Question
Find the smallest number which, when divided by 7, 9 and
12 leaves a remainder of 5 in each case.Solution
Let the required number be N. Then N − 5 is divisible by 7, 9, and 12. So N − 5 = LCM(7, 9, 12) × k Find LCM(7, 9, 12): 7 is prime; 9 = 3²; 12 = 2² × 3 LCM = 2² × 3² × 7 = 4 × 9 × 7 = 252 For smallest positive such N, take k = 1: N − 5 = 252 N = 252 + 5 = 257 Answer: 257.
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