1989 AIME Problems/Problem 9

Revision as of 20:46, 10 March 2019 by Shawn316 (talk | contribs) (Solution 1)


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 1

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,

$3 + 0 + 4 + 2 \equiv n\pmod{5}$
$4 \equiv n\pmod{5}$

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 $\boxed{144}$.

Solution 2

We can cheat a little bit and approximate, since we are dealing with such large numbers. As above, $n^5\equiv n\pmod{5}$, and it is easy to see that $n^5\equiv n\pmod 2$. Therefore, $133^5+110^5+84^5+27^5\equiv 3+0+4+7\equiv 4\pmod{10}$, so the last digit of $n$ is 4.

We notice that $133,110,84,$ and $27$ are all very close or equal to multiples of 27. We can rewrite $n^5$ as approximately equal to $27^5(5^5+4^5+3^5+1^5) = 27^5(4393)$. This means $\frac{n}{27}$ must be close to $4393$.

134 will obviously be too small, so we try 144. $\left(\frac{144}{27}\right)^5=\left(\frac{16}{3}\right)^5$. Bashing through the division, we find that $\frac{1048576}{243}\approx 4315$, which is very close to $4393$. It is clear that 154 will not give any closer of an answer, given the rate that fifth powers grow, so we can safely assume that $\boxed{144}$ is the answer.

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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