Find the largest positive integer $n$ such that $n$ is divisible by all the positive integers less than $\sqrt[3]{n}$.
Problem
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Tags: Divisibility Theory
16.10.2007 10:32
$ n=420<8^3$. It is easy to chek, that $ k|420,k\le 7$. Let $ x_n=lcm(1,2,...,n)=\prod_{p\le n}p^{[ln(n)/ln(p)]}$. $ ln(x_n)=\sum_{p\le n} lnp[\frac{ln n}{ln p}]=n+o(n)>3ln(n+1)$, when n>10. $ x_8=840>9^3,x_9=x_{10}=2520>11^3$.
28.08.2009 12:11
Could you give the more specific solution?I can't understand the above solution
28.08.2009 16:58
a more elementary solution. Let $ x$ be the largest integer less than $ \sqrt [3]{n}$. Than $ (x + 1)^3\geq n$. We have $ x|n$, $ x - 1|n$, $ x - 2|n$ and $ x - 3|n$. But than $ lcm(x,x - 1,x - 2,x - 3)|n$. But therefore: \[ \frac {x(x - 1)(x - 2)(x - 3)}{6}\leq lcm(x,x - 1,x - 2,x - 3)\leq n\leq (x + 1)^3\] which is false for $ x\geq13$. so $ x\leq 12$. Now denote $ a_n = lcm(1,2...,n)$. We have: $ a_{12} = a_{11} = 2^3\cdot 3^2 \cdot 5 \cdot 7\cdot 11 = (2^2\cdot3^2)(5\cdot 7)(2\cdot 11) > 13^3$ $ a_9 = a_{10} = 2^3\cdot 3^2 \cdot 5 \cdot 7 = (2^2\cdot 3)(3\cdot 5)(2\cdot 7) > 11^3$ $ a_8 = 2^3\cdot 3\cdot 5 \cdot 7 = 840 > 9^3$ Therefore $ x\leq7$ and we have $ 420|n$ ($ a_7 = 420$) and $ n < 8^3 = 512$ so the solution is $ n = 420$