What is the largest integer divisible by all positive integers less than its cube root.
Solution
Answer: 420.
Let N be a positive integer satisfying the condition and let n be the largest integer not exceeding its cube root. If n = 7, then 3·4·5·7 = 420 must divide N. But N cannot exceed 83 - 1 = 511, so the largest such N is 420.
If n ≥ 8, then 3·8·5·7 = 840 divides N, so N > 729 = 93. Hence 9 divides N, and hence 3·840 = 2520 divides N. But we show that no N > 2000 can satisfy the condition.
Note that 2(x - 1)3 > x3 for any x > 4. Hence [x]3 > x3/2 for x > 4. So certainly if N > 2000, we have n3 > N/2. Now let pk be the highest power of k which does not exceed n. Then pk > n/k. Hence p2p3p5 > n3/30 > N/60. But since N > 2000, we have 7 < 11 < n and hence p2, p3, p5, 7, 11 are all ≤ n. But 77 p2p3p5 > N, so N cannot satisfy the condition.
© John Scholes
jscholes@kalva.demon.co.uk
11 Apr 2002