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=\ | + | 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 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 |