < b^2$. Thus, $x + y \leq b - 1$.
The above equation can be reorganized as
\[ \left( x + y \right) \left( x + y - 1 \right) = \left( b - 1 \right) x . \]
Denote $z = x + y$ and $b' = b - 1$.
Thus, we have
\[ z \left( z - 1 \right) = b' x , \hspace{1cm} (1) \]
where $z \in \left\{ 2, 3, \cdots , b' \right\}$ and $x \in \left\{ 1, 2, \cdots , b' \right\}$.
Next, for each $b'$, we solve Equation (1).
We write $b'$ in the prime factorization form as $b' = \Pi_{i=1}^n p_i^{k_i}$.
Let $\left(A, \bar A \right)$ be any ordered partition of $\left\{ 1, 2, \cdots , n \right\}$ (we allow one set to be empty).
Denote $P_A = \Pi_{i \in A} p_i^{k_i}$ and $P_{\bar A} = \Pi_{i \in \bar A} p_i^{k_i}$.
Because ${\rm gcd} \left( z, z-1 \right) = 1$, there must exist such an ordered partition, such that $P_A | z$ and $P_{\bar A} | z-1$.
Next, we prove that for each ordered partition $\left( A, \bar A \right)$, if a solution of $z$ exists, then it must be unique.
Suppose there are two solutions of $z$ under partition $\left( A, \bar A \right)$: $z_1 = c_1 P_A$, $z_1 - 1 = d_1 P_{\bar A}$, and $z_2 = c_2 P_A$, $z_2 - 1 = d_2 P_{\bar A}$.
W.L.O.G., assume $c_1 < c_2$.
Hence, we have
\[ \left( c_2 - c_1 \right) P_A = \left( d_2 - d_1 \right) P_{\bar A} . \]
Because ${\rm gcd} \left( P_A, P_{\bar A} \right) = 1$ and $c_1 < c_2$, there exists a positive integer $m$, such that $c_2 = c_1 + m P_{\bar A}$ and $d_2 = d_1 + m P_A$.
Thus,
\begin{align*}
z_2 & = z_1 + m P_A P_{\bar A} \\
& = z_1 + m b' \\
& > b' .
\end{align*}
However, recall $z_2 \leq b'$. We get a contradiction.
Therefore, under each ordered partition for $b'$, the solution of $z$ is unique.
Note that if $b'$ has $n$ distinct prime factors, the number of ordered partitions is $2^n$.
Therefore, to find a $b'$ such that the number of solutions of $z$ is more than 10, the smallest $n$ is 4.
With $n = 4$, the smallest number is $2 \cdot 3 \cdot 5 \cdot 7 = 210$.
Now, we set $b' = 210$ and check whether the number of solutions of $z$ under this $b'$ is more than 10.
We can easily see that all ordered partitions (except $A = \emptyset$) guarantee feasible solutions of $z$.
Therefore, we have found a valid $b'$.
Therefore, $b = b' + 1 = \boxed{\textbf{(211) }}$.
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Answer: 211

<｜begin▁of▁sentence｜><｜User｜>Let \(b\ge 2\) be an integer. Call a positive integer \(n\) \(b\text-\textit{eautiful}\) if it has exactly two digits when expressed in base \(b\)  and these two digits sum to \(\sqrt n\). For example, \(81\) is \(13\text-\textit{eautiful}\) because \(81  = \underline{6} \ \underline{3}_{13} \) and \(6 + 3 =  \sqrt{81}\). Find the least integer \(b\ge 2\) for which there are more than ten \(b\text-\textit{eautiful}\) integers.<｜Assistant｜><think>

Okay, so I need to find the smallest base b (where b is at least 2) such that there are more than ten b-beautiful numbers. Let me start by understanding the problem again to make sure I get all the details right.

A number n is called b-beautiful if two conditions are met:
1. When n is written in base b, it has exactly two digits.
2. The sum of these two digits equals the square root of n.

The example given is 81 being 13-beautiful because in base 13, 81 is written as 63 (since 6*13 + 3 = 81), and 6 + 3 = 9, which is the square root of 81. That makes sense.

So, my task is to find the minimal b where there are more than ten such numbers. Let me think about how to approach this.

First, let's formalize the problem. Let me denote the two digits in base b as d1 and d0, so the number n can be expressed as:

n = d1 * b + d0

Since it's a two-digit number in base b, d1 must be between 1 and b-1 (inclusive), and d0 must be between 0 and b-1 (inclusive). The second condition is that the sum of the digits equals the square root of n:

d1 + d0 = sqrt(n)

But since n = d1*b + d0, substituting that in:

d1 + d0 = sqrt(d1*b + d0)

Let me square both sides to eliminate the square root:

(d1 + d0)^2 = d1*b + d0

Expanding the left side:

d1^2 + 2*d1*d0 + d0^2 = d1*b + d0

Let me rearrange terms to bring everything to one side:

d1^2 + 2*d1*d0 + d0^2 - d1*b - d0 = 0

Hmm, this equation might be a bit complicated. Maybe there's a better way to approach this. Let me think again.

Alternatively, since d1 and d0 are digits in base b, their values are constrained. Let me denote S = d1 + d0, so the equation becomes:

S = sqrt(n) => n = S^2

But n is also equal to d1*b + d0. Therefore:

d1*b + d0 = S^2

But since S = d1 + d0, substitute that in:

d1*b + d0 = (d1 + d0)^2

So, rearranged:

(d1 + d0)^2 - d1*b - d0 = 0

Hmm, same equation as before. Let me see if I can express this in terms of variables that can be more easily handled.

Let me let S = d1 + d0, so then the equation becomes:

S^2 = d1*b + d0

But since S = d1 + d0, we can write d0 = S - d1. Substitute that into the equation:

S^2 = d1*b + (S - d1)

Simplify the right-hand side:

S^2 = d1*b + S - d1 = d1*(b - 1) + S

Therefore, rearranged:

S^2 - S = d1*(b - 1)

So, S(S - 1) = d1*(b - 1)

Hmm, this seems promising. Let me note that d1 is an integer between 1 and b-1, and S is the sum of the digits, so S = d1 + d0. Since d0 is between 0 and b-1, S can range from 1 (if d1=1 and d0=0) up to (b-1) + (b-1) = 2b - 2. But actually, since d1 is at least 1 and d0 is at most b-1, the maximum S is (b-1) + (b-1) = 2b - 2, but the minimum is 1 + 0 = 1.

But in the equation S(S - 1) = d1*(b - 1), since d1 is at least 1 and at most b-1, and S must be an integer because it's the sum of digits. So S must be an integer such that S(S - 1) is divisible by (b - 1), and then d1 would be S(S - 1)/(b - 1). But d1 must also be an integer between 1 and b-1, so that fraction must result in an integer in that range.

Alternatively, perhaps we can think of this as for each base b, we can iterate over possible S and see if there's a corresponding d1 and d0 that satisfy the equation, then count the numbers n = S^2 that are two-digit numbers in base b.

Alternatively, maybe it's better to approach this by considering possible S and b, and see what d1 and d0 would be.

Alternatively, let me think of the equation again:

S^2 = d1*b + d0, and S = d1 + d0.

So substituting d0 = S - d1 into the first equation gives:

S^2 = d1*b + S - d1

Which simplifies to S^2 - S = d1*(b - 1)

Therefore, d1 = (S^2 - S)/(b - 1)

Since d1 must be an integer between 1 and b-1, inclusive, then (S^2 - S) must be divisible by (b - 1), and the result must be between 1 and b-1.

Therefore, for a given base b, the possible S must satisfy that (S^2 - S) is divisible by (b - 1), and then d1 is (S^2 - S)/(b - 1). Then, once d1 is known, d0 is S - d1, and we must also have that d0 is between 0 and b-1, and that n = S^2 must be a two-digit number in base b, which requires that n is between b (the smallest two-digit number, which is 10_b = b in decimal) and b^2 - 1 (the largest two-digit number, which is (b-1)(b) + (b-1) = b^2 -1).

Therefore, S^2 must be between b and b^2 -1. Since S is the sum of two digits, which are each less than b, so S <= 2b - 2, so S^2 <= (2b - 2)^2. But since n = S^2 must be less than b^2, so (2b - 2)^2 < b^2? Wait, that might not hold for all b. Let me check for b=2: (2*2 -2)^2 = 2^2 =4, which is equal to b^2=4. Hmm, so maybe the upper limit is okay.

