Competition problem — Number theory

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Number theory Difficulty 5.1 AIME, harder Prove it Philippines

Problem:

Find the largest positive integer $x$ such that $2017!$ is divisible by $19^{x}$ .

Solution

Solution:

There are $\left\lfloor\frac{2017}{19}\right\rfloor = 106$ numbers from $1$ to $2017$ which are divisible by $19$ .

Among these, five numbers, $19^{2}, 2 \cdot 19^{2}, 3 \cdot 19^{2}, 4 \cdot 19^{2}, 5 \cdot 19^{2}$ , are divisible by $19^{2}$ .

Therefore, $19^{111} \mid 2017!$ and so the largest possible $x$ is $111$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.