The Power Game

Number Theory Level 3

If four distinct positive integers $u$ , $v$ , $w$ and $x$ satisfies the following equation

$u=v^2=w^3=x^4.$

The minimum value of $u$ is

This is one part of 1+1 is not = to 3 .

The answer is 4096.

3 solutions

Kenneth Tan
Jul 14, 2014

$u$ can't be 1 as this would make $u=v=w=x=1$ which is a contradiction. Suppose $u=2^n$ , because $v$ , $w$ and $x$ are integers, to make $u$ divisable by $v$ , $w$ and $x$ , we take $n$ as the L.C.M. of the powers of $v$ , $w$ and $x$ which is 12.

Hence, $u=2^n=2^{12}=4096$

A conclusion should also be made in the solution that even if $u=x^n~,~x\gt 2;x,n\in \mathbb{Z^+}$ satisfies all the conditions for some values of $x$ and $n$ , the value we will get will not be the least since the solution grows at an exponential rate with increase in $x$ and $n$ .

Prasun Biswas - 6 years, 5 months ago

It should be 64.

Kevin Patel - 6 years, 11 months ago

If $u=64$ , then there will be no such integer that satisfies $x^4=64$ . If there are only 3 distinct positive integers $u$ , $v$ and $w$ which is $u=v^2=w^3$ , then you are correct, in this case, the minimum value of $u$ is 64.

Kenneth Tan - 6 years, 11 months ago

Curtis Clement
Jan 14, 2015

${LCM}$ { ${1,2,3,4}$ } = 12, so 12 will be the highest exponent which we must apply to the smallest integer that is more than 1. That is $2^{12}$ = $\boxed{4096}$

Pragjyotish Bhuyan Gogoi
Oct 25, 2014

the condition, x^4=v^2=u, does not put any restriction on the value of 'x'; for every integer x^4, there will always be a 'x^2' and x, so the only expression that matters is, w^3=x^4, so, w=x^(4/3), and since w is an integer and different from x, x at least has to be 2^3, i.e x=8, u=4096.

The Power Game

This is one part of 1+1 is not = to 3 .

The answer is 4096.

3 solutions

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