Multiple choice

Find the sum of all the three digit numbers which leave remainder $2$ when divided by $5$.

  1. $86889$
  2. $78442$
  3. $85339$
  4. $98910$
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D Correct answer
Explanation

Three-digit numbers leaving remainder 2 when divided by 5 form an arithmetic progression: 102, 107, ..., 997. The number of terms n is found via 997 = 102 + (n-1)5, giving n = 180. The sum is (n/2)(first + last) = (180/2)(102 + 997) = 90 * 1099 = 98910.

AI explanation

The three-digit numbers leaving a remainder of 2 when divided by 5 form an arithmetic progression starting at 102 and ending at 997 with a common difference of 5. The number of terms is (997 minus 102) divided by 5, plus 1, which gives 180. Using the arithmetic series sum formula, the sum equals 180 divided by 2, multiplied by the sum of the first and last terms (102 plus 997). This calculates to 90 multiplied by 1099, giving a total sum of 98910.

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