Difference between revisions of "1989 AIME Problems/Problem 9"
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<center><math>0 \equiv n\pmod{3}</math></center> | <center><math>0 \equiv n\pmod{3}</math></center> | ||
| − | Thus, <math>n</math> is divisible by three and leaves a remainder of four when divided by 5. It's obvious that <math>n>133</math>, so the only possibilities are <math>n = 144</math> or <math>n \geq 174</math>. It quickly becomes apparent that 174 is much too large, so <math>n</math> must be <math>144</math>. | + | Thus, <math>n</math> is divisible by three and leaves a remainder of four when divided by 5. It's obvious that <math>n>133</math>, so the only possibilities are <math>n = 144</math> or <math>n \geq 174</math>. It quickly becomes apparent that 174 is much too large, so <math>n</math> must be <math>\boxed{144}</math>. |
== See also == | == See also == | ||
Revision as of 14:09, 17 April 2011
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 | ||
