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Find the smallest number which when …

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Find the smallest number which when increased by 17 is exactly divisible by both 520 and 468

    • Posted by Aryan Raj 6 years, 8 months ago

    CBSE > Class 10 > Mathematics
    • 2 answers

    Yogita Ingle 6 years, 8 months ago

    Last line is
    Smallest number which when increased by 17 is exactly divisible by both 520 and 468 = 4680 - 17 = 4663.

    0Thank You

    Yogita Ingle 6 years, 8 months ago

    The given numbers are 520 and 468.

    The smallest number which when increased by 17 is exactly divisible by both 520 and 468 is obtained by subtracting 17 from the LCM of 520 and 468.

    Prime factorisation of 520 = 2 × 2 × 2 × 5 × 13

    Prime factorisation of 468 = 2 × 2 × 3 × 3 × 13

    LCM of  520 and 468 = 2 × 2 × 2 × 3 × 3 × 5 × 13 = 4680.

    Smallest number which when increased by 17 is exactly divisible by both 520 and 468 = 4680 × 17 = 4663.

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