Good luck expanding!
Find the absolute value of the sum of all integral x and y for which the equation
x 3 + ( x + 1 ) 3 + ( x + 2 ) 3 + ( x + 3 ) 3 + ⋯ + ( x + 7 ) 3 = y 3
is satisfied. As an explicit example, if the solutions were ( 3 , 7 ) and ( 9 , 9 ) , you would be asked to find 3 + 7 + 9 + 9 .
The answer is 14.
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3 solutions
Yes! A legit solution! Thanks. :D
awesome observation !!
This can also be written as 4 ( x + 7 ) 2 ( x + 8 ) 2 − 4 ( x − 1 ) 2 x 2 = y 3 = > 8 x 3 + 8 4 x 2 + 4 2 0 x + 7 8 4 = y 3 Possible integral values of (x,y) are ( − 5 , − 6 ) , ( − 4 , − 4 ) , ( − 3 , 4 ) , ( − 2 , 6 ) S u m = − 5 − 6 − 4 − 4 − 3 + 4 − 2 + 6 = − 1 4 ∣ S u m ∣ = 1 4
Could you explain some of your steps in getting to the solution? Thanks. :D
The expansion can also be written as
i = 1 ∑ i = x + 7 i 3 − i = 1 ∑ i = x − 1 i 3 = y 3
= > 8 x 3 + 8 4 x 2 + 4 2 0 x + 7 8 4 = y 3 Then checking for integers b/w -10 and +10 for x, I got (x,y) as above.
Note that this is not a solution, which will be left for the reader to find.
The solutions are ( − 2 , 6 ) , ( − 3 , 4 ) , ( − 4 , − 4 ) , and − 5 , − 6 ) . Thus the answer is ∣ − 2 + 6 − 3 + 4 − 4 − 4 − 5 − 6 ∣ = 1 4 .
@Daniel Liu try this! :D
Woah @Victor Loh did you actually get this or just guessed? Could you post a solution? Thanks. :D
Oops I guessed this :P I hope to see a solution some time later too :D
What is the need of taking modulus |-14| ? Sum can be both positive or negative ...
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Modulus? I don't understand. :P
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Modulus is another word for absolute value.
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@Michael Tang – Aha. To answer your question @Vinay Sipani it was just to keep the mood positive. :D
please can anyone post a proper solution to this problem.
@Calvin Lin you decreased the rating? Why? It's a perfectly hard problem! D:
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As a guideline, if I see an immediate approach, it's not "perfectly hard". The (possibly) hard version of the problem, would be if we had only 3 or 4 terms on the LHS.
Because you have 8 terms on the LHS, it is very easy to bound it with a perfect cube, in particular ( 2 x + 7 ) 3 . This then becomes a standard bounding exercise, and having a few small cases to check. Assuming that you did it correctly, I see that the cases are all very small and easy to obtain.
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Yes but look at the ratio--11 to 34 people. It seems pretty tough. :P
Yes that is the method, but it's easy to forget one or two cases.
Plus, nobody's gotten it right so far.
First note that 8 x 3 < y 3 < 8 ( x + 7 ) 3 , so 2 x < y < 2 x + 1 4 .
Write y = 2 x + a ; so 0 < a < 1 4 . Expand and simplify to get ( 8 4 − 1 2 a ) x 2 + ( 4 2 0 − 6 a 2 ) x + ( 7 8 4 − a 3 ) = 0 . Looking mod 6 , we get a 3 ≡ 4 mod 6 , so the only possibility is a ≡ 4 mod 6 . This leaves only two possibilities, a = 4 , 1 0 .
In the first case we get x = − 4 , − 5 and in the second case we get x = − 3 , − 2 . The solutions are ( − 4 , − 4 ) , ( − 5 , − 6 ) , ( − 3 , 4 ) , ( − 2 , 6 ) , and the sum of the coordinates is − 1 4 , so the answer is 1 4 .