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Saudi Arabian Mathematical Competitions

Number theory Difficulty 4.7 AIME Prove it Saudi Arabia

Let nn be a positive integer such that 201120112011^{2011} divides n!n!. Prove that 201120122011^{2012} divides n!n!.

Solution

Since 20112011 is a prime and 201120112011^{2011} divides n!n!, it follows that in n!n! we have at least 20112011 multiples of 20112011. These are
2011,2⋅2011,3⋅2011,…,2011⋅2011. 2011, 2 \cdot 2011, 3 \cdot 2011, \ldots, 2011 \cdot 2011.

Therefore 2011!⋅201120112011! \cdot 2011^{2011} divides n!n!, hence 201120122011^{2012} divides n!n!.

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