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

Revision as of 22:12, 30 November 2009

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 $144$ .

@@ Line 3: / Line 3: @@
 == Solution ==
-Note that <math>n</math> is even, since the <math>RHS</math> consists of two odd and two even numbers. By [[Fermat's Little Theorem]], we know <math>{n^{5}}</math> is congruent to <math>n</math> [[modulo]] 5.  Hence,
+Note that <math>n</math> is even, since the <math>LHS</math> consists of two odd and two even numbers. 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>

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

Revision as of 22:12, 30 November 2009

Problem

Solution

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