Question

What is the sum of all two-digit numbers that give a remainder of 3 when they are divided by 7?

• A. 666
• B. 676
• C. 683
• D. 777

Hint:

When dealing with sequences based on modular arithmetic, express the numbers as \(7k + 3\) and use the formula for the sum of an arithmetic series.

Answer 1

The Correct Option is B

We need to find the sum of all two-digit numbers that give a remainder of 3 when divided by 7. Step 1: The numbers that satisfy this condition are in the form \( 7k + 3 \), where \( k \) is an integer. The smallest two-digit number is 10, and the largest is 99. \[ 7k + 3 \geq 10 \Rightarrow k \geq \frac{7}{7} = 1 \] \[ 7k + 3 \leq 99 \Rightarrow k \leq \frac{96}{7} \approx 13 \] Thus, the values of \( k \) range from 1 to 13. The numbers are: \[ 7(1) + 3 = 10, \, 7(2) + 3 = 17, \, 7(3) + 3 = 24, \dots, 7(13) + 3 = 94 \] Step 2: Calculate the sum of these numbers: \[ 10 + 17 + 24 + \dots + 94 \] This is an arithmetic sequence with the first term 10, the common difference 7, and 13 terms. The sum of the sequence is given by: \[ S = \frac{n}{2} \times (\text{first term} + \text{last term}) = \frac{13}{2} \times (10 + 94) = 676 \] Thus, the answer is: 676.