What is the remainder when $10!$ is divided by $2^{7}$? (Reminder: If $n$ is a positive integer, then $n!$ stands for the product $1\cdot 2\cdot 3\cdot \cdots \cdot (n-1)\cdot n$.)
10! = 1 * 2 * 3 * (2 * 2) * 5 * ( 3 * 2) * 7 * ( 2^3) * 9 * (5 * 2) / 2^7 =
2^8 ( 3 * 5 * 3 * 7 * 9 * 5)
________________________ = 2 ( 3^2 * 5^2 * 7 * 9) = 28350
2^7
(Remainder = 0 )
10! = [2^8 * 3^4 * 5^2 * 7] mod 2^7 ==0 - The remainder.
