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1989 AIME Problems/Problem 9

Revision as of 07:49, 7 August 2021 by MRENTHUSIASM (talk | contribs) (→‎Solution 2: LaTeX the solution.)

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 1

Note that $n$ is even, since the LHS consists of two odd and two even numbers. By Fermat's Little Theorem, we know $n^5\equiv n\pmod{5}.$ Hence, \[n\equiv3+0+4+2\equiv4\pmod{5}.\] Continuing, we examine the equation modulo $3,$ \[n\equiv1-1+0+0\equiv0\pmod{5}.\] 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 $\boxed{144}.$

~Azjps (Solution)

~MRENTHUSIASM (Reformatting)

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 $4393$.

Note that $134$ 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.

Solution 3

In this solution we take advantage of the large numbers and utilize parity properties to give us a very good guess at the answer. The units digit of each of the four numbers is 3, 0, 4, and 7, respectively. This means the units digit of $n^5$ is 4. This tells us $n$ is even. Since we are dealing with enormous numbers, $n$ should not be that far from 133. $n$'s unit digit is 0, 2, 4, 6, or 8. When to the power of 5 they each give 0, 2, 4, 6, and 8 as unit digits. This further clues us that $n$ ends in 4.

$n$ is obviously larger than 133, so we start with 134. Now we need a way of distinguishing between numbers with unit digit 4. This can be done by simply solving up to the hundreds digit of $133^5$, $110^5$, $84^5$, and $27^5$, which isn't that difficult. For 133, all that has to be done is square it and take the last three digits, 689, and raise them to the power of 2 again, 721, then multiply this by 133. This gives us 893. Doing this for each tells us $n^5$ ends in 224. Testing 134 the same way we did with 133 gives us 424, 144 gives us 224, 154 gives us 024, 164 gives us 824, 624, 424, and finally 194 also gives 224.

By simply using our judgement it's hard to imagine the sum of those powers of 5 to be $194^5$. The answer is $\boxed{144}$.


-jackshi2006

See also

1989 AIME (Problems • Answer Key • Resources)
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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