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

Revision as of 21:24, 17 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 $128+256=\boxed{384}$ .

@@ Line 13: / Line 13: @@
 ==Solution==
-{{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.
+<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>.
 ==See also==
 {{CYMO box|year=2006|l=Lyceum|num-b=4|num-a=6}}

2006 Cyprus MO, Lyceum (Problems)
Preceded by Problem 4	Followed by Problem 6
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Difference between revisions of "2006 Cyprus MO/Lyceum/Problem 5"

Revision as of 21:24, 17 October 2007

Problem

Solution

See also