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Math Help - remainder problem

  1. #1
    Newbie cUe?'s Avatar
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    remainder problem

    Wat will be the remainder when

    21^3+22^3+23^3+24^3 is divided by 90..
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  2. #2
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    Hello, cUe?!

    I assume we aren't allowed calculators . . .


    What is remainder when <br />
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.

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