LCM and HCF


  1. A number p is divisible by 7. When this number is divided by 8, 12 and 16. It leaves a remainder 3 in each case. The least value of p is:









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    We get LCM of 8, 12 and 16 = 48
    ∴ Required number = 48a + 3 which is divisible by 7.
    ∴ p = 48a + 3 = (7 × 6a) + (6a + 3) which is
    divisible by 7.
    i.e. 6a + 3 is divisible by 7.
    When a = 3, 6a + 3 = 18 + 3 = 21 , which is divisible by 7.

    Correct Option: D

    We get LCM of 8, 12 and 16 = 48
    ∴ Required number = 48a + 3 which is divisible by 7.
    ∴ p = 48a + 3 = (7 × 6a) + (6a + 3) which is
    divisible by 7.
    i.e. 6a + 3 is divisible by 7.
    When a = 3, 6a + 3 = 18 + 3 = 21 , which is divisible by 7.
    ∴ p = 48 × 3 + 3 = 144 + 3 = 147
    Hence , The least value of p is 147 .


  1. The greatest four digit number which is exactly divisible by each one of the numbers 12, 18, 21 and 28 is









  1. View Hint View Answer Discuss in Forum

    First of all , we find the LCM of 12 , 18 , 21 and 28

    ∴ LCM = 2 × 2 × 3 × 3 × 7 = 252
    The largest 4-digit number = 9999

    Here , Remainder = 171
    ∴ Required number = The largest 4-digit number - Remainder

    Correct Option: A

    First of all , we find the LCM of 12 , 18 , 21 and 28

    ∴ LCM = 2 × 2 × 3 × 3 × 7 = 252
    The largest 4-digit number = 9999

    Here , Remainder = 171
    ∴ Required number = The largest 4-digit number - Remainder
    Hence , Required number = 9999 – 171 = 9828



  1. Let p be the least number, which when divided by 5, 6, 7 and 8 leaves a remainder 3 in each case but when divided by 9 leaves no remainder. The sum of digits of p is









  1. View Hint View Answer Discuss in Forum

    We find LCM of 5, 6, 7 and 8 = 840

    ∴ LCM = 2 × 5 × 3 × 7 × 4 = 840
    ∴ Required number = 840p + 3 which is divisible by 9 for a certain least value of p.
    Now, 840p + 3 = 93p × 9 + 3p + 3
    ∴ 3p + 3 , is divisible by 9 for p = 2

    Correct Option: C

    We find LCM of 5, 6, 7 and 8 = 840

    ∴ LCM = 2 × 5 × 3 × 7 × 4 = 840
    ∴ Required number = 840p + 3 which is divisible by 9 for a certain least value of p.
    Now, 840p + 3 = 93p × 9 + 3p + 3
    3p + 3 , is divisible by 9 for p = 2
    ∴ Required number = 840 × 2 + 3 = 1680 + 3 = 1683
    ∴ Sum of digits = 1 + 6 + 8 + 3 = 18


  1. The number between 3000 and 4000 which is exactly divisible by 30, 36 and 80 is









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    Firstly, we find the LCM of 30, 36 and 80.

    ∴ LCM = 2 × 2 × 3 × 5 × 3 × 4 = 720

    Correct Option: D

    Firstly, we find the LCM of 30, 36 and 80.

    ∴ LCM = 2 × 2 × 3 × 5 × 3 × 4 = 720
    ∴ Required number = Multiple of 720 = 720 × 5 = 3600 ; because 3000 < 3600 < 4000



  1. HCF and LCM of two numbers are 7 and 140 respectively. If the numbers are between 20 and 45, the sum of the numbers is :









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    Let the numbers be 7p and 7q where p and q are co-prime.
    Now, LCM of 7p and 7q = 7pq
    ∴ 7pq = 140

    ⇒ pq =
    140
    = 20
    7

    Now, required values of p and q whose product is 20 and are co-prime, will be 4 and 5.

    Correct Option: C

    Let the numbers be 7p and 7q where p and q are co-prime.
    Now, LCM of 7p and 7q = 7pq
    ∴ 7pq = 140

    ⇒ pq =
    140
    = 20
    7

    Now, required values of p and q whose product is 20 and are co-prime, will be 4 and 5.
    ∴ Numbers are 28 and 35 which lie between 20 and 45.
    ∴ Required sum = 28 + 35 = 63.