Given n, we wish to know how many numbers m there are such that n = tau(m), and also n divides m. The focus in this document is on the case where n is squarefree. Lemma: the number of prime divisors of tau(m) counted with multiplicity is greater than or equal to the number of distinct prime divisors of m. Let the factorization of m be Prod_{i=1}^j p_i^k_i, where j is then the number of distinct prime divisors of m. Then tau(m) = Prod_{i=1}^j (k_i+1). Each k_i contributes at least one prime, counted with multiplicity, to tau(m). So the number of prime divisors of tau(m) with multiplicity is the sum of j numbers each at least 1, hence it is >= j. Q.E.D. Now let squarefree n = tau(m) = Prod_{i=1}^j p_i. Since n divides m, each p_i must also divide m, so m has at least j distinct prime factors. But by the lemma, the number of distinct prime factors of m is >= j. Hence the primes dividing m are precisely the primes dividing n. Now, let the factorization of m = Prod_{i=1}^j p_i^k_i. For tau(m) to equal n, each k_i+1 must be prime. In fact, k_i+1 must be one of the prime divisors of n. I.E. m = Prod_{i=1}^j p_i^(q_i-1), where the q_i are a permutation of the p_i. Conversely, if the q_i+1 are a permutation of the p_i, tau(m) will equal n. Hence, if n is squarefree, there are omega(n)! m that are multiples of n and have tau(m) = n.