math prob help?

Question

If n^3 is a divisor of 10!, what is the greatest possible integer value of n?

Answer 1

12. Since from 10x9x8x7x6x5x4x3x2x1, you can get 8x8x27, which all have integers as cubic roots. Therefore, 2x2x3 = 12.


XR

Answer 2

Prime factorize 10! first.

10! = 2^8 * 3^4 * 5^2 * 7

obviously n can't be a factor of 5 or 7, otherwise n^3 won't divide 10!.

so we must have n = 2^2 * 3 = 12, since then we have n^3 = 2^6 * 3^3, which divides 10!.

Answer 3

10!=2^8x3^4x5^2x7
n^3=2^6x3^3=1728
n=12