What is the smallest positive number that leaves a remainder of 3 when divided by 5, by 6 and by 8?
Answer and explanation
Correct answer: A. 123
- Step 1: The number minus 3 must be a multiple of 5, 6 and 8, so it is a multiple of lcm(5, 6, 8) = 120.
- Step 2: The smallest positive number is 120 + 3 = 123.
Answer
123.
Why not the others: 63 and 27 fail one of the divisors, and 243 is 2 x 120 + 3, a valid but larger number.
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