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Math Help - Real no.

  1. #1
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    Real no.

    ques-find the smallest no. which when divided by 28 gives remainder 8,and when divided by32 gives remainder 12.
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  2. #2
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    Hello, sumit2009!

    This can be solved with Modulo Arithmetic,
    . . but here is an algebraic solution.


    Find the smallest number which when divided by 28 gives remainder 8,
    and when divided by 32 gives remainder 12.
    Let N = the number.

    Then: . \begin{array}{cccc}N &=& 28a + 9 & {\color{blue}[1]} \\ N &=& 32b + 12 & {\color{blue}[2]} \end{array}\;\text{ for integers }a,b


    Equate [1] and [2]: . 28a + 8 \:=\:32b + 12 \quad\Rightarrow\quad 7a \:=\:8b + 1

    . . Then: . a \:=\:\frac{8b+1}{7} \quad\Rightarrow\quad a \:=\:b + \frac{b+1}{7}


    Since a is an integer, b+1 must be a multiple of 7.

    . . The first time this happens is: . b \:=\:6


    Substitute into [2]: . N \:=\:32(6) + 12 \quad\Rightarrow\quad\boxed{ N \:=\:204}

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