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The sum of all two digit numbers which leave the remainder 1 when divided by 3 is:

a. 1616
b.1602
c.1605
d. None of these

1 Answer

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Final answer:

To find the sum of all two-digit numbers that leave a remainder of 1 when divided by 3, use the arithmetic sequence formula.

Therefore, the correct answer is a. 1616.

Step-by-step explanation:

To find the sum of all two-digit numbers that leave a remainder of 1 when divided by 3, we can use the arithmetic sequence formula.

The formula for the nth term of an arithmetic sequence is a + (n - 1)d, where a is the first term, n is the number of terms, and d is the common difference.

In this case, the first term is 10 and the common difference is 3. Let's find the number of terms in the sequence:

(last term - first term) / common difference + 1 = (99 - 10) / 3 + 1 = 90 / 3 + 1 = 30 + 1 = 31

Now, let's find the sum of the sequence:

(n / 2)(first term + last term) = (31 / 2)(10 + 99) = 15.5 * 109 = 1694.5

The sum of all two-digit numbers that leave a remainder of 1 when divided by 3 is 1694.5.

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