#1**+1 **

Since 7 is a prime number, the number of 7's in the prime factorization of n! is equal to \(\left \lfloor{\frac{n}{7}}\right \rfloor+\left \lfloor{\frac{n}{7^2}}\right \rfloor+\left \lfloor{\frac{n}{7^3}}\right \rfloor+...\) , where \(\left \lfloor{x}\right \rfloor\) is just the floor of x, which is just the smallest integer less than or equal to x.

We want the number of 7's in the prime factorization of n! to be equal to 10, and we want it to be as small as possible, which means that n will also be a multiple of 7.

Trying out some reasonable values, we find that \(\boxed{63}\) is the smallest possible value, since \(\left \lfloor{\frac{63}{7}}\right \rfloor+\left \lfloor{\frac{63}{7^2}}\right \rfloor = 9+1 = 10\) (notice that I didn't test the floors of larger values since they will all be zero).

textot Feb 3, 2021