Difference between revisions of "1989 AIME Problems/Problem 9"
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==Solution 3== | ==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 digit of each of the four numbers is 3, 0, 4, and 7, respectively. This means the units digit of <math>n^5</math> is 4. This tells us <math>n</math> is even. Since we are dealing with enormous numbers, <math>n</math> should not be that far from 133. <math>n</math>'s unit digit is 0, 2, 4, 6, or 8. When to the power of 5 they each give 0, 2, 4, 6, and 8 as unit digits. This further clues us that <math>n</math> ends in 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 digit of each of the four numbers is 3, 0, 4, and 7, respectively. This means the units digit of <math>n^5</math> is 4. This tells us <math>n</math> is even. Since we are dealing with enormous numbers, <math>n</math> should not be that far from 133. <math>n</math>'s unit digit is 0, 2, 4, 6, or 8. When to the power of 5 they each give 0, 2, 4, 6, and 8 as unit digits. This further clues us that <math>n</math> ends in 4. | ||
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<math>n</math> is obviously larger than 133, so we start with 134. Now we need a way of distinguishing between numbers with unit digit 4. 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 133, all that has to be done is square it and take the last three digits, 689, and raise them to the power of 2 again, 721, then multiply this by 133. This gives us 893. Doing this for each tells us <math>n^5</math> ends in 224. Testing 134 the same way we did with 133 gives us 424, 144 gives us 224, 154 gives us 024, 164 gives us 824, 624, 424, and finally 194 also gives 224. | <math>n</math> is obviously larger than 133, so we start with 134. Now we need a way of distinguishing between numbers with unit digit 4. 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 133, all that has to be done is square it and take the last three digits, 689, and raise them to the power of 2 again, 721, then multiply this by 133. This gives us 893. Doing this for each tells us <math>n^5</math> ends in 224. Testing 134 the same way we did with 133 gives us 424, 144 gives us 224, 154 gives us 024, 164 gives us 824, 624, 424, and finally 194 also gives 224. | ||
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By simply using our judgement it's hard to imagine the sum of those powers of 5 to be <math>194^5</math>. The answer is <math>\boxed{144}</math>. | By simply using our judgement it's hard to imagine the sum of those powers of 5 to be <math>194^5</math>. The answer is <math>\boxed{144}</math>. | ||
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== See also == | == See also == |
Revision as of 18:41, 3 September 2020
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.
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 digit of each of the four numbers is 3, 0, 4, and 7, respectively. This means the units digit of is 4. This tells us is even. Since we are dealing with enormous numbers, should not be that far from 133. 's unit digit is 0, 2, 4, 6, or 8. When to the power of 5 they each give 0, 2, 4, 6, and 8 as unit digits. This further clues us that ends in 4.
is obviously larger than 133, so we start with 134. Now we need a way of distinguishing between numbers with unit digit 4. This can be done by simply solving up to the hundreds digit of , , , and , which isn't that difficult. For 133, all that has to be done is square it and take the last three digits, 689, and raise them to the power of 2 again, 721, then multiply this by 133. This gives us 893. Doing this for each tells us ends in 224. Testing 134 the same way we did with 133 gives us 424, 144 gives us 224, 154 gives us 024, 164 gives us 824, 624, 424, and finally 194 also gives 224.
By simply using our judgement it's hard to imagine the sum of those powers of 5 to be . 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 |
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