Difference between revisions of "1989 AIME Problems/Problem 9"
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== Solution == | == Solution == | ||
| − | {{ | + | 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>4 \equiv n\pmod{5}</math></center> | ||
| + | |||
| + | Continuing, we examine the equation modulo 3, | ||
| + | <center><math>-1 + 1 + 0 + 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 = 174</math>. It quickly becomes apparent that 174 is much too large, so <math>n</math> must be <math>144</math>. | ||
== See also == | == See also == | ||
| − | + | {{AIME box|year=1989|num-b=8|num-a=10}} | |
| − | + | ||
| − | + | [[Category:Intermediate Number Theory Problems]] | |
Revision as of 22:17, 26 February 2007
Problem
One of Euler's conjectures was disproved in then 1960s by three American mathematicians when they showed there was a positive integer such that 
. Find the value of 
.
Solution
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 | ||
