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1989 AIME Problems/Problem 9

Revision as of 11:32, 8 August 2021 by MRENTHUSIASM (talk | contribs) (→‎Solution 3)

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 $144$ and get $\left(\frac{144}{27}\right)^5=\left(\frac{16}{3}\right)^5.$ 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 digits of the four numbers are $3, 0, 4,$ 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 $133.$ Note that $n$'s units digit is $0, 2, 4, 6,$ or $8.$ When to the power of $5,$ they each give $0, 2, 4, 6,$ and $8$ as the units digits. This further clues us that $n$ ends in $4.$

Clearly, $n$ is larger than $133,$ so we start with $134.$ Now we need a way of distinguishing between numbers with units 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 $824; \ 174,$ gives $624; \ 184$ gives us $424,$ and finally $194$ also gives $224.$

By observations, $n=194$ is obviously an overestimate. So, the answer is $n=\boxed{144}.$

-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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