Wait, actually, n must be a two-digit number in base b, so n must be less than b^2. Since S^2 = n, so S^2 < b^2, so S < b. Wait, that's an important point!

Wait, if n is a two-digit number in base b, then n must satisfy b ≤ n ≤ b^2 -1. Therefore, S^2 must be between b and b^2 -1. Therefore, S must satisfy sqrt(b) ≤ S ≤ sqrt(b^2 -1). Since S is an integer, so S must be at least ceiling(sqrt(b)) and at most floor(sqrt(b^2 -1)). But sqrt(b^2 -1) is slightly less than b, so floor(sqrt(b^2 -1)) is b -1. Therefore, S must be between ceiling(sqrt(b)) and b -1, inclusive.

Wait, that's a key point. So S must be an integer between ceiling(sqrt(b)) and b -1. Because S^2 must be at least b (so S ≥ sqrt(b)), and S must be less than b (since S is the sum of two digits, each less than b, so maximum S is 2b - 2, but actually S can be up to 2b - 2, but in order for S^2 < b^2, S must be less than b. Wait, hold on, that might not hold. Let me think again.

Wait, if S is the sum of two digits in base b, then each digit is less than b, so the maximum S is (b-1) + (b-1) = 2b - 2. However, n = S^2 must be a two-digit number in base b, so n must be less than b^2. Therefore, S^2 < b^2 ⇒ S < b. Therefore, even though the digit sum can be up to 2b - 2, in order for n to be a two-digit number, S must be less than b. Therefore, S must be between 1 and b -1. But also, n must be at least b, so S^2 ≥ b ⇒ S ≥ ceil(sqrt(b)).

Therefore, the possible S values for a given base b are integers S where ceil(sqrt(b)) ≤ S ≤ b -1.

Therefore, for each base b, the possible S values are from ceil(sqrt(b)) up to b-1. For each such S, we can check whether (S^2 - S) is divisible by (b -1), and then d1 = (S^2 - S)/(b -1) must be an integer between 1 and b-1, and then d0 = S - d1 must be between 0 and b-1. If all these conditions are met, then n = S^2 is a b-beautiful number.

Therefore, the strategy is:

1. For each base b starting from 2 upwards, compute the number of b-beautiful numbers.
2. Find the smallest b where this count exceeds 10.

To do this efficiently, perhaps we can find a way to compute the number of solutions S for each b, given the constraints above.

Let me formalize the conditions again:

Given b, for each S in [ceil(sqrt(b)), b-1], check:

- (S^2 - S) must be divisible by (b -1). Let me denote D = b -1, so the condition becomes (S^2 - S) mod D == 0.

Then, d1 = (S^2 - S)/D must satisfy 1 ≤ d1 ≤ b -1 (since d1 is a digit in base b, so up to b-1). Also, d0 = S - d1 must satisfy 0 ≤ d0 ≤ b -1.

But since S ≤ b -1 and d1 is part of S (since d1 = S - d0), perhaps the constraints on d0 are automatically satisfied? Let me see.

Wait, d0 = S - d1. Since d1 = (S^2 - S)/D, so substituting D = b -1,

d0 = S - (S^2 - S)/(b -1)

Hmm, maybe it's better to check the constraints step by step.

Alternatively, perhaps once d1 is computed as (S^2 - S)/(b -1), then since d1 must be an integer between 1 and b-1, and d0 = S - d1 must be between 0 and b-1, so we can check those conditions.

Alternatively, maybe the constraints are automatically satisfied if S is in the correct range and (S^2 - S) is divisible by (b-1). Let me see.

Suppose that (S^2 - S) is divisible by (b -1), so d1 is an integer. Since S ≤ b -1, then d1 = (S(S -1))/(b -1). Since S ≤ b -1, then S -1 ≤ b -2, so the numerator is at most (b -1)(b -2). The denominator is b -1, so d1 ≤ (b -1)(b -2)/(b -1) = b -2. Wait, so d1 would be at most b -2? Hmm, but d1 can be up to b -1. Wait, maybe not. Let me see with an example.

Take the example given: b =13, S = 9 (since 6 + 3 =9). Then D =12. Then d1 = (81 -9)/12 = 72/12 =6, which is between 1 and 12. Then d0 = 9 -6 =3, which is okay.

Another example: Let me pick b=2. Then D=1. Let's see possible S: ceil(sqrt(2))=2, but S must be ≤1 (since b-1=1). Wait, that can't be. So for b=2, the possible S must be between ceil(sqrt(2))≈2 and 1, which is impossible. Therefore, there are no 2-beautiful numbers. That makes sense because in base 2, two-digit numbers are 2 (10_2) up to 3 (11_2). The possible S would be 1+0=1 (for 2) and 1+1=2 (for 3). Then sqrt(n) would have to equal S. For n=2, sqrt(2) ≈1.414≠1. For n=3, sqrt(3)≈1.732≠2. So indeed, no 2-beautiful numbers.

Similarly, for b=3, D=2. The possible S must be between ceil(sqrt(3))=2 and 2 (since b-1=2). So S=2. Then check if (4 -2)/2 = 2/2=1=d1. Then d0 =2 -1=1. So n=S^2=4. Check in base 3: 4 is 11_3, which are digits 1 and 1, sum to 2, sqrt(4)=2. So yes, 4 is 3-beautiful. So for b=3, there is 1 such number.

Continuing this way, perhaps we can see a pattern.

But doing this manually for each b until we find one with more than 10 numbers is going to take a long time. Let me see if I can find a smarter approach.

First, let me note that for each b, the number of b-beautiful numbers is equal to the number of integers S in [ceil(sqrt(b)), b-1] such that (S^2 - S) is divisible by (b -1), and the resulting d1 and d0 are valid digits (d1 between 1 and b-1, d0 between 0 and b-1). But as we saw in the example with b=13 and S=9, the d0 is S - d1, so if d1 is between 1 and b-1, then d0 would be between S - (b-1) and S -1. Since S ≤ b-1, the lower bound could be negative, but we need d0 ≥0. Therefore, we must have S - d1 ≥0 ⇒ d1 ≤ S. Since d1 = (S^2 - S)/(b -1), so (S^2 - S)/(b -1) ≤ S ⇒ S^2 - S ≤ S*(b -1) ⇒ S^2 - S - Sb + S ≤0 ⇒ S^2 - Sb ≤0 ⇒ S(S - b) ≤0. Since S ≤ b-1, then S - b ≤ -1, so S(S - b) is ≤0. Therefore, the inequality holds, so d1 ≤ S. Therefore, d0 = S - d1 ≥0. So the only constraints we need to check are:

1. (S^2 - S) divisible by (b -1)
2. d1 = (S^2 - S)/(b -1) must be an integer between 1 and b-1
3. n = S^2 must be a two-digit number in base b, which as we saw requires S^2 ≥ b and S < b, so S is between ceil(sqrt(b)) and b-1.

Therefore, the main conditions are the divisibility and the range of d1.

Alternatively, since d1 must be at least 1, so (S^2 - S)/(b -1) ≥1 ⇒ S^2 - S ≥ b -1 ⇒ S^2 - S - (b -1) ≥0.

Hmm, but perhaps it's better to proceed step by step.

Let me consider that for each b, the number of solutions S is the number of integers S in [ceil(sqrt(b)), b-1] such that (S^2 - S) mod (b-1) =0, and (S^2 - S)/(b-1) is between 1 and b-1.

Alternatively, since D = b-1, so for each b, D = b-1, so the equation becomes S^2 - S ≡0 mod D, so S(S -1) ≡0 mod D.

Therefore, for each D (since D = b-1, so b = D+1), we can rephrase the problem as finding S such that S(S -1) ≡0 mod D, and S is between ceil(sqrt(D+1)) and D, and then d1 = S(S-1)/D must be between 1 and D, and then n = S^2 must be a two-digit number in base b = D+1, which requires S^2 < (D+1)^2, which is already satisfied since S ≤ D.

Wait, perhaps rephrasing in terms of D might help. Let me try that.

Let D = b -1, so b = D +1. Then the problem becomes: For each D ≥1 (since b ≥2), find the number of integers S in [ceil(sqrt(D+1)), D] such that S(S -1) ≡0 mod D, and (S(S -1))/D is between 1 and D. Then, the count for each D is the number of such S, and we need to find the smallest D+1 (i.e., b) where this count exceeds 10.

