SOLUTION: Find the smallest number which is divisible by 11 and has 24 factors.

It is well known that if the prime factorization of n is

n = p1a1p2a2p3a3···

Then the number of divisors is the product 

(a1+1)(a2+1)(a3+1)···

The prime factorization of 24 is 2·2·2·3

That's equal to (1+1)(1+1)(1+1)(2+1)

So any product of four primes with the exponents 1,1,1,2 

p11p21p31p42···

will have 24 factors.  We picked all the prime factors of 24 
to make the exponents as small as possible.

We pick the smallest prime, 2, to have the largest exponent, 2,
and then we pick the next smallest possible primes to have the 
1 exponents.  We would pick 3, 5, and 7, to have the 1 exponents.
but we are required to pick 11, so the smallest 4 primes we can 
pick to have the 1 exponents are 3,5 and 11.

So the answer is 223151111 = 660.

Edwin