https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Shawn316&feedformat=atomAoPS Wiki - User contributions [en]2024-03-29T04:50:23ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=1989_AIME_Problems/Problem_9&diff=1042461989 AIME Problems/Problem 92019-03-11T00:46:04Z<p>Shawn316: /* Solution 1 */</p>
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<div>== Problem ==<br />
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>.<br />
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== Solution 1 ==<br />
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,<br />
<center><math>3 + 0 + 4 + 2 \equiv n\pmod{5}</math></center><br />
<center><math>4 \equiv n\pmod{5}</math></center><br />
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Continuing, we examine the equation modulo 3,<br />
<center><math>1 - 1 + 0 + 0 \equiv n\pmod{3}</math></center><br />
<center><math>0 \equiv n\pmod{3}</math></center><br />
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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>.<br />
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== Solution 2 ==<br />
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.<br />
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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>.<br />
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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.<br />
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== See also ==<br />
{{AIME box|year=1989|num-b=8|num-a=10}}<br />
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[[Category:Intermediate Number Theory Problems]]<br />
{{MAA Notice}}</div>Shawn316