Without code... We want the coefficient of $x^{35}$ in the series expansion of $\begin{aligned} (x + x^2 + x^3 + x^4 + x^5 + x^6)^{10} & = \; x^{10}(1 -x^6)^{10}(1 - x)^{-10} \; = \; x^{10}\left(\sum_{m=0}^{10}(-1)^m \binom{10}{m}x^{6m}\right)\left(\sum_{n=0}^\infty \binom{9+n}{9}x^n\right) \\ & = \;\sum_{m=0}^{10} \sum_{n=0}^\infty (-1)^m \binom{10}{m}\binom{9+n}{9} x^{10+6m+n} \end{aligned}$ which is $X \; = \; \sum_{m=0}^4 (-1)^m\binom{10}{m}\binom{34-6m}{9} \; = \; 4395456$ making the probability equal to $\frac{X}{6^{10}} \; = \; \frac{7631}{104976}$ which gives the answer $7631 + 104976 = \boxed{112607}$ .

It works with Windows:

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#include <iostream>
using namespace std;
long long int solution=0;
long long int digit(long long int a, int b);    /// Get the b-th digit
long long int power(long long int a);           /// return: 10^a
int main()
{
    long long int a=1111111111;
    /**
    a is the dice. Initially each die is 1. The program will run
    until 6,6,6,6,6,1,1,1,1,1, because 6+6+6+6+6+1+1+1+1+2>35
    **/
    int alpha=0;
    for(;a<=6666611111;a++){
        long long int theta=1;                  /// If theta==0, then we should restart the function
        for(long long int b=1;b<10&&theta;b++)  /// For all digits
            if(digit(a,b)>6){                   /// If one of the dice is >6
                a-=power(b-1)*6;                /// Delete the digit
                a+=power(b);                    /// Add one to the next digit
                theta=0;                        /// Restart the function ---+
            }                                   ///                         |
        if(!theta){                             /// <-----------------------+
            a--;
            continue;
        }
        long long int sum=0;
        for(long long int b=1;b<=10;b++)
            sum+=digit(a,b);
        if(sum==35) solution++;                 /// If the sum is 35, then +1 solution exist
    }
    cout <<solution;
    return 0;
}
long long int digit(long long int a, int b){    /// Get the b-th digit
    long long int c=a%power(b);
    c=c/power(b-1);
    return c;
}
long long int power(long long int a){           /// return: 10^a
    long long int to_return=1;
    for(long long int c=0;c<a;c++)    to_return*=10;
    return to_return;
}

Output:

1

4395456

Solution:

$\cfrac{4395456}{6^{10}}=\cfrac{7631}{104976}\implies p+q=7631+104976=112607$

I used the generating function method

$g(x) = (x + x^2 + x^3 + x^4 + x^5 + x^6)^{10}$

You can convince yourself that the coefficient of $x^k$ in $g(x)$ is exactly the number of ways to get a sum of $k$ on the $10$ dice.

I used simple code to find the coefficients of the resulting polynomial $g(x)$ . The coefficient of $x^{35}$ is $4395456$

Therefore, the probability is $P = \dfrac{4395456}{6^{10}} = \dfrac{7631}{104976}$ in smallest terms.

And this makes the answer $7631 + 104976 = 112607$

A018901

$\frac{4395456}{6^{10}}$

Answer $112607$ .

Since code is used, I addressed directly as follows using the code environment available on this Brilliant website.

Then $\dfrac pq = \dfrac {4395456}{6^{10}} = \dfrac {7631}{104976}$ and $p+q = 7631 + 104976 = \boxed{112607}$ .