2006 Cyprus MO/Lyceum/Problem 5

Problem

If both integers $\alpha,\beta$ are bigger than 1 and satisfy $a^7=b^8$ , then the minimum value of $\alpha+\beta$ is

$\mathrm{(A)}\ 384\qquad\mathrm{(B)}\ 2\qquad\mathrm{(C)}\ 15\qquad\mathrm{(D)}\ 56\qquad\mathrm{(E)}\ 512$

Solution

Since $b$ is greater than $1$ and therefore not equal to zero, we can divide both sides of the equation by $b^7$ to obtain $a^7/b^7=b$ , or $\left(\frac{a}{b}\right)^7=b$ Since $b$ is an integer, we must have $a/b$ is an integer. So, we can start testing out seventh powers of integers.

$a/b=1$ doesn't work, since $a$ and $b$ are defined to be greater than $1$ . The next smallest thing we try is $a/b=2$ .

This gives $b=(a/b)^7=2^7=128$ , so $a=2b=2(128)=256$ . Thus, our sum is $128+256=384$ , and the answer is $\mathrm{(A)}$ .

2006 Cyprus MO, Lyceum (Problems)
Preceded by Problem 4	Followed by Problem 6
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30

Page

Toolbox

Search

2006 Cyprus MO/Lyceum/Problem 5

Problem

Solution

See also