Difference between revisions of "1989 AIME Problems/Problem 9"
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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 | + | 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 4. |
− | <math>n</math> is | + | 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</math>, <math>110^5</math>, <math>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> |
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+ | By observations, <math>n=194</math> is obviously an overestimate. So, the answer is <math>n=\boxed{144}.</math> | ||
-jackshi2006 | -jackshi2006 |
Revision as of 11:31, 8 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 Find the value of .
Solution 1
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's 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 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
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 3
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 and respectively. This means 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 4.
Clearly, is larger than so we start with Now we need a way of distinguishing between numbers with units digit This can be done by simply solving up to the hundreds digit of , , , and , which isn't that difficult. For all that has to be done is square it and take the last three digits, and raise them to the power of again, then multiply this by This gives us Doing this for each tells us ends in Testing the same way we did with gives us gives us gives us gives us gives gives us and finally also gives
By observations, is obviously an overestimate. So, the answer is
-jackshi2006
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.