Difference between revisions of "2006 Cyprus MO/Lyceum/Problem 5"
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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 | ||
− | + | <math>\mathrm{(A)}\ 384\qquad\mathrm{(B)}\ 2\qquad\mathrm{(C)}\ 15\qquad\mathrm{(D)}\ 56\qquad\mathrm{(E)}\ 512</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. | ||
− | + | <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=384</math>, and the answer is <math>\mathrm{(A)}</math>. | |
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==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 11:03, 27 April 2008
Problem
If both integers are bigger than 1 and satisfy
, then the minimum value of
is
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
, and the answer 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 |