Power Play

Number Theory Level 3

The smallest number $n$ greater than 1 such that $\large \sqrt{n}$ , $\large \sqrt[3]{n}$ , $\large \sqrt[4]{n}$ , $\large \sqrt[5]{n}$ , $\large \sqrt[6]{n}$ , $\large \sqrt[7]{n}$ , $\large \sqrt[8]{n}$ , $\large \sqrt[9]{n}$ , $\large \sqrt[10]{n}$ are all integers can be expressed in the form $a^{b}$ , where $a$ and $b$ are both positive integers and with $a$ as small as possible.

Find $a+b$ .

The answer is 2522.

1 solution

William Whitehouse
Jul 21, 2016

Clearly if the $m$ th root of a number is integral then the number can be expressed as $x^{m}$ . $n$ can be written as $x_{1}^{2}$ , $x_{2}^{3}$ etc. With all $x$ being integral. This means that the exponent must be divisible by all of the powers. I.e. 2,3,4,5,6,7,8,9,10. The smallest number divisible by all of these is 2520. The base must be as small as possible (but as specified in the question cannot be 1). It is therefore 2.

So a=2 b=2520 and a+b=2522.

The initial claim is not true. For example, $4^ 2$ has an integer fourth root, even though $\frac{2}{4}$ is not an integer. You have to figure out what is the important characteristic to focus on, in order to present a complete solution.

Calvin Lin Staff - 4 years, 10 months ago

That is true but the fact that the exponent must be a multiple of 2520 doesn't change, the claim I made was that it can be written as x^m, not that that was the most simple way of writing it.

William Whitehouse - 4 years, 10 months ago

The wording is very unclear

Esp Ter - 4 years, 7 months ago

Power Play

The answer is 2522.

1 solution

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