What is the least number which when divided by 3, 15 & 18 leave a remainder 6 in each case & it is also divisible by 12?
Number which when divided by 3, 15 & 18 leaves a remainder 6 = LCM (3, 15 & 18)k+6 = 90k+6, Now it should be also divisible by 12. So (90k+6)/12 = (84k+6k+6)/12 , it means (6k+6) should be divisible by 12. So when k = 1 then it will be 6×1+6= 12 which is divisible by 12. Hence final answer = 90k+6 = 90 × 1 + 6 = 96.
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