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?

Difficulty: Easy

Correct Answer: 676

Explanation:


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:

First term: 10 (since 10 = 7*1 + 3).Last term: 94 (since 94 = 7*13 + 3 ≤ 99).Number of terms: n = ((94 − 10)/7) + 1 = 12 + 1 = 13.Sum S = 13/2 * (10 + 94) = 13/2 * 104 = 13 * 52 = 676.


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:

676

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