The expression 120!/(60!*60!) is an integer. What is the largest integer n such that 7^n divides this integer?
+307 Solution: \(1\)
Answer:
We are looking for the number of sevens there are in the prime factorization of the number.
60! has 9 sevens in its prime factorization, because there are eight multiples of seven in 1x2x3x4...x60, and one of them is 7².
Because the 60! is being square, the number of sevens in its prime factorization doubles to 18.
120! has 19 sevens in its prime factorizations, because there are 17 multiples of seven in 1x2x3x4....x120, and two of them are multiples of 49.
\(x \cdot7^{18}\over y \cdot 7^{19}\)\(=x/y \cdot 7\)
where x and y are two numbers that aren't multiples of 7 (and so therefore neither is their quotient).
There is only one seven the prime factorization of this number, which means that the largest integer n such that 7^n that divides it is just \(1\).
