Difference between revisions of "1989 AIME Problems/Problem 9"
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== Problem == | == 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 < | + | One of Euler's conjectures was disproved in the 1960s by three American mathematicians when they showed there was a positive integer such that <cmath>133^5+110^5+84^5+27^5=n^{5}.</cmath> Find the value of <math>n</math>. |
− | == Solution 1 == | + | == Solution 1 (FLT, CRT, Inequalities) == |
− | + | Taking the given equation modulo <math>2,3,</math> and <math>5,</math> respectively, we have | |
− | < | + | <cmath>\begin{align*} |
− | < | + | n^5&\equiv0\pmod{2}, \\ |
+ | n^5&\equiv0\pmod{3}, \\ | ||
+ | n^5&\equiv4\pmod{5}. | ||
+ | \end{align*}</cmath> | ||
+ | By either <b>Fermat's Little Theorem (FLT)</b> or inspection, we get | ||
+ | <cmath>\begin{align*} | ||
+ | n&\equiv0\pmod{2}, \\ | ||
+ | n&\equiv0\pmod{3}, \\ | ||
+ | n&\equiv4\pmod{5}. | ||
+ | \end{align*}</cmath> | ||
+ | By either the <b>Chinese Remainder Theorem (CRT)</b> or inspection, we get <math>n\equiv24\pmod{30}.</math> | ||
− | + | It is clear that <math>n>133,</math> so the possible values for <math>n</math> are <math>144,174,204,\ldots.</math> Note that | |
− | < | + | <cmath>\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*}</cmath> | ||
+ | from which <math>\left(\frac{n}{133}\right)^5<3.</math> | ||
− | + | If <math>n\geq174,</math> then | |
+ | <cmath>\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*}</cmath> | ||
+ | which arrives at a contradiction. Therefore, we conclude that <math>n=\boxed{144}.</math> | ||
+ | |||
+ | ~MRENTHUSIASM | ||
== Solution 2 == | == 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> | + | Note that <math>n</math> is even, since the LHS consists of two odd and two even numbers. By Fermat's Little Theorem, we know <math>n^5\equiv n\pmod{5}.</math> Hence, <cmath>n\equiv3+0+4+2\equiv4\pmod{5}.</cmath> |
+ | Continuing, we examine the equation modulo <math>3,</math> <cmath>n\equiv1-1+0+0\equiv0\pmod{3}.</cmath> | ||
+ | Thus, <math>n</math> is divisible by three and leaves a remainder of four when divided by <math>5.</math> It is 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 <math>174</math> is much too large, so <math>n</math> must be <math>\boxed{144}.</math> | ||
+ | |||
+ | ~Azjps (Solution) | ||
+ | |||
+ | ~MRENTHUSIASM (Reformatting) | ||
+ | |||
+ | == Solution 3 == | ||
+ | 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 <math>4.</math> | ||
+ | |||
+ | We notice that <math>133,110,84,</math> and <math>27</math> are all very close or equal to multiples of <math>27.</math> 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^5}{27^5}</math> must be close to <math>4393.</math> | ||
+ | |||
+ | Note that <math>134</math> will obviously be too small, so we try <math>144</math> and get <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 <math>154</math> 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. | ||
+ | |||
+ | ==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 <math>133^5, 110^5, 84^5, 27^5</math> are <math>3, 0, 4, 7,</math> respectively, so 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> | ||
− | We | + | 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> We can do this by finding the last three digits for each of <math>133^5, 110^5, 84^5,</math> and <math>27^5,</math> which is not that difficult. For <math>133^5,</math> we have |
+ | <cmath>\begin{align*} | ||
+ | 133^5&=133^2\cdot133^2\cdot133 \\ | ||
+ | &\equiv689\cdot689\cdot133 \\ | ||
+ | &\equiv721\cdot133 \\ | ||
+ | &\equiv893\pmod{1000}. | ||
+ | \end{align*}</cmath> | ||
+ | By the same 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> | ||
− | + | ~jackshi2006 (Solution) | |
+ | ~MRENTHUSIASM (Revisions and <math>\LaTeX</math> Adjustments) | ||
− | ==Solution | + | ==Solution 5== |
− | |||
− | <math>n</math> | + | First, we take mod <math>2</math> on both sides to get <math>n^5\equiv 0\pmod{2}\implies n\equiv 0\pmod{2}</math>. Mod <math>3</math> gives <math>n^5\equiv 0\pmod{3}\implies n\equiv 0\pmod{3}</math>. Also, mod <math>5</math> gives <math>n^5\equiv -1\pmod{5}\implies n\equiv -1\pmod{5}</math> (by FLT). Finally, note that mod <math>7</math> gives <math>n^5\equiv 2\pmod{7}\implies n^{-1}\equiv 2\pmod{7}\implies n\equiv 4\pmod{7}</math>. Thus, |
+ | <cmath>\begin{align*} | ||
+ | n&\equiv 0\pmod{2}, \\ | ||
+ | n&\equiv 0\pmod{3}, \\ | ||
+ | n&\equiv -1\pmod{5}, \\ | ||
+ | n&\equiv 4\pmod{7}. | ||
+ | \end{align*}</cmath> | ||
+ | By CRT, <math>n\equiv 144\pmod{210}</math>, so <math>n</math> is one of <math>144, 354, ...</math>. However, <math>133^5 + 110^5 + 84^5 + 27^5 < 4\cdot 133^5 < (2\cdot 133)^5 < 354^5</math>, so <math>n < 354</math>. Thus, <math>n = \boxed{144}</math>. | ||
− | + | ~brainfertilzer | |
+ | ==Solution 6== | ||
− | + | We have <cmath>n^5 = 133^5 + 110^5 + 84^5 +27^5 = 61917364224,</cmath> for which <math>n = \sqrt [5]{61917364224} = \boxed{144}.</math> | |
== See also == | == See also == |
Latest revision as of 14:49, 11 August 2023
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 (FLT, CRT, Inequalities)
Taking the given equation modulo and respectively, we have By either Fermat's Little Theorem (FLT) or inspection, we get By either the Chinese Remainder Theorem (CRT) or inspection, we get
It is clear that so the possible values for are Note that from which
If then which arrives at a contradiction. Therefore, we conclude that
~MRENTHUSIASM
Solution 2
Note that is even, since the LHS consists of two odd and two even numbers. By Fermat's Little Theorem, we know Hence, Continuing, we examine the equation modulo Thus, is divisible by three and leaves a remainder of four when divided by It is obvious that so the only possibilities are or It quickly becomes apparent that is much too large, so must be
~Azjps (Solution)
~MRENTHUSIASM (Reformatting)
Solution 3
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
We notice that and are all very close or equal to multiples of We can rewrite as approximately equal to This means must be close to
Note that will obviously be too small, so we try and get Bashing through the division, we find that which is very close to It is clear that will not give any closer of an answer, given the rate that fifth powers grow, so we can safely assume that 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 are respectively, so the units digit of is This tells us is even. Since we are dealing with enormous numbers, should not be that far from Note that 's units digit is or When to the power of they each give and as the units digits. This further clues us that ends in
Clearly, so we start with Now we need a way of distinguishing between numbers with units digit We can do this by finding the last three digits for each of and which is not that difficult. For we have By the same reasoning, we get Note that By observations, is obviously an overestimate. So, the answer is
~jackshi2006 (Solution)
~MRENTHUSIASM (Revisions and Adjustments)
Solution 5
First, we take mod on both sides to get . Mod gives . Also, mod gives (by FLT). Finally, note that mod gives . Thus, By CRT, , so is one of . However, , so . Thus, .
~brainfertilzer
Solution 6
We have for which
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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