Suppose that $26$ integers are chosen from the set $S=\{1,2,\ldots,50\}$.
By writing these numbers as $2^k m$ with $m$ odd, prove that one of the chosen numbers is a multiple of another of the chosen numbers.
I'm not sure how to set pigeons and pigeonholes here. So this is my attempt
The case with a repeated integer is trivial as each integer divides itself. So assume the integers are distinct. There are $\binom{50}{26}$ sets that have $26$ distinct integers.
Now, if the integer is odd, then $k=0$. If the integer is a power of $2$, then $m=1$. Also, if the integer is prime, then $k=0$.
However, I'm not sure what to do now
The $25$ holes are the odd integers $1,3,\ldots,49$. The $26$ pigeons are the selected integers, written with all powers of $2$ factored out, as you noted.
Two of those pigeons must be $2^km$ and $2^jm$ for the same $m$. Suppose $k\ge j$; then the first of these two pigeons is a multiple of the second.