find the smallest no which when divided by 28 and 32 leaves remainder 8 and 12 respectively

Question

find the smallest no which when divided by 28 and 32 leaves remainder 8 and 12 respectively

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Savannah 2 weeks 2021-09-13T22:35:08+00:00 1 Answer 0 views 0

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    2021-09-13T22:36:24+00:00

    Answer:

    Given that the smallest number when divided by 28 and 32 leaves remainder 8 and 12 respectively.

    28 – 8 = 20 and 32 – 12 = 20 are divisible by the required numbers.

    Therefore the required number will be 20 less than the LCM of 28 and 32.

    Prime factorization of 28 = 2 * 2 * 7

    Prime factorization of 32 = 2 * 2 * 2 * 2 * 2

    LCM(28,32) = 2 * 2 * 2 * 2 * 2 * 7

                        = 224.

    Therefore the required smallest number = 224 – 20

                                                                       = 204.

    Verification:

    204/28 = 28 * 7 = 196.

                = 204 – 196    

     

                = 8

    204/32 = 32 * 6 = 192

                = 204 – 192

                = 12.

    Hope this helps!

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