1989 AIME Problems/Problem 9

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 (FLT, CRT, Inequalities)

Taking the given equation modulo $2,3,$ and $5,$ respectively, we have $\begin{align*} n^5&\equiv0\pmod{2}, \\ n^5&\equiv0\pmod{3}, \\ n^5&\equiv4\pmod{5}. \end{align*}$ By either Fermat's Little Theorem (FLT) or inspection, we get $\begin{align*} n&\equiv0\pmod{2}, \\ n&\equiv0\pmod{3}, \\ n&\equiv4\pmod{5}. \end{align*}$ By either the Chinese Remainder Theorem (CRT) or inspection, we get $n\equiv24\pmod{30}.$

It is clear that $n>133,$ so the possible values for $n$ are $144,174,204,\ldots.$ Note that $\begin{align*} n^5&=133^5+110^5+84^5+27^5 \\ &<133^5+110^5+(84+27)^5 \\ &=133^5+110^5+111^5 \\ &<3\cdot133^5, \end{align*}$ from which $\left(\frac{n}{133}\right)^5<3.$

If $n\geq174,$ then $\begin{align*} \left(\frac{n}{133}\right)^5&>1.3^5 \\ &=1.3^2\cdot1.3^2\cdot1.3 \\ &>1.6\cdot1.6\cdot1.3 \\ &=2.56\cdot1.3 \\ &>2.5\cdot1.2 \\ &=3, \end{align*}$ which arrives at a contradiction. Therefore, we conclude that $n=\boxed{144}.$

~MRENTHUSIASM

Solution 2

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 $3,$ $n\equiv1-1+0+0\equiv0\pmod{3}.$ Thus, $n$ is divisible by three and leaves a remainder of four when divided by $5.$ It is 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 3

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 $\frac{n^5}{27^5}$ 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 4

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 $133^5, 110^5, 84^5, 27^5$ are $3, 0, 4, 7,$ respectively, so 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, $n>133,$ so we start with $134.$ Now we need a way of distinguishing between numbers with units digit $4.$ We can do this by finding the last three digits for each of $133^5, 110^5, 84^5,$ and $27^5,$ which is not that difficult. For $133^5,$ we have $\begin{align*} 133^5&=133^2\cdot133^2\cdot133 \\ &\equiv689\cdot689\cdot133 \\ &\equiv721\cdot133 \\ &\equiv893\pmod{1000}. \end{align*}$ By the same reasoning, we get $\begin{align*} n^5&=133^5+110^5+84^5+27^5 \\ &\equiv893+0+424+907 \\ &\equiv224\pmod{1000}. \end{align*}$ Note that $\begin{align*} 134^5&\equiv424\pmod{1000}, \\ 144^5&\equiv224\pmod{1000}, \\ 154^5&\equiv024\pmod{1000}, \\ 164^5&\equiv824\pmod{1000}, \\ 174^5&\equiv624\pmod{1000}, \\ 184^5&\equiv424\pmod{1000}, \\ 194^5&\equiv224\pmod{1000}. \end{align*}$ By observations, $n=194$ is obviously an overestimate. So, the answer is $n=\boxed{144}.$

~jackshi2006 (Solution)

~MRENTHUSIASM (Revisions and $\LaTeX$ Adjustments)

Solution 5

First, we take mod $2$ on both sides to get $n^5\equiv 0\pmod{2}\implies n\equiv 0\pmod{2}$ . Mod $3$ gives $n^5\equiv 0\pmod{3}\implies n\equiv 0\pmod{3}$ . Also, mod $5$ gives $n^5\equiv -1\pmod{5}\implies n\equiv -1\pmod{5}$ (by FLT). Finally, note that mod $7$ gives $n^5\equiv 2\pmod{7}\implies n^{-1}\equiv 2\pmod{7}\implies n\equiv 4\pmod{7}$ . Thus, $\begin{align*} n&\equiv 0\pmod{2}, \\ n&\equiv 0\pmod{3}, \\ n&\equiv -1\pmod{5}, \\ n&\equiv 4\pmod{7}. \end{align*}$ By CRT, $n\equiv 144\pmod{210}$ , so $n$ is one of $144, 354, ...$ . However, $133^5 + 110^5 + 84^5 + 27^5 < 4\cdot 133^5 < (2\cdot 133)^5 < 354^5$ , so $n < 354$ . Thus, $n = \boxed{144}$ .

~brainfertilzer

Solution 6 (Brute Force)

We have $n^5 = 133^5 + 110^5 + 84^5 +27^5 = 61917364224,$ for which $n = \sqrt [5]{61917364224} = \boxed{144}.$

1989 AIME (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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