Difference between revisions of "2006 Cyprus MO/Lyceum/Problem 5"
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==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== | ==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}} |
Revision as of 22:24, 17 October 2007
Problem
If both integers are bigger than 1 and satisfy
, then the minimum value of
is
A.
B.
C.
D.
E.
Solution
Since is greater than
and therefore not equal to zero, we can divide both sides of the equation by
to obtain
, or
Since
is an integer, we must have
is an integer. So, we can start testing out seventh powers of integers.
doesn't work, since
and
are defined to be greater than
. The next smallest thing we try is
. This gives
, so
. Thus, our sum is
.
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 |