LCM and HCF

LCM and HCF

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








  1. View Hint View Answer Discuss in Forum

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








  1. View Hint View Answer Discuss in Forum

    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 H.C.F. and L.C.M. of two numbers are 44 and 264 respectively. If the first number is divided by 2, the quotient is 44. The other number is








  1. View Hint View Answer Discuss in Forum

    Given , The H.C.F. and L.C.M. of two numbers are 44 and 264 .
    According to question ,
    First number = 2 × 44 = 88
    As we know that ,
    ∴ First number × Second number = H.C.F. × L.C.M.
    ⇒ 88 × Second number = 44 × 264

    Correct Option: C

    Given , The H.C.F. and L.C.M. of two numbers are 44 and 264 .
    According to question ,
    First number = 2 × 44 = 88
    As we know that ,
    ∴ First number × Second number = H.C.F. × L.C.M.
    ⇒ 88 × Second number = 44 × 264

    ⇒ Second number =
    44 × 264
    		 = 132
    88

  1. The L.C.M. of three different numbers is 120. Which of the following cannot be their H.C.F.?








  1. View Hint View Answer Discuss in Forum

    Given that , The L.C.M. of three different numbers is 120.
    LCM = 2 × 2 × 2 × 3 × 5
    Hence, HCF = 4, 8, 12 or 24

    Correct Option: D

    LCM = 2 × 2 × 2 × 3 × 5
    Hence, HCF = 4, 8, 12 or 24
    According to question ,
    35 cannot be H.C.F. of 120.
    Hence , required answer is 35.

  1. The HCF of two numbers is 23 and the other two factors of their LCM are 13 and 14. The larger of the two numbers is :








  1. View Hint View Answer Discuss in Forum

    Let the numbers be 23p and 23q .where p and q are co-prime.
    ∴ LCM = 23pq
    According to question ,
    23pq = 23 × 13 × 14
    ∴ p = 13, q = 14

    Correct Option: E

    Let the numbers be 23p and 23q .where p and q are co-prime.
    ∴ LCM = 23pq
    According to question ,
    23pq = 23 × 13 × 14
    ∴ p = 13, q = 14
    ∴ The larger number = 23q = 23 × 14 = 322