1989 AIME Problems/Problem 9

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 or

If then which arrives at a contradiction. Therefore, we conclude that

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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'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 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 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

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 (Solution)

~MRENTHUSIASM (Minor Revisions and Adjustments)

See also

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