Difference between revisions of "1989 AIME Problems/Problem 9"
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== Solution == | == Solution == | ||
− | Note that <math>n</math> is even, since the <math> | + | Note that <math>n</math> is even, since the <math>LHS</math> consists of two odd and two even numbers. By [[Fermat's Little Theorem]], we know <math>{n^{5}}</math> is congruent to <math>n</math> [[modulo]] 5. Hence, |
<center><math>3 + 0 + 4 + 7 \equiv n\pmod{5}</math></center> | <center><math>3 + 0 + 4 + 7 \equiv n\pmod{5}</math></center> | ||
<center><math>4 \equiv n\pmod{5}</math></center> | <center><math>4 \equiv n\pmod{5}</math></center> |
Revision as of 22:12, 30 November 2009
Problem
One of Euler's conjectures was disproved in the 1960s by three American mathematicians when they showed there was a positive integer such that . Find the value of .
Solution
Note that is even, since the consists of two odd and two even numbers. By Fermat's Little Theorem, we know is congruent to modulo 5. Hence,
Continuing, we examine the equation modulo 3,
Thus, is divisible by three and leaves a remainder of four when divided by 5. It's obvious that , so the only possibilities are or . It quickly becomes apparent that 174 is much too large, so must be .
See also
1989 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |