Integers, Sequences and a Property!

Number Theory Level 4

If $n$ is a positive integer, let $a_n$ be the smallest positive number for which $(a_n)!$ is divisible by $n$ . Then determine the sum of all positive integers $n$ satisfying:

$\Large{\dfrac{a_n}{n} = \dfrac23 }$

The answer is 9.

1 solution

Patrick Corn
Aug 17, 2015

Suppose $n$ is such an integer. Since $3|2n$ , we need $3|n$ . So let $n = 3k$ . Note that $3k | (2k)!$ but $3k \nmid k!$ . This is only possible if $k \le 3$ , since otherwise both $3$ and $k$ appear separately in $k!$ . So $n = 3,6,9$ are the only possibilities. It's easy to check that $n = 9$ is the only one that works, so the answer is $\fbox{9}$ .

Integers, Sequences and a Property!

The answer is 9.

1 solution

0 pending reports