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.
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Codeforces (c) Copyright 2010-2024 Михаил Мирзаянов
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Время на сервере: 02.06.2024 12:10:47 (k3).
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1344
Let n = p1α1 * p2α2 * ... * pkαk
To maximize number of divisors, according to count of divisors formula, you have to find maximum value of (α1 + 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.
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.