LCM and HCF


  1. The least number which when divided by 4, 6, 8, 12 and 16 leaves a remainder of 2 in each case is :









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    Here , Remainder ( r ) = 2
    As we know that When a number is divided by a, b , c or d leaving same remainder ‘r’ in each case then that number must be k + r where k is LCM of a, b , c and d.
    So , L.C.M. of 4, 6, 8, 12 and 16 = k = 48

    Correct Option: C

    Here , Remainder ( r ) = 2
    As we know that When a number is divided by a, b , c or d leaving same remainder ‘r’ in each case then that number must be k + r where k is LCM of a, b , c and d.
    So , L.C.M. of 4, 6, 8, 12 and 16 = k = 48
    ∴ Required number = k + r = 48 + 2 = 50


  1. The least number, which when divided by 12, 15, 20 or 54 leaves a remainder of 4 in each case, is :









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    Here , Remainder ( r ) = 4
    As we know that When a number is divided by a, b , c or d leaving same remainder ‘r’ in each case then that number must be k + r where k is LCM of a, b , c and d.
    So , LCM of 15, 12, 20, 54 ( k ) = 540

    Correct Option: D

    Here , Remainder ( r ) = 4
    As we know that When a number is divided by a, b , c or d leaving same remainder ‘r’ in each case then that number must be k + r where k is LCM of a, b , c and d.
    So , LCM of 15, 12, 20, 54 ( k ) = 540
    Hence , Then number = k + r = 540 + 4 = 544



  1. Find the greatest number of five digits which when divided by 3, 5, 8, 12 have 2 as remainder :









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    Here , Remainder ( r ) = 2
    As we know that When a number is divided by a, b , c or d leaving same remainder ‘r’ in each case then that number must be k + r where k is LCM of a, b , c and d.
    The greatest number of five digits is 99999.
    LCM of 3, 5, 8 and 12

    ∴ LCM = 2 × 2 × 3 × 5 × 2 = 120
    After dividing 99999 by 120, we get 39 as remainder
    99999 – 39 = 99960 = ( 833 × 120 )

    Correct Option: D

    Here , Remainder ( r ) = 2
    As we know that When a number is divided by a, b , c or d leaving same remainder ‘r’ in each case then that number must be k + r where k is LCM of a, b , c and d.
    The greatest number of five digits is 99999.
    LCM of 3, 5, 8 and 12

    ∴ LCM = 2 × 2 × 3 × 5 × 2 = 120
    After dividing 99999 by 120, we get 39 as remainder
    99999 – 39 = 99960 = ( 833 × 120 )
    99960 is the greatest five digit number divisible by the given divisors.
    In order to get 2 as remainder in each case we will simply add 2 to 99960.
    ∴ Greatest number = 99960 + 2 = 99962


  1. The least perfect square, which is divisible by each of 21, 36 and 66 is









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    LCM of 21, 36 and 66
    ∴ LCM = 3 × 2 × 7 × 6 × 11
    ⇒ LCM = 3 × 3 × 2 × 2 × 7 × 11
    ∴ Required number = 32 × 22 × 72 × 112

    Correct Option: C

    LCM of 21, 36 and 66
    ∴ LCM = 3 × 2 × 7 × 6 × 11
    ⇒ LCM = 3 × 3 × 2 × 2 × 7 × 11
    ∴ Required number = 32 × 22 × 72 × 112
    Required number = 213444



  1. Let the least number of six digits which when divided by 4, 6, 10, 15 leaves in each case same remainder 2 be N. The sum of digits in N is :









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    LCM of 4, 6, 10, 15 = 60
    Least number of 6 digits = 100000
    The least number of 6 digits which is exactly divisible by 60 = 100000 + (60 – 40) = 100020
    ∴ Required number (N) = 100020 + 2 = 100022

    Correct Option: B

    LCM of 4, 6, 10, 15 = 60
    Least number of 6 digits = 100000
    The least number of 6 digits which is exactly divisible by 60 = 100000 + (60 – 40) = 100020
    ∴ Required number (N) = 100020 + 2 = 100022
    Hence, the sum of digits = 1 + 0 + 0 + 0 + 2 + 2 = 5