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a^3+b^3

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by joyseychow » Fri Jun 05, 2009 1:46 am
My guess is A?
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by anksgupta » Fri Jun 05, 2009 6:55 am
I hope there is an easy way out for this solution....

Burnt my head on this, still no clues... Finally got this solution from net...

If a is an integer, then a^3 gives remainder 0, 1, or 6 when divided by 7 (just consider the
remainders of 0^3, 1^3, 2^3, 3^3, 4^3, 5^3, and 6^3 when divided by 7). Then a^3 + b^3 gives remainder 0, 1, 2, 5, or 6 when divided by 7. Well, 7077283 gives remainder 3 when divided by 7. (The
remaining numbers can be represented as a sum of two cubes of integers, 700056 = 403 + 863,
707713 = 143 + 893, 7000639 = 323 + 1913, and 7077915 = 33 + 1923.)

:(
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by joyseychow » Sat Jun 13, 2009 3:26 am
anksgupta wrote:I hope there is an easy way out for this solution....

Burnt my head on this, still no clues... Finally got this solution from net...

If a is an integer, then a^3 gives remainder 0, 1, or 6 when divided by 7 (just consider the
remainders of 0^3, 1^3, 2^3, 3^3, 4^3, 5^3, and 6^3 when divided by 7). Then a^3 + b^3 gives remainder 0, 1, 2, 5, or 6 when divided by 7. Well, 7077283 gives remainder 3 when divided by 7. (The
remaining numbers can be represented as a sum of two cubes of integers, 700056 = 403 + 863,
707713 = 143 + 893, 7000639 = 323 + 1913, and 7077915 = 33 + 1923.)

:(
Interesting approach. M curious. Is there any other approach than this? I tried using the last digit method, but I realize that my initial answer A cannot be the answer. Can someone bring some light to this?
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