Grades 11-12 Video Solutions 2013-Levels 11&12 Video Solutions 2013 problem27

Levels 11&12 Video Solutions 2013 problem27

Video Transcription

Video Summary

The problem involves finding the number \( n \) for which the sum of the first \( n \) positive integers equals a three-digit number with identical digits. The sum is expressed as \( \frac{n(n+1)}{2} \), equating to a multiple of 111, specifically \( 111k \) where \( 1 \leq k \leq 9 \). Solving the equation with the factorization leads to \( n = 36 \). The sum of 36's digits, 3 and 6, is 9. Therefore, the answer is 9.

Keywords

sum of integers

three-digit identical digits

factorization

n equals 36

sum of digits