Numerical answer
What is the sum of all three-digit natural numbers that leave a remainder of 2 when divided by 7?
Show answer and solution
Answer: 70692
Since 100 = 7 × 14 + 2, the first such number is 100.
The largest three-digit number is 999 = 7 × 142 + 5, so the last one with remainder 2 is 996.
The numbers 100, 107, …, 996 form an AP with common difference 7.
Number of terms = (996 − 100)/7 + 1 = 128 + 1 = 129.
Sum = 129 × (100 + 996)/2 = 129 × 548 = 70,692.
Tip: Always find the first and last terms directly before applying n = (l − a)/d + 1.
The largest three-digit number is 999 = 7 × 142 + 5, so the last one with remainder 2 is 996.
The numbers 100, 107, …, 996 form an AP with common difference 7.
Number of terms = (996 − 100)/7 + 1 = 128 + 1 = 129.
Sum = 129 × (100 + 996)/2 = 129 × 548 = 70,692.
Tip: Always find the first and last terms directly before applying n = (l − a)/d + 1.
Saved on this device. Sign in or create a free account to keep it on every device.