Hmm, perhaps this is a better way. Let me see.

So, for each D from 1 upwards, compute the number of S in [ceil(sqrt(D+1)), D] such that D divides S(S-1), and (S(S-1))/D is between 1 and D. Then, when this count exceeds 10, the corresponding b is D+1, which is the answer.

Therefore, perhaps we can think in terms of divisors. Since D divides S(S-1), which are two consecutive integers, so S and S-1 are coprime. Therefore, D must divide the product of two coprime numbers, so D must be a product of factors dividing S and factors dividing S-1. Since S and S-1 are coprime, then D must be a product of a divisor of S and a divisor of S-1. Therefore, the divisors of D must be split between S and S-1.

Therefore, for each D, the solutions S are numbers such that S ≡0 or 1 mod the factors of D? Hmm, maybe not straightforward.

Alternatively, since S and S-1 are coprime, then D must divide S*(S -1), so D can be written as D1*D2 where D1 divides S and D2 divides S -1, and D1 and D2 are coprime. Since D1 divides S and D2 divides S -1, then S ≡0 mod D1 and S ≡1 mod D2. Since D1 and D2 are coprime, by the Chinese Remainder Theorem, there is a unique solution modulo D1*D2 = D. Therefore, for each pair (D1, D2) such that D1*D2 = D and gcd(D1,D2)=1, there is a unique solution S modulo D. Therefore, the number of solutions S modulo D is equal to the number of such pairs (D1,D2). However, since S must be in the range [ceil(sqrt(D+1)), D], the actual count might be different.

Wait, perhaps this is getting too abstract. Let me think of specific examples.

Take D=12 (so b=13, the example given). Then D=12. We need S in [ceil(sqrt(13))=4, 12]. So S from 4 to 12. We need S(S-1) divisible by 12.

Let me check for S=9: 9*8=72, which is divisible by 12, so yes. That's the example. Let's see how many such S there are for D=12.

Check each S from 4 to 12:

S=4: 4*3=12 divisible by 12? 12/12=1, yes. So d1=12/12=1. Then d0=S -d1=4-1=3. Then n=S^2=16. Check in base 13: 16 is 13 +3, so 1 3, digits 1 and 3, sum 4, sqrt(16)=4. So that works. So S=4 is valid.

S=5: 5*4=20. 20 mod12=8, not divisible.

S=6:6*5=30 mod12=6, not divisible.

S=7:7*6=42 mod12= 42-36=6, no.

S=8:8*7=56 mod12= 56-48=8, no.

S=9:9*8=72 mod12=0, yes. So that's the example.

S=10:10*9=90 mod12= 90-84=6, no.

S=11:11*10=110 mod12= 110-108=2, no.

S=12:12*11=132 mod12=0, yes. So S=12.

Check S=12: d1=(12*11)/12=11. Then d0=12 -11=1. Then n=12^2=144. In base 13, 144 is 11*13 +1= 143 +1=144, so digits 11 and 1. Sum is 12, sqrt(144)=12. That works. So S=12 is also valid.

Therefore, for D=12 (b=13), there are three solutions: S=4,9,12. Wait, but in the example, only 81 (S=9) is mentioned, but there are two others? Wait, let me check n=16 (S=4):

In base 13, 16 is 13 +3, so digits 1 and 3. 1+3=4, sqrt(16)=4. That works. So that's another one. And n=144 (S=12): digits 11 and 1, sum 12, sqrt(144)=12. So that's three numbers. So the count for b=13 is 3. So the example is one of them.

Therefore, for D=12, there are three solutions. So the count is 3. So we need to find D where the count exceeds 10, so 11 or more.

Hmm, so perhaps we need to find D such that the number of S in [ceil(sqrt(D+1)), D] with S(S-1) divisible by D is at least 11. Then the minimal D+1 is the answer.

This seems like a number theory problem now. Let me think of how to compute the number of such S for a given D.

Since S and S-1 are coprime, D must divide S*(S-1). Since they are coprime, D must be a product of two coprime numbers, one dividing S and the other dividing S-1. So for each divisor pair (d1, d2) of D with d1*d2 = D and gcd(d1,d2)=1, there is a unique solution S modulo D such that S ≡0 mod d1 and S≡1 mod d2. Therefore, the number of solutions modulo D is equal to the number of such divisor pairs. However, the actual solutions S in the range [ceil(sqrt(D+1)), D] may vary.

Wait, so the number of solutions modulo D is equal to the number of ways to write D as d1*d2 with d1 and d2 coprime. Each such pair gives a unique solution S mod D. Therefore, the total number of solutions modulo D is equal to 2^{k}, where k is the number of prime factors of D? Wait, no, actually, the number of coprime divisor pairs (d1,d2) with d1*d2=D is 2^{ω(D)}, where ω(D) is the number of distinct prime factors of D. Because each prime factor can be assigned to d1 or d2, but not both. So for each prime factor, two choices, hence 2^{ω(D)}.

Wait, but actually, since the order matters (d1 and d2), but since d1*d2=D, each coprime pair (d1,d2) with d1*d2=D corresponds to a subset of the prime factors assigned to d1, and the rest to d2. So the number is indeed 2^{ω(D)}. However, since (d1,d2) and (d2,d1) would give different solutions unless d1=d2, but since d1 and d2 must be coprime and multiply to D, they can only be equal if D=1. So for D>1, all coprime pairs (d1,d2) with d1*d2=D come in pairs unless d1=1 or d2=1. Hmm, maybe I need to think again.

Actually, the number of coprime pairs (d1,d2) with d1*d2=D is exactly 2^{ω(D)}. Because each prime factor can go to d1 or d2, so for each prime, 2 choices, so total 2^{ω(D)}. For example, if D is prime, then ω(D)=1, so 2^1=2 pairs: (1,D) and (D,1). But since d1 and d2 must be positive integers, so yes.

Therefore, the number of solutions S mod D is 2^{ω(D)}. However, each solution corresponds to a unique S modulo D. Therefore, in the interval [1, D], there are exactly 2^{ω(D)} solutions. But we need S in [ceil(sqrt(D+1)), D]. So the number of solutions in that interval would be roughly 2^{ω(D)} minus the number of solutions less than ceil(sqrt(D+1)). But this might complicate.

Alternatively, perhaps for larger D, the number of solutions S in [ceil(sqrt(D+1)), D] is approximately 2^{ω(D)} minus the number of solutions below the lower bound. However, this might not be straightforward.

Alternatively, maybe for certain D with many prime factors, the number of solutions S can be large. Since each prime factor contributes to more divisors, hence more solutions. So perhaps D needs to be a number with many prime factors, so that 2^{ω(D)} is large. To get more than 10 solutions, we need 2^{ω(D)} >10, so ω(D) ≥4, since 2^4=16. So if D has 4 distinct prime factors, then there are 16 solutions mod D, so potentially up to 16 solutions in [1,D], but we need those in [ceil(sqrt(D+1)), D].

Wait, but even if there are 16 solutions mod D, the number in the desired interval might be less. However, perhaps when D is a product of the first few primes, the number of solutions S in the upper half (near D) could be sufficient.

Alternatively, maybe the number of solutions S in [ceil(sqrt(D+1)), D] is equal to the number of coprime divisor pairs (d1,d2) of D where the corresponding S is in that interval. Since each coprime pair gives a unique S mod D, which is in [1,D], so exactly one in each interval of length D. So for each coprime pair (d1,d2), the solution S is unique mod D, so exactly one solution in [1,D].

Therefore, the total number of solutions in [1,D] is exactly 2^{ω(D)}. Therefore, the number of solutions in [ceil(sqrt(D+1)), D] is equal to 2^{ω(D)} minus the number of solutions less than ceil(sqrt(D+1)).

To maximize the number of solutions in the upper interval, we need as many of the 2^{ω(D)} solutions as possible to lie in the upper part. Since the solutions are spread out modulo D, perhaps roughly half are in the upper half, but this is not precise.

Alternatively, maybe for certain D, many of the solutions S are close to D. For example, when D is a factorial or a primorial, which have many small factors, perhaps the solutions S can be arranged such that many are in the upper range.

Alternatively, perhaps the number of solutions S in the required interval is equal to the number of divisors d of D such that d <= sqrt(D) or something like that? Not sure.

Alternatively, perhaps instead of getting bogged down in theory, I can try to compute the number of solutions for increasing D until I find one with more than 10.

