Difference between revisions of "1989 AIME Problems/Problem 9"

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== Solution ==
 
== Solution ==
{{solution}}
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By [[Fermat's Little Theorem]], we know <math>{n^{5}}</math> is congruent to <math>n</math> [[modulo]] 5.  Hence,
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<center><math>3 + 0 + 4 + 7 \equiv n\pmod{5}</math></center>
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<center><math>4 \equiv n\pmod{5}</math></center>
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Continuing, we examine the equation modulo 3,
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<center><math>-1 + 1 + 0 + 0 \equiv n\pmod{3}</math></center>
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<center><math>0 \equiv n\pmod{3}</math></center>
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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 ==
* [[1989 AIME Problems/Problem 10|Next Problem]]
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{{AIME box|year=1989|num-b=8|num-a=10}}
* [[1989 AIME Problems/Problem 8|Previous Problem]]
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* [[1989 AIME Problems]]
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[[Category:Intermediate Number Theory Problems]]

Revision as of 21:17, 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 $133^5+110^5+84^5+27^5=n^{5}_{}$. Find the value of $n^{}_{}$.

Solution

By Fermat's Little Theorem, we know ${n^{5}}$ is congruent to $n$ modulo 5. Hence,

$3 + 0 + 4 + 7 \equiv n\pmod{5}$
$4 \equiv n\pmod{5}$

Continuing, we examine the equation modulo 3,

$-1 + 1 + 0 + 0 \equiv n\pmod{3}$
$0 \equiv n\pmod{3}$

Thus, $n$ is divisible by three and leaves a remainder of four when divided by 5. It's obvious that $n>133$, so the only possibilities are $n = 144$ or $n = 174$. It quickly becomes apparent that 174 is much too large, so $n$ must be $144$.

See also

1989 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
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All AIME Problems and Solutions
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