Track / Stage 5 / 20 of 400 #620 of 2000

Problem 620

AIME late

Number theory Difficulty 5.1 Find the answer

For example, $1 k$ is a natural number, and $\frac{1001 \cdot 1002 \cdot \cdots \cdot 2005 \cdot 2006}{11^{k}}$ is an integer, what is the maximum value of $k$ ?

Your answer

Official solution

Solve $\frac{1001 \cdot 1002 \cdot \cdots \cdot 2005 \cdot 2006}{11^{k}}=\frac{2006!}{11^{k} \cdot 1000!}$ .
The highest power of the prime factor 11 in 2006! is
$\left[\frac{2006}{11}\right]+\left[\frac{2006}{11^{2}}\right]+\left[\frac{2006}{11^{3}}\right]=182+16+1=199 \text {. }$
The highest power of the prime factor 11 in 1000! is
$\left[\frac{1000}{11}\right]+\left[\frac{1000}{11^{2}}\right]=90+8=98 \text {. }$

Therefore, the maximum value of $k$ is $199-98=101$ .

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.