Vasiljko's blog

By Vasiljko, history, 7 years ago, In English

I was doing one problem and thought of maximum number of divisors. What I want to say is : We iterate from 1 to N (N<=10^9) and define S(i) number of divisors of number i. What is maximum S we can get? I think it is about 100 or maybe less but if someone can prove.

  • Vote: I like it
  • 0
  • Vote: I do not like it

»
7 years ago, # |
  Vote: I like it +7 Vote: I do not like it

1344

»
7 years ago, # |
Rev. 2   Vote: I like it +14 Vote: I do not like it

Let n = p1α1 * p2α2 * ... * pkαk

To maximize number of divisors, according to count of divisors formula, you have to find maximum value of 1 + 1) * (α2 + 1) * ... * (αk + 1) with constraints n ≤ 109. Answer is n = 931170240 = 26 * 32 * 51 * 71 * 111 * 131 * 171 * 191 with 1344 divisors.

UPD: As, nickitat said, answer is 1344.

»
7 years ago, # |
  Vote: I like it +2 Vote: I do not like it

Related: the maximum number of divisors for any n-digit number, along with the numbers with this much divisors (1, 2), are very useful when constructing maximal test cases in certain problems, either problemsetting or hacking.