Triangles within triangles

Well, this question looks difficult(at least to me) at first sight but it is very easy(including that easy guess).

Well, we will not first solve the sum.

In Pascal's triangle, rows consist of numbers of the form $C(n-1,k)$ for row $n$ and for $k = [0,n-1]$ .

In $(S+1)$ th row, the $n-1$ would be just equal to $S$ .

Next, we have to find $C(S,k) mod 37$ for $k = [0,S]$ (hence $S+1$ elements)

Now, Lucas Theorem says $C(n,r) mod p$ is the product of combination operators of the corresponding coefficients of the base $p$ expansion of $n$ and $r$ .

$S = 9*{37}^{37} + 9*{37}^{36} +..... + 9*37 + 0*37$

and $k = {n}_{37}{p}^{37} + {n}_{36}{p}^{36} + .... + {n}_{1}p + {n}_{0}$

Hence, $C(S,k) = C(9,{n}_{37})*C(9,{n}_{36})*C(9,{n}_{35})....C(9,{n}_{1})*C(0,{n}_{0}) mod 37$

Now the remainder side will be non-zero if ${n}_{k} \leq 9$ (except for ${n}_{0}$ when it has to be equal to $0$ only because if it is greater than $0$ , the operator becomes $0$ and i f it is less than $0$ , it again becomes $0$ . So,

${n}_{k} = [0,1,2,3,4,5,6,7,8,9]$ (for $k = [1, 37]$ )

and ${n}_{0} = 0$

As a result, we can use Law of Multiplication of counting and get the final answer as - ${10}^{37}$ which will be followed by $log {10}^{37} = \boxed{37}$