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Find the LCM of the given numbers.

Mathematics

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16 steps

1. Find the LCM of the given numbers.

  • i) 12 and 30
  1. Find the prime factorization of each number.

    $$
    12 = 2^2 \times 3
    $$
    $$
    30 = 2 \times 3 \times 5
    $$
    
  2. Multiply the highest powers of all prime factors.

    $$
    \text{LCM}(12, 30) = 2^2 \times 3 \times 5 = 4 \times 3 \times 5 = \boxed{\text{60}}
    $$
    
    • ii) 16 and 50
  3. Find the prime factorization of each number.

    $$
    16 = 2^4
    $$
    $$
    50 = 2 \times 5^2
    $$
    
  4. Multiply the highest powers of all prime factors.

    $$
    \text{LCM}(16, 50) = 2^4 \times 5^2 = 16 \times 25 = \boxed{\text{400}}
    $$
    
    • iii) 18, 20 and 36
  5. Find the prime factorization of each number.

    $$
    18 = 2 \times 3^2
    $$
    $$
    20 = 2^2 \times 5
    $$
    $$
    36 = 2^2 \times 3^2
    $$
    
  6. Multiply the highest powers of all prime factors.

    $$
    \text{LCM}(18, 20, 36) = 2^2 \times 3^2 \times 5 = 4 \times 9 \times 5 = \boxed{\text{180}}
    $$
    
    • iv) 55, 70 and 77
  7. Find the prime factorization of each number.

    $$
    55 = 5 \times 11
    $$
    $$
    70 = 2 \times 5 \times 7
    $$
    $$
    77 = 7 \times 11
    $$
    
  8. Multiply the highest powers of all prime factors.

    $$
    \text{LCM}(55, 70, 77) = 2 \times 5 \times 7 \times 11 = \boxed{\text{770}}
    $$
    
    • v) 45, 60, 90 and 120
  9. Find the prime factorization of each number.

    $$
    45 = 3^2 \times 5
    $$
    $$
    60 = 2^2 \times 3 \times 5
    $$
    $$
    90 = 2 \times 3^2 \times 5
    $$
    $$
    120 = 2^3 \times 3 \times 5
    $$
    
  10. Multiply the highest powers of all prime factors.

    $$
    \text{LCM}(45, 60, 90, 120) = 2^3 \times 3^2 \times 5 = 8 \times 9 \times 5 = \boxed{\text{360}}
    $$
    

    2. Find the smallest number which on adding 11 becomes exactly divisible by 24, 36 and 60.

  11. Let the required number be NN.

    If N+11N+11 is exactly divisible by 24, 36, and 60, then N+11N+11 must be the LCM of these numbers.

  12. Find the LCM of 24, 36, and 60.

    $$
    24 = 2^3 \times 3
    $$
    $$
    36 = 2^2 \times 3^2
    $$
    $$
    60 = 2^2 \times 3 \times 5
    $$
    $$
    \text{LCM}(24, 36, 60) = 2^3 \times 3^2 \times 5 = 8 \times 9 \times 5 = 360
    $$
    
  13. Set up the equation and solve for NN.

    $$
    N + 11 = 360
    $$
    $$
    N = 360 - 11
    $$
    $$
    N = \boxed{\text{349}}
    $$
    

    3. Find the smallest number which leaves a remainder 7 when divided by 20, 50 and 75.

  14. Let the required number be NN.

    If NN leaves a remainder 7 when divided by 20, 50, and 75, then N−7N-7 must be exactly divisible by 20, 50, and 75.

  15. Find the LCM of 20, 50, and 75.

    $$
    20 = 2^2 \times 5
    $$
    $$
    50 = 2 \times 5^2
    $$
    $$
    75 = 3 \times 5^2
    $$
    $$
    \text{LCM}(20, 50, 75) = 2^2 \times 3 \times 5^2 = 4 \times 3 \times 25 = 300
    $$
    
  16. Set up the equation and solve for NN.

    $$
    N - 7 = 300
    $$
    $$
    N = 300 + 7
    $$
    $$
    N = \boxed{\text{307}}
    $$
    
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