Multiple choice

The sum of all two-digit numbers which when divided by 4 leave 1 as remainder is

  1. 1012

  2. 1201

  3. 1212

  4. 1210
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D Correct answer
Explanation

Two-digit numbers leaving a remainder of 1 when divided by 4 follow the sequence 13, 17, 21, ..., 97. This is an arithmetic progression with first term a=13, last term l=97, and common difference d=4. The number of terms n is (97-13)/4 + 1 = 22. The sum is (n/2)(a+l) = (22/2)(13+97) = 11 * 110 = 1210.

AI explanation

The two-digit numbers leaving a remainder of 1 when divided by 4 form an arithmetic progression: 13, 17, 21, up to 97. The first term is 13, the last term is 97, and the common difference is 4. The number of terms is found by (97 minus 13) divided by 4, plus 1, which gives 22 terms. Using the sum formula for an arithmetic progression, the sum is 22 divided by 2, multiplied by the sum of the first and last terms (13 + 97), yielding 11 times 110 = 1210.

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