text: 64, divisible by 2,4,8,16,32.\n\nThats actually easy to prove. If you write out any number n as a product of primes:\n2^(a1) * 3^(a2) * 5^(a3)...\nthen how many factors does it have? For each factor, we need to choose a power of 2 between 0 and a1, a power of 3 between 0 and a2.., ie a total of (a1+1)(a2+1)(a3+1)... choices.\n\nIn this case, we need exactly 7 factors (5, plus 1 and itself), and the only way to make 7 out of a product like that is to make a1 = 6 (since we want the smallest, no need to choose a2 etc).\n\n(Note that it is *NOT* 2^(n+1) for n positive integers in general, as one answerer said. Take n = 7, say (where we want 9 factors). Its much better to choose a1 = 2, a2 = 2 to get 36, as opposed to taking a1 = 8 to get 256.)\n\nReplying to Jud: Sorry, I made a mistake. a1 = 2, a2 = 2 gives 36 as the smallest number with 7 factors other than 1: 2,3,4,6,9,12,18.\n256 is quite a bit higher than that :)