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 $133^5+110^5+84^5+27^5=n^{5}$. Find the value of $n$.

Solution

Note that $n$ is even, since the $LHS$ consists of two odd and two even numbers. 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 $n \geq 174$. It quickly becomes apparent that 174 is much too large, so $n$ must be $\boxed{144}$.

See also

1989 AIME (ProblemsAnswer KeyResources)
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
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