Find the smallest positive integer n such that n has exactly 144 distinct positive divisors and there are 10 consecutive integers among them (Note: 1 and n are both divisors of n)

To have exactly 144 distinct positive divisors, the number must factor into the squares of each of two different prime numbers, and the first power of each of four other prime numbers. That way each of the 144 factors can be formed by multiplying 0, 1 or 2 occurrences of each of the squared primes, and 0 or 1 each of the first-power primes.

With 2^2 * 3^2 * 5 * 7 * 11 * 13, there can't be a divisor of 8, though all other divisors from 1 through 15 work, so some other set of primes needs to be considered (unless there's a sequence of 10 that's higher up using this set).

Posted by Charlie on 2006-02-20 10:32:24