[ Decimal numeral system ]

$[$ Division with remainder $]$

Find all such two-digit numbers that when multiplied by some integer, the resulting number has a second-to-last digit of 5.
First, note that numbers divisible by 20 do not work. Indeed, after multiplying by any integer, we get a number that is divisible by 20. Therefore, the last digit of this number will be zero, and the second-to-last digit will be even.

We will prove that for any number $N$ not divisible by 20, there exists a number such that when $N$ is multiplied by it, the resulting number has a second-to-last digit equal to 5.

Lemma. Let $d = \text{GCD}(m, n)$. Then the set of remainders of numbers of the form $kn$ when divided by $m$ is the set of remainders that are multiples of $d$.

Proof. According to problem $\underline{60488}$ b) $d = am + bn$ for some integers $a$ and $b$. Multiplying both sides of the equation by $l$, we get $ld = alm + bnl$, which means that the number $bnl$ gives a remainder of $ld$ when divided by $m$. On the other hand, a remainder not divisible by $d$ cannot occur.

It remains to consider different values of $d = \text{GCD}(100, N)$. If the number $d$ is one of the numbers $1, 2, 5, 10, 25, 50$, then among numbers of the form $Nk$ there will be a number ending in 50. If $d = 4$, then among these numbers there will be a number ending in 52.

## Answer

All numbers not divisible by 20.