These numbers look perfect

It is easy to see that any natural number (in the decimal system) with the last two digits of 15 is divisible by 5, but not divisible by 25.

Hence, in it's prime factorisation, 5 will have the power of 1. Therefore, the number of its divisors will have a factor of (1 + 1) = 2.

This means, that any natural number ending in 15 has an even number of divisors, so none of them can have exactly 15 (odd number) divisors.

Only square numbers have an odd number of factors, so for a number to have $15$ positive divisors, it must be the square of an integer. As shown below, the last digit of the square root of a number which satisfies must also end in $5$ .

Now the last $k$ digits of a number $r$ are the only digits that affect the last $k$ digits of $r^2$ . Therefore, to end in $15$ , we need to consider the last $2$ digits of the square root.

Let a number $n$ have the last digits ' $a5$ '. When we write this in decimal form, and square it we get, $(10a + 5)^2 = 100a^2 + 100a + 25$

The first two terms of this result are multiples of $100$ , so won't affect the last two digits of $n^2$ . Therefore, we are left with the constant term of $25$ which means that the square of all integers which end in $5$ , ends in $25$ . Hence, there are no positive integer with $15$ positive divisors which ends in $15$ .