Find the sum of all two-digit numbers which leave a remainder 2 when they are divided by 6.
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 840
Numbers that leave remainder 2 on division by 6 are of the form \(6k+2\): they rise in steps of 6, so the two-digit ones form an arithmetic progression, and its sum is the number of terms times the average of the first and last terms.
First two-digit term: \(6\times2+2=14\); last: \(6\times16+2=98\)
Number of terms: \(\frac{98-14}{6}+1=15\)
Sum \(=\frac{15}{2}\times(14+98)=15\times56=840\)