Given that the problem asks for the minimal b (so minimal D = b-1), starting from D=1 (b=2) upwards, compute the count for each D until the count exceeds 10.

This might be time-consuming, but perhaps manageable with some observations.

First, let me note that for D with more prime factors, the number of solutions (2^{ω(D)}) increases, so we can focus on numbers D with high ω(D). The minimal D with ω(D)=4 is 2*3*5*7=210, but that might be too big. Wait, but maybe smaller numbers with multiple prime factors?

Wait, let me see:

Let me start with small D and compute counts:

Starting with D=1 (b=2):

- D=1, so b=2. The interval for S is [ceil(sqrt(2))=2, 1], which is empty. So count=0.

D=2 (b=3):

- S must be in [2,2]. So S=2.

Check if 2*1 divisible by 2? 2 divides 2, yes. Then d1=(4-2)/2=1. So d1=1, d0=1. n=4. Which is valid. So count=1.

D=3 (b=4):

S in [2,3].

Check S=2: 2*1=2 divisible by 3? No.

S=3: 3*2=6 divisible by 3? Yes. Then d1=(9 -3)/3=6/3=2. d0=3-2=1. n=9. In base 4, 9 is 21_4, digits 2+1=3, sqrt(9)=3. Valid. So count=1.

D=4 (b=5):

S in [2,4].

Check S=2: 2*1=2 mod4? 2 mod4≠0. No.

S=3: 3*2=6 mod4=2≠0.

S=4:4*3=12 mod4=0. Yes. Then d1=(16-4)/4=12/4=3. d0=4-3=1. n=16. In base5, 16 is 31_5 (3*5 +1=16). 3+1=4, sqrt(16)=4. Valid. So count=1.

D=5 (b=6):

S in [3,5].

S=3: 3*2=6 mod5=1≠0.

S=4:4*3=12 mod5=2≠0.

S=5:5*4=20 mod5=0. Yes. d1=(25 -5)/5=20/5=4. d0=5-4=1. n=25. In base6: 41_6 (4*6 +1=25). 4+1=5, sqrt(25)=5. Valid. Count=1.

D=6 (b=7):

S in [3,6].

Check divisibility of S(S-1) by 6.

S=3: 3*2=6 divisible by6? Yes. So d1=(9-3)/6=6/6=1. d0=3-1=2. n=9. In base7: 12_7 (1*7 +2=9). 1+2=3, sqrt(9)=3. Valid.

S=4:4*3=12 mod6=0. Yes. d1=(16-4)/6=12/6=2. d0=4-2=2. n=16. In base7: 22_7 (2*7 +2=16). 2+2=4, sqrt(16)=4. Valid.

S=5:5*4=20 mod6=2≠0.

S=6:6*5=30 mod6=0. Yes. d1=(36-6)/6=30/6=5. d0=6-5=1. n=36. In base7: 51_7 (5*7 +1=36). 5+1=6, sqrt(36)=6. Valid.

Thus, S=3,4,6. So count=3.

So for D=6 (b=7), count=3.

Continuing:

D=7 (b=8):

S in [3,7].

Check S(S-1) divisible by7.

S=3: 3*2=6 mod7≠0.

S=4:4*3=12 mod7=5≠0.

S=5:5*4=20 mod7=6≠0.

S=6:6*5=30 mod7=2≠0.

S=7:7*6=42 mod7=0. Yes. Then d1=(49 -7)/7=42/7=6. d0=7-6=1. n=49. In base8: 61_8 (6*8 +1=49). 6+1=7, sqrt(49)=7. Valid. So count=1.

D=8 (b=9):

S in [3,8].

Check S(S-1) divisible by8.

S=3: 3*2=6 mod8≠0.

S=4:4*3=12 mod8=4≠0.

S=5:5*4=20 mod8=4≠0.

S=6:6*5=30 mod8=6≠0.

S=7:7*6=42 mod8=2≠0.

S=8:8*7=56 mod8=0. Yes. d1=(64 -8)/8=56/8=7. d0=8-7=1. n=64. In base9: 71_9 (7*9 +1=64). 7+1=8, sqrt(64)=8. Valid. So count=1.

D=9 (b=10):

S in [3,9].

Check S(S-1) divisible by9.

S=3:3*2=6 mod9≠0.

S=4:4*3=12 mod9=3≠0.

S=5:5*4=20 mod9=2≠0.

S=6:6*5=30 mod9=3≠0.

S=7:7*6=42 mod9=6≠0.

S=8:8*7=56 mod9=2≠0.

S=9:9*8=72 mod9=0. Yes. d1=(81 -9)/9=72/9=8. d0=9-8=1. n=81. In base10: 81, which is 8*10 +1. 8+1=9, sqrt(81)=9. Valid. Count=1.

Hmm, so D=9 gives count=1.

D=10 (b=11):

S in [4,10].

Check S(S-1) divisible by10.

S=4:4*3=12 mod10=2≠0.

S=5:5*4=20 mod10=0. Yes. d1=(25-5)/10=20/10=2. d0=5-2=3. n=25. In base11: 23_11 (2*11 +3=25). 2+3=5, sqrt(25)=5. Valid.

S=6:6*5=30 mod10=0. Yes. d1=(36-6)/10=30/10=3. d0=6-3=3. n=36. In base11: 33_11 (3*11 +3=36). 3+3=6, sqrt(36)=6. Valid.

S=7:7*6=42 mod10=2≠0.

S=8:8*7=56 mod10=6≠0.

S=9:9*8=72 mod10=2≠0.

S=10:10*9=90 mod10=0. Yes. d1=(100 -10)/10=90/10=9. d0=10-9=1. n=100. In base11: 91_11 (9*11 +1=100). 9+1=10, sqrt(100)=10. Valid.

Thus, S=5,6,10. Count=3.

Continuing to D=12 (b=13):

Earlier, we saw count=3 (S=4,9,12). Wait, let me confirm again:

Wait D=12 (b=13):

S ranges from ceil(sqrt(13))=4 to 12.

Check S(S-1) divisible by12.

S=4:4*3=12 divisible by12. Yes. d1=(16-4)/12=12/12=1. d0=4-1=3. n=16. In base13: 13 +3=16, digits 1 and3. 1+3=4. Valid.

S=5:5*4=20 mod12=8≠0.

S=6:6*5=30 mod12=6≠0.

S=7:7*6=42 mod12=6≠0.

S=8:8*7=56 mod12=8≠0.

S=9:9*8=72 mod12=0. Yes. d1=(81-9)/12=72/12=6. d0=9-6=3. n=81. Valid.

S=10:10*9=90 mod12=6≠0.

S=11:11*10=110 mod12= 110-96=14 mod12=2≠0.

S=12:12*11=132 mod12=0. Yes. d1=(144-12)/12=132/12=11. d0=12-11=1. n=144. Valid.

Thus, S=4,9,12. Count=3.

Hmm, so D=12 gives count=3.

Continuing to D=16 (b=17):

Wait maybe I need to jump to higher D with more factors.

Alternatively, let's try D=24 (b=25). Let me see:

Wait, perhaps first D=16 (b=17):

D=16 (b=17):

S ranges from ceil(sqrt(17))=5 to 16.

Check S(S-1) divisible by16.

Looking for S where S(S-1) divisible by16.

Since 16 is 2^4. Since S and S-1 are consecutive, one of them must be even, but to get divisibility by 16, either:

- S is divisible by 16, then S-1 is odd, so S must be 16.

- S-1 divisible by 16, so S=1 mod16, but S ≤16, so S=1, but S starts at 5.

Alternatively, since 16=16, so S or S-1 must be divisible by 16, or one divisible by 8 and the other by 2, etc.

Wait, let's think in terms of factors:

To have S(S-1) divisible by 16, since S and S-1 are coprime, one must be divisible by 16, or one divisible by 8 and the other by 2, but since they are consecutive, they can't both be even. So the even one must be divisible by 16, or the even one divisible by 8 and the other divisible by 2, but since they are consecutive, the other is odd. Wait, perhaps:

Let me check possible S:

S must be even or odd. Let me see:

If S is even, then S must be divisible by 16, or S divisible by 8 and S-1 divisible by 2 (but S-1 would be odd, so that doesn't help). Alternatively, S divisible by 16, then S-1 is odd. So S=16.

Alternatively, S-1 divisible by 16, so S=1 mod16, but S ≥5, so S=17, which is beyond 16.

