# Thread: Finding the smallest positive integral value of n

1. ## Finding the smallest positive integral value of n

"Find the smallest positive integral value of $n$ for which the following product is an integer.

$(1\frac{1}{36})$ $(1\frac{1}{37})$ $...$ $(1\frac{1}{n})$

If anybody could give me a hint on how to find this I would appreciate it. Thanks!

2. The product is equal to

$\frac{37}{36}\cdot\frac{38}{37}\cdot\ldots\cdot\fr ac{n}{n-1}\cdot\frac{n+1}{n}=\frac{n+1}{36}$

The smallest multiple of 36, greater than 36, is 72.

Then, $n+1=72\Rightarrow n=71$