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Difference between revisions of "1989 AIME Problems/Problem 9"

(Prioritize the comprehensive solution. Let me know if anyone is unhappy about the order of the solutions.)
(→‎Solution 4: Made the solution easier to read (Translated words to math.))
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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 <math>3, 0, 4,</math> and <math>7,</math> respectively. This means the units digit of <math>n^5</math> is <math>4.</math> This tells us <math>n</math> is even. Since we are dealing with enormous numbers, <math>n</math> should not be that far from <math>133.</math> Note that <math>n</math>'s units digit is <math>0, 2, 4, 6,</math> or <math>8.</math> When to the power of <math>5,</math> they each give <math>0, 2, 4, 6,</math> and <math>8</math> as the units digits. This further clues us that <math>n</math> ends in <math>4.</math>  
 
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 <math>3, 0, 4,</math> and <math>7,</math> respectively. This means the units digit of <math>n^5</math> is <math>4.</math> This tells us <math>n</math> is even. Since we are dealing with enormous numbers, <math>n</math> should not be that far from <math>133.</math> Note that <math>n</math>'s units digit is <math>0, 2, 4, 6,</math> or <math>8.</math> When to the power of <math>5,</math> they each give <math>0, 2, 4, 6,</math> and <math>8</math> as the units digits. This further clues us that <math>n</math> ends in <math>4.</math>  
  
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Clearly, <math>n</math> is larger than <math>133,</math> so we start with <math>134.</math> Now we need a way of distinguishing between numbers with units digit <math>4.</math> This can be done by simply solving up to the hundreds digit of <math>133^5, 110^5, 84^5,</math> and <math>27^5,</math> which isn't that difficult. For <math>133,</math> all that has to be done is square it and take the last three digits, <math>689,</math> and raise them to the power of <math>2</math> again, <math>721,</math> then multiply this by <math>133.</math> This gives us <math>893.</math> Doing this for each tells us <math>n^5</math> ends in <math>224.</math> Testing <math>134</math> the same way we did with <math>133</math> gives us <math>424; \ 144</math> gives us <math>224; \ 154</math> gives us <math>024; \ 164</math> gives us <math>824; \ 174,</math> gives <math>624; \ 184</math> gives us <math>424,</math> and finally <math>194</math> also gives <math>224.</math>
+
Clearly, <math>n>133,</math> so we start with <math>134.</math> Now we need a way of distinguishing between numbers with units digit <math>4.</math> This can be done by simply solving up to the hundreds digit of <math>133^5, 110^5, 84^5,</math> and <math>27^5,</math> which isn't that difficult. For <math>133^5,</math> we have
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<cmath>\begin{align*}
 +
133^5&=133^2\cdot133^2\cdot133 \\
 +
&\equiv689\cdot689\cdot133 \\
 +
&\equiv721\cdot133 \\
 +
&\equiv893\pmod{1000}.
 +
\end{align*}</cmath>
 +
By the similar reasoning, we get
 +
<cmath>\begin{align*}
 +
n^5&=133^5+110^5+84^5+27^5 \\
 +
&\equiv893+0+424+907 \\
 +
&\equiv224\pmod{1000}.
 +
\end{align*}</cmath>
 +
Note that
 +
<cmath>\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*}</cmath>
 
By observations, <math>n=194</math> is obviously an overestimate. So, the answer is <math>n=\boxed{144}.</math>
 
By observations, <math>n=194</math> is obviously an overestimate. So, the answer is <math>n=\boxed{144}.</math>
  
 
-jackshi2006 (Solution)
 
-jackshi2006 (Solution)
  
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~MRENTHUSIASM (Minor Revisions and <math>\LaTeX</math> Adjustments)
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~MRENTHUSIASM (Revisions and <math>\LaTeX</math> Adjustments)
  
 
== See also ==
 
== See also ==

Revision as of 05:09, 9 August 2021

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 $144,174,204,\cdots.$

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*} or $\left(\frac{n}{133}\right)^5<3.$

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 $3,$ \[n\equiv1-1+0+0\equiv0\pmod{5}.\] 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 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 $\frac{n}{27}$ 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 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 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, $n>133,$ 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 $133^5,$ we have \begin{align*} 133^5&=133^2\cdot133^2\cdot133 \\ &\equiv689\cdot689\cdot133 \\ &\equiv721\cdot133 \\ &\equiv893\pmod{1000}. \end{align*} By the similar 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)

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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