Alternatively, S even, so S divisible by 2, and S-1 divisible by 8? Since 2*8=16. Since S and S-1 are coprime, so if S is even (divisible by 2) and S-1 divisible by 8, then their product is divisible by 16.

So S ≡1 mod8, and S even? Wait, S-1 divisible by8 ⇒ S ≡1 mod8. But then S is 1 mod8, so S could be 1,9,17,... but S must be even? Wait, no, S can be odd or even. Wait, if S is 1 mod8, then S can be odd or even depending on the modulus. For example, 9 is 1 mod8 and odd, 25 is 1 mod8 and odd, etc. Wait, actually 1 mod8 is odd. So S must be odd in this case. But then S is odd, so S-1 is even. Wait, this is getting confusing.

Alternatively, let me check S from 5 to16:

S=5:5*4=20 mod16=4≠0.

S=6:6*5=30 mod16=14≠0.

S=7:7*6=42 mod16=10≠0.

S=8:8*7=56 mod16=8≠0.

S=9:9*8=72 mod16=8≠0.

S=10:10*9=90 mod16=10≠0.

S=11:11*10=110 mod16= 110-96=14≠0.

S=12:12*11=132 mod16= 132-128=4≠0.

S=13:13*12=156 mod16= 156-144=12≠0.

S=14:14*13=182 mod16= 182-176=6≠0.

S=15:15*14=210 mod16= 210-208=2≠0.

S=16:16*15=240 mod16=0. Yes. So only S=16.

Thus, count=1.

Hmm, not helpful.

Let me try D=12 (already did that). Maybe D=24 (b=25):

D=24 (b=25):

S ranges from ceil(sqrt(25))=5 to24.

We need S(S-1) divisible by24.

24 factors into 2^3*3. Since S and S-1 are coprime, one must be divisible by 8 and the other by 3, or one divisible by 3 and the other by 8, or one divisible by 24.

Alternatively, since 24=8*3, with gcd(8,3)=1, so the coprime pairs (d1,d2) are (8,3), (3,8), (24,1), (1,24). So there are 2^2=4 divisor pairs (since ω(24)=2 (primes 2 and3)), so 2^2=4 solutions mod24.

Wait, ω(24)=2 (prime factors 2 and3), so number of coprime divisor pairs is 2^2=4. So there are 4 solutions mod24. Let me find them:

The coprime pairs (d1,d2) where d1*d2=24:

(1,24), (3,8), (8,3), (24,1).

For each pair:

1. (d1=1, d2=24):

S ≡0 mod1 (always true) and S≡1 mod24 ⇒ S≡1 mod24. Since S ≤24, S=1. But S must be ≥5, so not in range.

2. (d1=3, d2=8):

S ≡0 mod3 and S≡1 mod8.

Find S ≡0 mod3 and S≡1 mod8.

Let S=8k+1. Then 8k+1 ≡0 mod3 ⇒8k ≡-1 mod3 ⇒2k≡2 mod3 ⇒k≡1 mod3. So k=3m+1. Thus S=8*(3m+1)+1=24m+9. Within 1-24, S=9.

Check S=9: 9 mod3=0, 9 mod8=1. Yes. So S=9.

3. (d1=8, d2=3):

S≡0 mod8 and S≡1 mod3.

Let S=8k. Then 8k≡1 mod3 ⇒2k≡1 mod3 ⇒k≡2 mod3. So k=3m+2. Thus S=8*(3m+2)=24m+16. Within 1-24, S=16.

Check S=16: 16 mod8=0, 16 mod3=1. Yes. So S=16.

4. (d1=24, d2=1):

S≡0 mod24 and S≡1 mod1 (always true). So S=24.

Thus, the solutions mod24 are S=9,16,24, and 1 (but 1 is too small). So in the range [5,24], the valid S are 9,16,24.

Wait, but 24 is included. Let me check each S:

S=9: 9*8=72 divisible by24? 72/24=3, yes. Then d1=(81 -9)/24=72/24=3. d0=9-3=6. n=81. In base25: 3*25 +6=81. Digits 3 and6, sum 9. sqrt(81)=9. Valid.

S=16:16*15=240 divisible by24? 240/24=10, yes. d1=(256 -16)/24=240/24=10. d0=16-10=6. n=256. In base25: 10*25 +6=256. 10+6=16, sqrt(256)=16. Valid.

S=24:24*23=552 divisible by24? 552/24=23, yes. d1=(576 -24)/24=552/24=23. d0=24-23=1. n=576. In base25: 23*25 +1=575+1=576. 23+1=24, sqrt(576)=24. Valid.

Thus, three solutions. So count=3.

Hmm, still low.

Wait, maybe D=60 (b=61). Let me see:

Wait, perhaps I need to find a D with more prime factors. Let's try D=2*3*5=30 (ω=3). So D=30 (b=31).

Then the number of solutions mod30 is 2^3=8. Let's see how many fall into [ceil(sqrt(31))=6, 30].

First, factor pairs (d1,d2) where d1*d2=30 and gcd(d1,d2)=1:

The coprime pairs are:

(1,30), (2,15), (3,10), (5,6), (6,5), (10,3), (15,2), (30,1).

Each gives a solution S mod30.

Let me compute each:

1. (1,30): S≡0 mod1 and S≡1 mod30 ⇒ S=1 mod30 ⇒ S=1 (too small).

2. (2,15): S≡0 mod2 and S≡1 mod15.

S even and S≡1 mod15. Let S=15k+1. Must be even: 15k+1 even ⇒15k odd ⇒k odd. Let k=1: S=16. 16 mod2=0, mod15=1. Valid. So S=16.

3. (3,10): S≡0 mod3 and S≡1 mod10.

S divisible by3, S≡1 mod10. Let S=10k+1. 10k+1 ≡1 mod3 ⇒10k ≡0 mod3 ⇒k≡0 mod3. So k=3m. Thus S=10*(3m)+1=30m+1. Within 1-30: S=1 (too small) or 31 (too big). Hmm, no solution here?

Wait, maybe another approach. Let me solve S≡0 mod3 and S≡1 mod10.

Let S=10k +1. Then 10k+1 ≡0 mod3 ⇒10k ≡-1 mod3 ⇒1k≡2 mod3 ⇒k≡2 mod3. So k=3m+2. Thus S=10*(3m+2)+1=30m+21. Within 1-30: S=21.

Check S=21: 21 mod3=0, mod10=1. Yes. So S=21.

4. (5,6): S≡0 mod5 and S≡1 mod6.

S divisible by5, S≡1 mod6. Let S=6k+1. 6k+1 ≡0 mod5 ⇒6k ≡-1 mod5 ⇒k≡2 mod5 (since 6≡1 mod5 ⇒k≡-1 mod5 ⇒k=5m+4). Wait:

Wait 6k ≡ -1 mod5 ⇒6k ≡4 mod5 ⇒k ≡4/6 mod5 ⇒since 6^{-1} mod5 is 1 (since 6≡1 mod5), so k≡4 mod5. Thus k=5m+4. Then S=6*(5m+4)+1=30m+25. Within 1-30: S=25.

Check S=25: 25 mod5=0, mod6=25 mod6=1. Yes. So S=25.

5. (6,5): S≡0 mod6 and S≡1 mod5.

S divisible by6, S≡1 mod5. Let S=6k. Then 6k≡1 mod5 ⇒6k≡1 mod5 ⇒k≡1 mod5 (since 6≡1 mod5 ⇒k≡1 mod5). So k=5m+1. Thus S=6*(5m+1)=30m+6. Within 1-30: S=6, 36 (too big). S=6.

Check S=6: 6 mod6=0, mod5=1. 6 mod5=1. Yes. So S=6.

6. (10,3): S≡0 mod10 and S≡1 mod3.

S divisible by10, S≡1 mod3. Let S=10k. 10k≡1 mod3 ⇒10≡1 mod3 ⇒k≡1 mod3. So k=3m+1. Thus S=10*(3m+1)=30m+10. Within 1-30: S=10,40 (too big). S=10.

Check S=10: 10 mod10=0, mod3=1. 10 mod3=1. Yes. So S=10.

7. (15,2): S≡0 mod15 and S≡1 mod2.

S divisible by15, and odd (since S≡1 mod2). So S=15k. 15k must be odd ⇒k odd. So S=15, 45 (too big). S=15.

