2、If $20 \times 21 \times 22 \times \ldots \times 2020=26^{k} \times m$, where $m$ is an integer, what is the maximum value of the integer $k$?
【Answer】165
【Analysis】Factor and Multiple
$26^{k}=2^{k} \times 13^{k}$, to make the equation hold and maximize $k$, it is evident that as many factors of 2 and 13 as possible should be extracted from the left side of the equation. Since the number of 2s far exceeds the number of 13s, we only need to find the number of factors of 13 in $1 \times 2 \times 3 \times \ldots \times 2020$, and subtract the number of factors of 13 in $1 \times 2 \times \ldots \times 19$.
In $1 \times 2 \times 3 \times \ldots \times 2020$, every 13 numbers contain 1 multiple of 13, every $13^{2}$ numbers can be broken down into 2 factors of 13, and so on:
$2020 \div 13=155$ (groups)...5 (numbers) are $13,26,39 \ldots 2015$
Among these 155 numbers, every 13 numbers can again be broken down into 1 factor of 13, for example, $169,338 \ldots$
$155 \div 13=11$ (groups)...12 (numbers) are $169,228 \ldots 1859$, totaling 11
These 11 numbers cannot be further broken down into 13, so in total there are 155+11=166 (numbers)
While 1-19 only have 1 factor of 13
$166-1=165$