+0

# I'm counting on you guys!!!

0
170
2
+196

Find the largest integer n for which 12^n  evenly divides 20!.

Aug 14, 2018

#1
+972
+1

$$12^n=(3\cdot4)^n=3^n\cdot4^n$$

$$20!=1\cdot\boxed{2}\cdot\boxed{3}\cdot\boxed{4}\cdot5\cdot\boxed{6}\cdot7\cdot\boxed{8}\cdot\boxed{9}\cdot\boxed{10}\cdot11\cdot\boxed{12}\cdot13\cdot\boxed{14}\cdot\boxed{15}\cdot\boxed{16}\cdot17\cdot\boxed{18}\cdot19\cdot\boxed{20}$$

Boxed numbers are multiples of 2 and 3:

$$2\cdot3\cdot4\cdot6\cdot8\cdot9\cdot10\cdot12\cdot14\cdot15\cdot16\cdot18\cdot20\Rightarrow x\cdot3^8\cdot4^{9}$$

Therefore, the largest integer n is 8.

I recall there being a better way to do this using modular arithmetic, but I don't know how to.

Aug 14, 2018
edited by GYanggg  Aug 14, 2018
#2
+1

But shouldn't 12^8 mod 20! = 0??!! 12^8 mod 20! =429,981,696 !!?

Aug 14, 2018