Check S=15: 15 mod15=0, mod2=1. 15 is odd. Yes. So S=15.

8. (30,1): S≡0 mod30 and S≡1 mod1 ⇒ S=30.

Thus, the solutions in [6,30] are S=6,10,15,16,21,25,30.

Wait let's list all solutions from above:

From pair (2,15): S=16

(3,10): S=21

(5,6): S=25

(6,5): S=6

(10,3): S=10

(15,2): S=15

(30,1): S=30

And also from (5,6):25, etc.

Wait total solutions in [6,30]:

S=6,10,15,16,21,25,30. That's 7 solutions.

Wait let me check each:

S=6: Check S(S-1)=6*5=30 divisible by30. Yes. d1=(36-6)/30=30/30=1. d0=6-1=5. n=36. In base31: 1*31 +5=36. 1+5=6, sqrt(36)=6. Valid.

S=10:10*9=90 divisible by30? 90/30=3. Yes. d1=(100-10)/30=90/30=3. d0=10-3=7. n=100. In base31: 3*31 +7=100. 3+7=10. Valid.

S=15:15*14=210 divisible by30? 210/30=7. Yes. d1=(225-15)/30=210/30=7. d0=15-7=8. n=225. In base31: 7*31 +8=217+8=225. 7+8=15. Valid.

S=16:16*15=240 divisible by30? 240/30=8. Yes. d1=(256-16)/30=240/30=8. d0=16-8=8. n=256. In base31:8*31 +8=248+8=256. 8+8=16. Valid.

S=21:21*20=420 divisible by30? 420/30=14. Yes. d1=(441-21)/30=420/30=14. d0=21-14=7. n=441. In base31:14*31 +7=434+7=441. 14+7=21. Valid.

S=25:25*24=600 divisible by30? 600/30=20. Yes. d1=(625-25)/30=600/30=20. d0=25-20=5. n=625. In base31:20*31 +5=620+5=625. 20+5=25. Valid.

S=30:30*29=870 divisible by30? 870/30=29. Yes. d1=(900-30)/30=870/30=29. d0=30-29=1. n=900. In base31:29*31 +1=899+1=900. 29+1=30. Valid.

Thus, total of 7 solutions. So count=7 for D=30 (b=31). Still not enough.

Hmm, need to go higher. Let's try D=2*3*5*7=210 (ω=4). That might be too big, but let's see.

Alternatively, D=60 (which has prime factors 2^2,3,5; so ω=3). Let me try D=60 (b=61):

Number of solutions mod60 is 2^3=8 (since ω=3). Let's see how many fall into [ceil(sqrt(61))=8,60].

First, factor pairs (d1,d2) with d1*d2=60 and gcd(d1,d2)=1:

The coprime pairs:

(1,60), (3,20), (4,15), (5,12), (12,5), (15,4), (20,3), (60,1).

Each gives a solution S mod60.

Let me compute each:

1. (1,60): S≡0 mod1 and S≡1 mod60 ⇒ S=1 (too small).

2. (3,20): S≡0 mod3 and S≡1 mod20.

S divisible by3, S≡1 mod20. Let S=20k+1. Then 20k+1 ≡0 mod3 ⇒20k ≡-1 mod3 ⇒2k≡2 mod3 ⇒k≡1 mod3. So k=3m+1. Thus S=20*(3m+1)+1=60m+21. Within 1-60: S=21, 81 (too big). So S=21.

3. (4,15): S≡0 mod4 and S≡1 mod15.

S divisible by4, S≡1 mod15. Let S=15k+1. 15k+1 ≡0 mod4 ⇒15k ≡-1 mod4 ⇒15≡3 mod4 ⇒3k≡-1 mod4 ⇒3k≡3 mod4 ⇒k≡1 mod4. So k=4m+1. Thus S=15*(4m+1)+1=60m+16. Within 1-60: S=16, 76 (too big). So S=16.

4. (5,12): S≡0 mod5 and S≡1 mod12.

S divisible by5, S≡1 mod12. Let S=12k+1. 12k+1 ≡0 mod5 ⇒12k ≡-1 mod5 ⇒12≡2 mod5 ⇒2k≡-1 mod5 ⇒2k≡4 mod5 ⇒k≡2 mod5. So k=5m+2. Thus S=12*(5m+2)+1=60m+25. Within 1-60: S=25, 85 (too big). So S=25.

5. (12,5): S≡0 mod12 and S≡1 mod5.

S divisible by12, S≡1 mod5. Let S=12k. 12k≡1 mod5 ⇒12≡2 mod5 ⇒2k≡1 mod5 ⇒k≡3 mod5 (since 2*3=6≡1 mod5). So k=5m+3. Thus S=12*(5m+3)=60m+36. Within 1-60: S=36, 96 (too big). So S=36.

6. (15,4): S≡0 mod15 and S≡1 mod4.

S divisible by15, S≡1 mod4. Let S=15k. 15k≡1 mod4 ⇒15≡3 mod4 ⇒3k≡1 mod4 ⇒k≡3 mod4 (since 3*3=9≡1 mod4). So k=4m+3. Thus S=15*(4m+3)=60m+45. Within 1-60: S=45, 105 (too big). So S=45.

7. (20,3): S≡0 mod20 and S≡1 mod3.

S divisible by20, S≡1 mod3. Let S=20k. 20k≡1 mod3 ⇒20≡2 mod3 ⇒2k≡1 mod3 ⇒k≡2 mod3. So k=3m+2. Thus S=20*(3m+2)=60m+40. Within 1-60: S=40, 100 (too big). So S=40.

8. (60,1): S≡0 mod60 and S≡1 mod1 ⇒ S=60.

Thus, the solutions in [8,60] are S=16,21,25,36,40,45,60.

Wait let me list all:

From pair (3,20):21

(4,15):16

(5,12):25

(12,5):36

(15,4):45

(20,3):40

(60,1):60

So total of 7 solutions. Wait, same as before. Hmm.

Wait let me check each S:

S=16: 16*15=240 divisible by60? 240/60=4. Yes. d1=(256-16)/60=240/60=4. d0=16-4=12. Wait, but in base61, digits must be less than 61. d0=12 is okay. n=256. In base61: 4*61 +12=244+12=256. 4+12=16. Valid.

S=21:21*20=420 divisible by60? 420/60=7. Yes. d1=(441-21)/60=420/60=7. d0=21-7=14. n=441. In base61:7*61 +14=427+14=441. 7+14=21. Valid.

S=25:25*24=600 divisible by60? 600/60=10. Yes. d1=(625-25)/60=600/60=10. d0=25-10=15. n=625. In base61:10*61 +15=610+15=625. 10+15=25. Valid.

S=36:36*35=1260 divisible by60? 1260/60=21. Yes. d1=(1296-36)/60=1260/60=21. d0=36-21=15. n=1296. In base61:21*61 +15=1281+15=1296. 21+15=36. Valid.

S=40:40*39=1560 divisible by60? 1560/60=26. Yes. d1=(1600-40)/60=1560/60=26. d0=40-26=14. n=1600. In base61:26*61 +14=1586+14=1600. 26+14=40. Valid.

S=45:45*44=1980 divisible by60? 1980/60=33. Yes. d1=(2025-45)/60=1980/60=33. d0=45-33=12. n=2025. In base61:33*61 +12=2013+12=2025. 33+12=45. Valid.

S=60:60*59=3540 divisible by60? 3540/60=59. Yes. d1=(3600-60)/60=3540/60=59. d0=60-59=1. n=3600. In base61:59*61 +1=3599+1=3600. 59+1=60. Valid.

Thus, 7 solutions again. Hmm, still not enough.

Hmm, maybe I need to try D with ω=4. Let's try D=2*3*5*7=210 (b=211). The number of solutions mod210 is 2^4=16. Let's see how many fall into [ceil(sqrt(211))=15, 210].

This might take time, but let's see.

First, the coprime divisor pairs (d1,d2) of 210 (prime factors 2,3,5,7):

The coprime pairs (d1,d2) where d1*d2=210:

Each prime can go to d1 or d2. So there are 2^4=16 pairs. Let me list them:

(1,210), (2,105), (3,70), (5,42), (6,35), (7,30), (10,21), (14,15),

(15,14), (21,10), (30,7), (35,6), (42,5), (70,3), (105,2), (210,1).

Each of these gives a solution S mod210. Let me compute each:

1. (1,210): S≡0 mod1 and S≡1 mod210 ⇒ S=1 (too small).

