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

ColdplayMX  Aug 14, 2018
 #1
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\(12^n=(3\cdot4)^n=3^n\cdot4^n\)

 

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

 

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. 

GYanggg  Aug 14, 2018
edited by GYanggg  Aug 14, 2018
 #2
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But shouldn't 12^8 mod 20! = 0??!! 12^8 mod 20! =429,981,696 !!?

Guest Aug 14, 2018

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