Multiple choice

What is the sum of all 3 digit numbers that leave a remainder of '2' when divided by 3?

  1. 145630

  2. 160320

  3. 164850

  4. 150230
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C Correct answer
Explanation

The numbers are 101, 104, ..., 998. This is an arithmetic progression with a = 101, d = 3, and last term l = 998. 998 = 101 + (n-1)3, so 897 = (n-1)3, n-1 = 299, n = 300. Sum = (n/2)(a + l) = (300/2)(101 + 998) = 150 * 1099 = 164850.

AI explanation

The sequence begins with the smallest three-digit number leaving a remainder of 2 when divided by 3, which is 101, and ends with the largest, which is 998. This forms an arithmetic progression with n = 300 terms, a first term of 101, and a last term of 998. Using the sum formula n(a + l)/2, the sum is 300 x (101 + 998)/2. This yields 300 x 1099/2, resulting in a total sum of 164850.

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