Question number 22, how many positive integers are multiples of 2013 and have exactly 2013 divisors, including one in the number itself? So let us begin by looking at a multiple of 2013. So we have a number that looks like 2013 times some number and we factor into prime factors. So there will be a 3 to some power, an 11 to some power, and then a 61 to some power, those being the prime factors of 2013. And then we factor A into prime numbers, let's call them P's with a subscript, and each of them will appear corresponding number of K times. And let's say there are N of them, and then there will be exactly 2013 divisors. So 3 times 11 times 61, that's 2013. That has to be the same thing as the product of X, Y and Z because 3, 11 and 61 must appear at least once. And then we multiply this also by the numbers appearing in the exponents of the prime numbers dividing A, but we increase each one by 1 to account for the possibility that there is a 0 in the exponent, like so. But then we see that we really can't have any of these factors here because we must have 3 numbers at least, the 3, the 11 and the 61 as choices for our X, Y and Z. So then what we do is we count the number of ways in which we can choose the X, Y and Z. There are 6 ways to choose the numbers X, Y and Z from the set of 3 numbers, mainly 3, 11 and 61. And so that is the number of positive integer multiples of 2013 that are divisible by exactly 2013 divisors.

positive integers

multiples of 2013

prime factorization

divisors condition

unique solutions