Multiple choice

How many 3 digit natural numbers, when divided by 5, 6 or 8, leave a remainder of 2 in each case.

  1. 6

  2. 8

  3. 9

  4. 10
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B Correct answer
Explanation

LCM(5, 6, 8) = 120. Numbers are of form 120k + 2. For 3-digit numbers, k can be 1, 2, 3, 4, 5, 6, 7, 8. This gives 122, 242, 362, 482, 602, 722, 842, 962. Total count is 8.

AI explanation

A number that leaves a remainder of 2 when divided by 5, 6, and 8 must be of the form LCM(5, 6, 8) plus 2. The least common multiple of 5, 6, and 8 is 120, so the number must be 120n plus 2, where n is a natural number. We need a three-digit number, so we find the values of n such that 100 is less than or equal to 120n + 2 is less than or equal to 999. Solving this inequality gives n ranging from 1 to 8, resulting in the numbers 122, 242, 362, 482, 602, 722, 842, and 962. There are exactly 8 such three-digit numbers.

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