Each of A, B and C is a positive integer, with A less than B, that satisfies this equation:
A3+B3=19C3
Determine all triplets (A,B,C) with A+B+C under 100,000
*** As an extra challenge only, solve this puzzle without using a computer program/spreadsheet aided method.
For all x in range:
(3x)^3+(5x)^3 = 19(2y)^3, so 152x^3 = 152y^3, since 152 = 19*2^3
(x)^3+(8x)^3 = 19*(3y)^3, so 513x^3 = 513y^3, since 513 = 19*3^3
(33x)^3+(92x)^3 = 19*(35y)^3, so 814625x^3 = 814625y^3, since 814625 = 19*5^3*7^3
(9613x)^3+(27323x)^3 = 19*(10386y)^3, so 21286220456664 x^3 = 21286220456664 y^3, since 21286220456664 = 19*2^3*3^6*577^3
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Posted by broll
on 2022-08-14 03:27:23 |
some thoughts
