What is the smallest number with 20 divisors?

Answer 1

"We want to find the smallest number with exactly 20 factors (or divisors). If a number's prime factorisation consists of n different prime numbers and r repeated primes, then the total number of factors is 2n x (r+1)." * This is confusing. If the p.f. is 2r * p1 * ... * pn , where p1 to pn are n distinct odd primes, then yes. But if our number is N and factors as N = p1e1 . . . pkek , then N has this many factors (or divisors): d(N) = (e1 + 1)(e2 + 1) . . . (ek + 1). This is the general case. "The best solution to 2n x (r+1) = 20 is n = 2, r = 4." * This is better than n = 1, r = 9; N = 29 * 3 = 1536 > 240. "So the smallest number with exactly twenty factors, including itself, is 24 x 3 x 5 = 240. Its factors are 1,3,5,15,2,6,10,30,4,12,20,60,8,24,40,120,16,48,80,240." "The smallest number with exactly 20 factors excluding itself is, I think, 221 = 2097152." * This is wrong on two levels. You want a number with 21 divisors, including itself. First, 221 has 22 factors (21 not including itself), so you meant N = 220 = 1048576. Second, you can do much better by breaking up 21 = 7 * 3 so by the above formula (adding 1 to each exponent in the p.f.), use exponents of 6 and 2; to get the smallest N use 26 * 32 = 576, a mite smaller! * For more info see my page on supercomposite numbers at www.dansmath.com