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Find the smallest number which when divided by 12 and 16 leaves no remainder, but leaves a remainder of 5 in each case when 5 is added.
More simply: find the smallest number that, when 5 is subtracted, is divisible by both 12 and 16.

Class 6HCF and LCMPlaying with Numbers
  • A58
  • B53
  • C106
  • D48

Step-by-Step Solution

Number - 5 must be divisible by 12 and 16, so it equals LCM(12,16) at minimum. , LCM=. Smallest number = 48 + 5 = 53.
Answer: B) 53

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