Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1989 AIME Problems/Problem 9"
(→‎Solution)
Line 2: Line 2:
 
One of Euler's conjectures was disproved in the 1960s by three American mathematicians when they showed there was a positive integer such that <math>133^5+110^5+84^5+27^5=n^{5}</math>. Find the value of <math>n</math>.
 
One of Euler's conjectures was disproved in the 1960s by three American mathematicians when they showed there was a positive integer such that <math>133^5+110^5+84^5+27^5=n^{5}</math>. Find the value of <math>n</math>.
  
−
== Solution ==
+
== Solution 1 ==
 
Note that <math>n</math> is even, since the <math>LHS</math> consists of two odd and two even numbers. By [[Fermat's Little Theorem]], we know <math>{n^{5}}</math> is congruent to <math>n</math> [[modulo]] 5.  Hence,
 
Note that <math>n</math> is even, since the <math>LHS</math> consists of two odd and two even numbers. By [[Fermat's Little Theorem]], we know <math>{n^{5}}</math> is congruent to <math>n</math> [[modulo]] 5.  Hence,
 
<center><math>3 + 0 + 4 + 7 \equiv n\pmod{5}</math></center>
 
<center><math>3 + 0 + 4 + 7 \equiv n\pmod{5}</math></center>
Line 12: Line 12:
  
 
Thus, <math>n</math> is divisible by three and leaves a remainder of four when divided by 5.  It's obvious that <math>n>133</math>, so the only possibilities are <math>n = 144</math> or <math>n \geq 174</math>.  It quickly becomes apparent that 174 is much too large, so <math>n</math> must be <math>\boxed{144}</math>.
 
Thus, <math>n</math> is divisible by three and leaves a remainder of four when divided by 5.  It's obvious that <math>n>133</math>, so the only possibilities are <math>n = 144</math> or <math>n \geq 174</math>.  It quickly becomes apparent that 174 is much too large, so <math>n</math> must be <math>\boxed{144}</math>.
 +
 +
 +
== Solution 2 ==
 +
We can cheat a little bit and approximate, since we are dealing with such large numbers. As above, <math>n^5\equiv n\pmod{5}</math>, and it is easy to see that <math>n^5\equiv n\pmod 2</math>. Therefore, <math>133^5+110^5+84^5+27^5\equiv 3+0+4+7\equiv 4\pmod{10}</math>, so the last digit of <math>n</math> is 4.
 +
 +
We notice that <math>133,110,84,</math> and <math>27</math> are all very close or equal to multiples of 27. We can rewrite <math>n^5</math> as approximately equal to <math>27^5(5^5+4^5+3^5+1^5) = 27^5(4393)</math>. This means <math>\frac{n}{27}</math> must be close to <math>4393</math>.
 +
 +
134 will obviously be too small, so we try 144. <math>\left(\frac{144}{27}\right)^5=\left(\frac{16}{3}\right)^5</math>. Bashing through the division, we find that <math>\frac{1048576}{243}\approx 4315</math>, which is very close to <math>4393</math>. 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 <math>\boxed{144}</math> is the answer.
  
 
== See also ==
 
== See also ==

Revision as of 21:12, 22 April 2017

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 ${n^{5}}$ is congruent to $n$ modulo 5. Hence,

$3 + 0 + 4 + 7 \equiv n\pmod{5}$
$4 \equiv n\pmod{5}$

Continuing, we examine the equation modulo 3,

$1 - 1 + 0 + 0 \equiv n\pmod{3}$
$0 \equiv n\pmod{3}$

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 $\boxed{144}$.


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 $4393$.

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.

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1989_AIME_Problems/Problem_9&oldid=85391"