Sum of two-digit numbers with remainder 3 on division by 7: What is the sum of all two-digit numbers that leave a remainder 3 when divided by 7?

Introduction / Context:Numbers congruent to 3 modulo 7 form an arithmetic sequence with common difference 7. Restricting to two-digit numbers gives a finite AP, whose sum we can compute with the AP sum formula.Given Data / Assumptions:

• Two-digit range: 10 to 99 inclusive.
• Condition: n ≡ 3 (mod 7).

Concept / Approach:Find the first and last two-digit numbers congruent to 3 mod 7, count terms, then sum using S_n = n/2 * (first + last).

Step-by-Step Solution:

Verification / Alternative check:Average term equals (first + last)/2 = 52. With 13 terms, total = 13 * 52 = 676. Confirms the AP sum result.

Why Other Options Are Wrong:

• 94: This is the last term, not the sum.
• 696: Too high; would imply 13 terms averaging about 53.5, which is not correct.
• None of these: Incorrect because 676 is attainable.

Common Pitfalls:Forgetting that 10 is valid (two-digit) and congruent to 3 mod 7, or miscounting terms by dropping the +1 in the count formula.

Final Answer: