Multiple choice

Find the sum of all three-digit whole numbers less than 500 that leave a remainder of 2 when they are divided by 3.

  1. 4963

  2. 39767

  3. 49634

  4. 39770
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B Correct answer
Explanation

Numbers are of the form 3k + 2. Smallest 3-digit number: 101 (3*33+2). Largest < 500: 497 (3*165+2). This is an AP: 101, 104, ..., 497. Number of terms n = (497 - 101)/3 + 1 = 396/3 + 1 = 132 + 1 = 133. Sum = n/2 * (first + last) = 133/2 * (101 + 497) = 133/2 * 598 = 133 * 299 = 39767.

AI explanation

The sequence of three digit whole numbers less than 500 that leave a remainder of 2 when divided by 3 begins at 101 and ends at 497. This forms an arithmetic progression with 133 terms, a first term of 101, and a last term of 497. Using the arithmetic series sum formula, the sum equals the number of terms multiplied by the sum of the first and last terms divided by 2, which is 133 times 299. This calculates to 39767.

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