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

Revision as of 14:52, 4 November 2012

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 $\boxed{144}$ .

@@ Line 8: / Line 8: @@
 Continuing, we examine the equation modulo 3,
-<center><math>-1 + 1 + 0 + 0 \equiv n\pmod{3}</math></center>
+<center><math>1 - 1 + 0 + 0 \equiv n\pmod{3}</math></center>
 <center><math>0 \equiv n\pmod{3}</math></center>

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

Revision as of 14:52, 4 November 2012

Problem

Solution

See also