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2006 Cyprus MO/Lyceum/Problem 5: Difference between revisions

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


A. <math>384</math>
<math>\mathrm{(A)}\ 384\qquad\mathrm{(B)}\ 2\qquad\mathrm{(C)}\ 15\qquad\mathrm{(D)}\ 56\qquad\mathrm{(E)}\ 512</math>


B. <math>2</math>
==Solution==
Since <math>b</math> is greater than <math>1</math> and therefore not equal to zero, we can divide both sides of the equation by <math>b^7</math> to obtain <math>a^7/b^7=b</math>, or
<cmath>
\left(\frac{a}{b}\right)^7=b
</cmath>
Since <math>b</math> is an integer, we must have <math>a/b</math> is an integer.  So, we can start testing out seventh powers of integers.


C. <math>15</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>.


D. <math>56</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=384</math>, and the answer is <math>\mathrm{(A)}</math>.
 
E. <math>512</math>
 
==Solution==
{{solution}}


==See also==
==See also==
{{CYMO box|year=2006|l=Lyceum|num-b=4|num-a=6}}
{{CYMO box|year=2006|l=Lyceum|num-b=4|num-a=6}}
[[Category:Introductory Algebra Problems]]

Latest revision as of 10:03, 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
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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