Question

Find the smallest number which when diminished by 6 is divisible by 12 15 20 and 27

Answer 1

Answer:

546

Step-by-step explanation:

12 = 2 x 2 x 3

15 = 3 x 5

20 = 2 x 2x 5

27 = 3 x 3x3

LCM = 2x3 x 5x2x 5x 3=540

THE REQUIRED SMALLEST NUMBER = 540+6=546

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Answer 2

Answer:

The smallest number when diminished by 6 is divisible by 12,15, 20, and 27 = 546

Step-by-step explanation:

To find,

The smallest number when diminished by 6 is divisible by 12,15, 20 and 27

Solution:

The smallest number divisible by 12, 15, 20, and 27 = LCM of 12, 15, 20 and 27

The smallest number which when diminished by 6 is divisible by 12,15, 20 and 27 = LCM ( 12, 15, 20, 27) + 6

Prime factorization of 12 = 2×2×3

Prime factorization of 15 = 3×5

Prime factorization of 20 =2×2×5

Prime factorization of 27 = 3×3×3

The least common multiple of 12,15,20 and 27 = 2×2×3×3×3×5 = 540

LCM (12,15,20,27) = 540

The smallest number which when diminished by 6 is divisible by 12,15, 20 and 27 = LCM (12,15,20,27) + 6

= 540 +6

= 546

∴ The smallest number when diminished by 6 is divisible by 12,15, 20, and 27 = 546

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