divisibility....

Number Theory Level 3

$(1363)^{3} + (1247)^{3} - (2610)^{3}$ is divisible by

This problem is a part of the sets - 3's & 4's & " N " for number theory

Follw me for more questions in the future.

3,13 3,7 29,43,47 7,13,3

1 solution

Sakanksha Deo
Mar 10, 2015

There is an identity,

$a^{3} + b^{3} + c^{3} - 3abc = ( a + b + c ) ( a^{2} + b^{2} + c^{2} - ab - bc - ac )$

When (a+ b + c) = 0,

$a^{3} + b^{3} + c^{3} = 3abc$

Now notice,

(1363) + (1247) + (-2610) = 0

Therefore,

$(1363)^{3} + (1247)^{3} - (2610)^{3} = 3\times1363\times1247\times2610 = 3\times3\times3\times2\times5\times29\times29\times29\times43\times47$