Sunday, 5 June 2016

Quants Question of the Day - Saturday, 4th June 2016

On 11:38:00 by Unknown in ,    1 comment
The remainder, when (17^3 + 19^3 + 21^3 + 23^3 ) divided by 80 is
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1 comment:

  1. UnknownThu Jun 09, 12:14:00 pm 2016

    Answer to the Question -

    (a^3 + b^3) = (a+b)(a^2- a*b + b^2)

    17^3 + 23^3 = 40 * (17^2 - 17*23 + 23^2)
    = 40 * (An Odd Number)

    Similarly 19^3 + 21^3 = 40 * (An Odd Number)

    Adding Both we get
    (17^3 + 19^3 + 21^3 + 23^3)= 40 * (An Odd Number + Another Odd Number)
    = 40 * An Even Number
    Hence it is divisible by 80 and the remainder will be zero.

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