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 20:12, 22 April 2017
Contents
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 . Find the value of .
Solution 1
Note that is even, since the consists of two odd and two even numbers. By Fermat's Little Theorem, we know is congruent to modulo 5. Hence,
Continuing, we examine the equation modulo 3,
Thus, is divisible by three and leaves a remainder of four when divided by 5. It's obvious that , so the only possibilities are or . It quickly becomes apparent that 174 is much too large, so must be .
Solution 2
We can cheat a little bit and approximate, since we are dealing with such large numbers. As above, , and it is easy to see that . Therefore, , so the last digit of is 4.
We notice that and are all very close or equal to multiples of 27. We can rewrite as approximately equal to . This means must be close to .
134 will obviously be too small, so we try 144. . Bashing through the division, we find that , which is very close to . 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 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.