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Difference between revisions of "2006 Cyprus MO/Lyceum/Problem 5"

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[[Category:Introductory Algebra Problems]]

Revision as of 22:01, 19 October 2007

Problem

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

A. $384$

B. $2$

C. $15$

D. $56$

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 $128+256=\boxed{384}$.

See also

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
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