If a number has an odd number of factors then that number is a perfect square. The prime factorization of 2013 is 3*11*61. So the multiples of 2013 are of the form k^2*3^2*11^2*61^2.
Writing things a bit more carefully the numbers must be of the form p1^e1 * p2^3e * ... * pk^ek * 3^(2+x) * 11^(2+y) * 61^(2+z)
Then the total number of factors is given by
(e1+1)*(e2+1)*...*(ek+1)*(x+3)*(y+3)*(z+3)
This is to equal 2013=3*11*61
But again, 2013 has exactly three prime factors so in order for things to work, all of e1, e2, ..., ek must be zero. Leaving (x+3)*(y+3)*(z+3) = 3*11*61.
So then this leaves us with six possible ordered triplets (x, y, z). Those are (0,8,58), (8,0,58), (0,58,8), (8,58,0), (58,0,8), and (58,8,0).
Then the six multiples of 2013 with 2013 factors are
3^2*11^10*61^60, 3^10*11^2*61^60, 3^2*11^60*61^10, 3^10*11^60*61^2, 3^60*11^2*61^10, and 3^60*11^10*61^2.
Solution