2. (2,105): S≡0 mod2 and S≡1 mod105.

S even and S≡1 mod105. Let S=105k+1. Must be even ⇒105k must be odd ⇒k odd. Let k=1: S=106. Check mod2=0, mod105=1. Valid. So S=106.

3. (3,70): S≡0 mod3 and S≡1 mod70.

S divisible by3, S≡1 mod70. Let S=70k+1. 70k+1 ≡0 mod3 ⇒70≡1 mod3 ⇒k+1≡0 mod3 ⇒k≡2 mod3. So k=3m+2. S=70*(3m+2)+1=210m+141. Within 1-210: S=141, 351 (too big). So S=141.

4. (5,42): S≡0 mod5 and S≡1 mod42.

S divisible by5, S≡1 mod42. Let S=42k+1. 42k+1 ≡0 mod5 ⇒42≡2 mod5 ⇒2k+1≡0 mod5 ⇒2k≡-1 mod5 ⇒2k≡4 mod5 ⇒k≡2 mod5. So k=5m+2. S=42*(5m+2)+1=210m+85. Within 1-210: S=85, 295 (too big). So S=85.

5. (6,35): S≡0 mod6 and S≡1 mod35.

S divisible by6, S≡1 mod35. Let S=35k+1. 35k+1 ≡0 mod6 ⇒35≡5 mod6 ⇒5k+1≡0 mod6 ⇒5k≡-1 mod6 ⇒5k≡5 mod6 ⇒k≡1 mod6. So k=6m+1. S=35*(6m+1)+1=210m+36. Within 1-210: S=36, 246 (too big). So S=36.

6. (7,30): S≡0 mod7 and S≡1 mod30.

S divisible by7, S≡1 mod30. Let S=30k+1. 30k+1 ≡0 mod7 ⇒30≡2 mod7 ⇒2k+1≡0 mod7 ⇒2k≡-1 mod7 ⇒2k≡6 mod7 ⇒k≡3 mod7. So k=7m+3. S=30*(7m+3)+1=210m+91. Within 1-210: S=91, 301 (too big). So S=91.

7. (10,21): S≡0 mod10 and S≡1 mod21.

S divisible by10, S≡1 mod21. Let S=21k+1. 21k+1 ≡0 mod10 ⇒21≡1 mod10 ⇒k+1≡0 mod10 ⇒k≡9 mod10. So k=10m+9. S=21*(10m+9)+1=210m+190. Within 1-210: S=190, 400 (too big). So S=190.

8. (14,15): S≡0 mod14 and S≡1 mod15.

S divisible by14, S≡1 mod15. Let S=15k+1. 15k+1 ≡0 mod14 ⇒15≡1 mod14 ⇒k+1≡0 mod14 ⇒k≡13 mod14. So k=14m+13. S=15*(14m+13)+1=210m+196. Within 1-210: S=196, 406 (too big). So S=196.

9. (15,14): S≡0 mod15 and S≡1 mod14.

S divisible by15, S≡1 mod14. Let S=14k+1. 14k+1 ≡0 mod15 ⇒14≡-1 mod15 ⇒-k+1≡0 mod15 ⇒-k≡-1 mod15 ⇒k≡1 mod15. So k=15m+1. S=14*(15m+1)+1=210m+15. Within 1-210: S=15, 225 (too big). So S=15.

10. (21,10): S≡0 mod21 and S≡1 mod10.

S divisible by21, S≡1 mod10. Let S=10k+1. 10k+1 ≡0 mod21 ⇒10k ≡-1 mod21 ⇒10k≡20 mod21 ⇒k≡2 mod21 (since 10*2=20≡-1 mod21). So k=21m+2. S=10*(21m+2)+1=210m+21. Within 1-210: S=21, 231 (too big). So S=21.

11. (30,7): S≡0 mod30 and S≡1 mod7.

S divisible by30, S≡1 mod7. Let S=30k. 30k≡1 mod7 ⇒30≡2 mod7 ⇒2k≡1 mod7 ⇒k≡4 mod7 (since 2*4=8≡1 mod7). So k=7m+4. S=30*(7m+4)=210m+120. Within 1-210: S=120, 330 (too big). So S=120.

12. (35,6): S≡0 mod35 and S≡1 mod6.

S divisible by35, S≡1 mod6. Let S=35k. 35k≡1 mod6 ⇒35≡5 mod6 ⇒5k≡1 mod6 ⇒k≡5 mod6 (since 5*5=25≡1 mod6). So k=6m+5. S=35*(6m+5)=210m+175. Within 1-210: S=175, 385 (too big). So S=175.

13. (42,5): S≡0 mod42 and S≡1 mod5.

S divisible by42, S≡1 mod5. Let S=42k. 42k≡1 mod5 ⇒42≡2 mod5 ⇒2k≡1 mod5 ⇒k≡3 mod5. So k=5m+3. S=42*(5m+3)=210m+126. Within 1-210: S=126, 336 (too big). So S=126.

14. (70,3): S≡0 mod70 and S≡1 mod3.

S divisible by70, S≡1 mod3. Let S=70k. 70k≡1 mod3 ⇒70≡1 mod3 ⇒k≡1 mod3. So k=3m+1. S=70*(3m+1)=210m+70. Within 1-210: S=70, 280 (too big). So S=70.

15. (105,2): S≡0 mod105 and S≡1 mod2.

S divisible by105, S odd. Since 105 is odd, S=105k must be odd ⇒k odd. Let k=1: S=105. Check mod2=1. Yes. So S=105.

16. (210,1): S≡0 mod210 and S≡1 mod1 ⇒ S=210.

Now, compiling all solutions in [15,210]:

From pair (2,105):106

(3,70):141

(5,42):85

(6,35):36

(7,30):91

(10,21):190

(14,15):196

(15,14):15

(21,10):21

(30,7):120

(35,6):175

(42,5):126

(70,3):70

(105,2):105

(210,1):210

Now let's list all these S values within the range [15,210]:

15,21,36,70,85,91,105,106,120,126,141,175,190,196,210.

Wait let me count them:

1. 15

2.21

3.36

4.70

5.85

6.91

7.105

8.106

9.120

10.126

11.141

12.175

13.190

14.196

15.210

That's 15 solutions! So for D=210 (b=211), there are 15 solutions, which is more than 10. But is this the minimal D?

Wait, but maybe there is a smaller D with ω=4. Let me check D= 2*3*5*7=210 is the first with four distinct primes. The next smallest number with four distinct prime factors is 2*3*5*7=210, so that's the minimal D with ω=4. So perhaps 210 is the minimal D where the count exceeds 10, so the answer would be b=211.

But wait, let me check D= 105 (which has prime factors 3,5,7; ω=3). Let me see if D=105 gives enough solutions.

Wait D=105 (b=106):

Number of solutions mod105 is 2^3=8. Let's see how many are in [ceil(sqrt(106))=11, 105].

The coprime pairs (d1,d2) with d1*d2=105:

(1,105),(3,35),(5,21),(7,15),(15,7),(21,5),(35,3),(105,1).

Calculating each solution:

1. (1,105): S=1 (too small).

2. (3,35): S≡0 mod3 and S≡1 mod35.

Let S=35k+1. 35k+1 ≡0 mod3 ⇒35≡2 mod3 ⇒2k+1≡0 mod3 ⇒2k≡-1 mod3 ⇒k≡1 mod3. So k=3m+1 ⇒ S=35*(3m+1)+1=105m+36. Within 1-105: S=36, 141 (too big). So S=36.

3. (5,21): S≡0 mod5 and S≡1 mod21.

S=21k+1. 21k+1 ≡0 mod5 ⇒21≡1 mod5 ⇒k+1≡0 mod5 ⇒k≡4 mod5. So k=5m+4 ⇒ S=21*(5m+4)+1=105m+85. Within 1-105: S=85, 190 (too big). So S=85.

4. (7,15): S≡0 mod7 and S≡1 mod15.

S=15k+1. 15k+1 ≡0 mod7 ⇒15≡1 mod7 ⇒k+1≡0 mod7 ⇒k≡6 mod7. So k=7m+6 ⇒ S=15*(7m+6)+1=105m+91. Within 1-105: S=91, 196 (too big). So S=91.

5. (15,7): S≡0 mod15 and S≡1 mod7.

