# remainder problem

• June 2nd 2008, 09:16 PM
cUe?
remainder problem
Wat will be the remainder when

21^3+22^3+23^3+24^3 is divided by 90..
• June 3rd 2008, 05:33 AM
Soroban
Hello, cUe?!

I assume we aren't allowed calculators . . .

Quote:

What is remainder when $
21^3+22^3+23^3+24^3$
is divided by 90 ?

$\text{We have: }\;N \;=\;\underbrace{(21^3 + 24^3)}_{\text{sum of cubes}} + \underbrace{(22^3 + 23^3)}_{\text{sum of cubes}}$

Factor: . $(21+24)(21^2 + 21\!\cdot\!24 + 24^2) + (22 +23)(22^2 + 22\!\cdot\!23 + 23^2)$

This is: . . . $(45)(\text{odd + even + even}) + (45)(\text{even + even + odd})$

. . . . . . . . . . . . $= \qquad\quad(45)(\text{odd}) + (45)(\text{odd})$

. . . . . . . . . . . . $= \qquad\qquad45(\text{odd + odd})$

. . . . . . . . . . . . $= \qquad\qquad\quad 45(\text{even})$

Therefore: $N$ is a multiple of 90 . . . The remainder is 0.