What is the smallest number which when divided by 12, 15, 20, and 54 leaves a remainder of 8 in each case?
Answer and explanation
Correct answer: C. 548
- Step 1: LCM of divisors: LCM(12, 15, 20, 54).
- Step 2: Prime factorizations: 12 = 2^2 3, 15 = 3 5, 20 = 2^2 5, 54 = 2 3^3.
- Step 3: LCM = 2^2 3^3 5 = 4 27 5 = 540.
- Step 4: Smallest number = LCM + remainder = 540 + 8 = 548.
Practise more HCF and LCM.
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