AIME 1989 · 第 9 题

Solution 1 (FLT, CRT, Inequalities)

Taking the given equation modulo $2,3,$ and $5,$ respectively, we have

$$\begin{aligned} n^5&\equiv0\pmod{2}, \\ n^5&\equiv0\pmod{3}, \\ n^5&\equiv4\pmod{5}. \end{aligned}$$ By either Fermat's Little Theorem (FLT) or inspection, we get

$$\begin{aligned} n&\equiv0\pmod{2}, \\ n&\equiv0\pmod{3}, \\ n&\equiv4\pmod{5}. \end{aligned}$$ 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

$$\begin{aligned} n^5&=133^5+110^5+84^5+27^5 \\ &<133^5+110^5+(84+27)^5 \\ &=133^5+110^5+111^5 \\ &<3\cdot133^5, \end{aligned}$$ from which $\left(\frac{n}{133}\right)^5<3.$

If $n\geq174,$ then

$$\begin{aligned} \left(\frac{n}{133}\right)^5&>1.3^5 \\ &=1.3^2\cdot1.3^2\cdot1.3 \\ &>1.6\cdot1.6\cdot1.3 \\ &=2.56\cdot1.3 \\ &>2.5\cdot1.2 \\ &=3, \end{aligned}$$ 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

$$\begin{aligned} 133^5&=133^2\cdot133^2\cdot133 \\ &\equiv689\cdot689\cdot133 \\ &\equiv721\cdot133 \\ &\equiv893\pmod{1000}. \end{aligned}$$ By the same reasoning, we get

$$\begin{aligned} n^5&=133^5+110^5+84^5+27^5 \\ &\equiv893+0+424+907 \\ &\equiv224\pmod{1000}. \end{aligned}$$ Note that

$$\begin{aligned} 134^5&\equiv424\pmod{1000}, \\ 144^5&\equiv224\pmod{1000}, \\ 154^5&\equiv024\pmod{1000}, \\ 164^5&\equiv824\pmod{1000}, \\ 174^5&\equiv624\pmod{1000}, \\ 184^5&\equiv424\pmod{1000}, \\ 194^5&\equiv224\pmod{1000}. \end{aligned}$$ 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,

$$\begin{aligned} n&\equiv 0\pmod{2}, \\ n&\equiv 0\pmod{3}, \\ n&\equiv -1\pmod{5}, \\ n&\equiv 4\pmod{7}. \end{aligned}$$ 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 $n = \sqrt [5]{61917364224} = \boxed{144}.$

Solution 7 (Cheese)

By approximation, $110 \approx \frac{5}{6}\cdot133$, $84 \approx \frac{7}{11}\cdot133$, and $23 \approx \frac{1}{5}\cdot133$. Adding the fifth power of each fraction of $133$ gives a total of roughly $1.506\cdot133^5$. By testing, we know that the fifth root of this value is less than $1.1\cdot133=146.3$, as $1.1^5=1.611$, meaning we have reduced $n$ to $133\le{n}\le{147}$. The last digit of all four numbers, in order, are $3$, $0$, $4$, and $7$, which is gained through writing out the cycles of the last digit up to the fifth power. This means that the answer has a units digit of $4$, reducing $n$ down to either $134$ or $144$. Since the previous cheesing shows obviously that the value is closer to $146.3$ than it is to $133$, we know our final answer of $\boxed{144}$ is correct.

Cheese solution by juwushu.

Solution 8 (Guesswork or Bash)

We know that $a^5 \equiv a \pmod{10}$ for all a. You can prove this later. So we know the number ends with 4. Since 133 is the biggest number, while everything else is small, we can safely assume that it is $\boxed{144}$, because 5th powers vary wildly. It is not 134 because $110^5$ plays a role in the sum. We can also just take $130^5$ and the other numbers $110^5, 80^5,20^5$, add them up, and we realize that $140^5$ is close to this. Then we also get $\boxed{144}$

~Aarav22

Remark

We could also take this expression mod $7,$ which would give $n\equiv4\pmod{7}.$ Then the only reasonable number that works is $144,$ so we could safely assume this is the answer.

~grogg007