S=7k+1. 7k+1 ≡0 mod15 ⇒7k≡-1 mod15 ⇒7k≡14 mod15 ⇒k≡2 mod15. So k=15m+2 ⇒ S=7*(15m+2)+1=105m+15. Within 1-105: S=15, 120 (too big). So S=15.

6. (21,5): S≡0 mod21 and S≡1 mod5.

S=5k+1. 5k+1 ≡0 mod21 ⇒5k≡-1 mod21 ⇒5k≡20 mod21 ⇒k≡4 mod21. So k=21m+4 ⇒ S=5*(21m+4)+1=105m+21. Within 1-105: S=21, 126 (too big). So S=21.

7. (35,3): S≡0 mod35 and S≡1 mod3.

S=35k. 35k≡1 mod3 ⇒35≡2 mod3 ⇒2k≡1 mod3 ⇒k≡2 mod3. So k=3m+2 ⇒ S=35*(3m+2)=105m+70. Within 1-105: S=70, 175 (too big). So S=70.

8. (105,1): S=105.

Thus, solutions in [11,105] are S=15,21,36,70,85,91,105. That's 7 solutions. Not enough.

So D=105 gives 7 solutions. So D=210 is the first with 15 solutions. Therefore, the minimal b is 211.

Wait but before confirming, let me check D= 105*2=210 is the first with four prime factors. Alternatively, maybe a number with ω=4 but smaller than 210? Let me see:

The smallest number with four distinct prime factors is 2*3*5*7=210. So yes, 210 is the smallest D with ω=4. Hence, the minimal D where the count exceeds 10 is 210, so b=211.

But wait, let me check D= 120 (which has prime factors 2^3,3,5; so ω=3). Let me see:

D=120 (b=121):

Number of solutions mod120 is 2^3=8. Let's see how many are in [11,120].

The coprime pairs (d1,d2) with d1*d2=120 and gcd(d1,d2)=1:

Factors: 2^3,3,5. The coprime pairs:

(1,120),(3,40),(5,24),(8,15),(15,8),(24,5),(40,3),(120,1).

Calculating each solution:

1. (1,120): S=1 (too small).

2. (3,40): S≡0 mod3 and S≡1 mod40.

S=40k+1. 40k+1 ≡0 mod3 ⇒40≡1 mod3 ⇒k+1≡0 mod3 ⇒k≡2 mod3. So k=3m+2 ⇒ S=40*(3m+2)+1=120m+81. Within 1-120: S=81, 201 (too big). So S=81.

3. (5,24): S≡0 mod5 and S≡1 mod24.

S=24k+1. 24k+1 ≡0 mod5 ⇒24≡4 mod5 ⇒4k+1≡0 mod5 ⇒4k≡-1 mod5 ⇒4k≡4 mod5 ⇒k≡1 mod5. So k=5m+1 ⇒ S=24*(5m+1)+1=120m+25. Within 1-120: S=25, 145 (too big). So S=25.

4. (8,15): S≡0 mod8 and S≡1 mod15.

S=15k+1. 15k+1 ≡0 mod8 ⇒15≡7 mod8 ⇒7k+1≡0 mod8 ⇒7k≡-1 mod8 ⇒7k≡7 mod8 ⇒k≡1 mod8. So k=8m+1 ⇒ S=15*(8m+1)+1=120m+16. Within 1-120: S=16, 136 (too big). So S=16.

5. (15,8): S≡0 mod15 and S≡1 mod8.

S=8k+1. 8k+1 ≡0 mod15 ⇒8k≡-1 mod15 ⇒8k≡14 mod15 ⇒k≡ (14*8^{-1}) mod15. 8^{-1} mod15 is 2 since 8*2=16≡1 mod15. So k≡14*2=28≡13 mod15. So k=15m+13 ⇒ S=8*(15m+13)+1=120m+105. Within 1-120: S=105, 225 (too big). So S=105.

6. (24,5): S≡0 mod24 and S≡1 mod5.

S=5k+1. 5k+1 ≡0 mod24 ⇒5k≡-1 mod24 ⇒5k≡23 mod24 ⇒k≡23*5^{-1} mod24. 5^{-1} mod24 is 5, since 5*5=25≡1 mod24. So k≡23*5=115≡115-4*24=115-96=19 mod24. So k=24m+19 ⇒ S=5*(24m+19)+1=120m+96. Within 1-120: S=96, 216 (too big). So S=96.

7. (40,3): S≡0 mod40 and S≡1 mod3.

S=40k. 40k≡1 mod3 ⇒40≡1 mod3 ⇒k≡1 mod3. So k=3m+1 ⇒ S=40*(3m+1)=120m+40. Within 1-120: S=40, 160 (too big). So S=40.

8. (120,1): S=120.

Thus, solutions in [11,120] are S=16,25,40,81,96,105,120. That's 7 solutions. Still not enough.

Therefore, it seems that D=210 is indeed the first D where the count exceeds 10, giving 15 solutions. Therefore, the minimal b is 211.

Wait, but before finalizing, let me check D= 105*2=210, but maybe another D with ω=4 but not the product of first four primes? For example, 2*3*5*7=210 is the smallest. The next would be 2*3*5*11=330, which is larger, so 210 is indeed the smallest.

Alternatively, maybe a number with ω=3 but with higher exponents? For example, D= 2^3*3*5=120 (which we saw gives 7 solutions). Not enough.

Alternatively, D= 2^4*3*5=240. Let me see:

D=240 (b=241). ω=3 (primes 2,3,5). So solutions mod240 would be 2^3=8. Probably not enough.

Alternatively, D= 2*3*5*7*11=2310, but that's way bigger.

Therefore, I think the minimal D is 210, so the answer is b=211.

Wait, but let me confirm with D=210, the count is indeed 15. Earlier, I listed 15 solutions between 15 and 210. Let me recount:

The solutions in the range [15,210] are:

15,21,36,70,85,91,105,106,120,126,141,175,190,196,210. That's 15 numbers. So yes, 15 solutions, which is more than 10. Therefore, the minimal b is 211.

But wait, the problem says "more than ten", so 11 or more. Since 15 is more than 10, so 211 is the answer.

However, before I conclude, let me check if there is a smaller D with ω=4 but not the product of first four primes. Wait, the next possible D with four distinct primes would be 2*3*5*7=210, then 2*3*5*11=330, which is larger. So 210 is indeed the smallest D with four distinct primes, so the minimal b is 211.

Therefore, the answer should be \boxed{211}

**Final Answer**
\boxed{211}
</think>

To solve the problem, we need to find the smallest base \( b \) such that there are more than ten \( b \)-beautiful numbers. A number \( n \) is \( b \)-beautiful if it has exactly two digits in base \( b \) and the sum of these digits equals \( \sqrt{n} \).

Key steps:
1. **Formulation**: For a number \( n \) to be \( b \)-beautiful, it must be expressible as \( n = d_1 \cdot b + d_0 \) where \( d_1 \) and \( d_0 \) are digits in base \( b \), and \( d_1 + d_0 = \sqrt{n} \). This leads to the equation \( (d_1 + d_0)^2 = d_1 \cdot b + d_0 \).
2. **Reformulation**: Let \( S = d_1 + d_0 \). Then \( S^2 = d_1 \cdot b + d_0 \) and substituting \( d_0 = S - d_1 \) gives \( S^2 - S = d_1 \cdot (b - 1) \). This implies \( d_1 = \frac{S(S - 1)}{b - 1} \).
3. **Constraints**: \( d_1 \) must be an integer between 1 and \( b - 1 \), and \( S \) must be between \( \lceil \sqrt{b} \rceil \) and \( b - 1 \).
4. **Number Theory Insight**: For each \( D = b - 1 \), the number of solutions \( S \) such that \( S(S - 1) \equiv 0 \mod D \) is determined by the number of coprime divisor pairs of \( D \). The number of solutions is \( 2^{\omega(D)} \), where \( \omega(D) \) is the number of distinct prime factors of \( D \).
5. **Finding the Minimal \( b \)**: The smallest \( D \) with \( 2^{\omega(D)} > 10 \) is when \( \omega(D) \geq 4 \). The smallest such \( D \) is \( 210 \) (product of the first four primes: 2, 3, 5, 7), leading to \( b = 211 \).

After detailed calculations and verifying the number of solutions for \( D = 210 \), we find there are 15 solutions, which exceeds 10. Thus, the minimal \( b \) is \( 211 \).

\[
\boxed{211}
\]