Show that for any set of 201 positive integers less than 300, there must be two whose quotient is a power of three (with no remainder)

Question

I guess we should not consider the zeroth power of 3 because it is equal to one. Any positive integer is a multiple of 1.

Lets define the set S3 of integers that are multiples of 3 strictly less than 300. The size of this set is 99, because 99 * 3 = 297.

Lets consider some subset of size 201. 98 integers are thrown out of the initial set. But we can choose 98 out of S3. So only one element of S3 is left. Where am I wrong?

Answer 1

HINT: Every integer in the set $\{1,2,\ldots,299\}$ can be written uniquely in the form $3^mn$, where $m\ge 0$, $n\ge 1$, and $n$ is not a multiple of $3$. You’ve already shown that there are $200$ possible choices for $n$. For each of these values of $n$ let

$$A_n=\{3^mn:m\ge 0\text{ and }3^mn<300\}\;;$$

the sets $A_n$ are a partition of $\{1,2,\ldots,299\}$ into $200$ parts. Now apply the pigeonhole principle.