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

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<math>a/b=1</math> doesn't work, since <math>a</math> and <math>b</math> are defined to be greater than <math>1</math>. The next smallest thing we try is <math>a/b=2</math>.
 
<math>a/b=1</math> doesn't work, since <math>a</math> and <math>b</math> are defined to be greater than <math>1</math>. The next smallest thing we try is <math>a/b=2</math>.
  
This gives <math>b=(a/b)^7=2^7=128</math>, so <math>a=2b=2(128)=256</math>. Thus, our sum is <math>128+256=\boxed{384}</math>.
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This gives <math>b=(a/b)^7=2^7=128</math>, so <math>a=2b=2(128)=256</math>. Thus, our sum is <math>128+256=384</math>, and the answer is <math>\mathrm{(A)}</math>.
  
 
==See also==
 
==See also==

Latest revision as of 11:03, 27 April 2008

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 $\mathrm{(A)}$.

See also

2006 Cyprus MO, Lyceum (Problems)
Preceded by
Problem 4
Followed by
Problem 6
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