Grades 11-12 Video Solutions 2011-11&12 Video Solutions 2011 problem21

11&12 Video Solutions 2011 problem21

Video Transcription

Video Summary

The problem involves finding all positive integers \( x \) less than 100 such that \( x^2 - 81 \) is a multiple of 100. Factoring \( x^2 - 81 \) as \((x - 9)(x + 9)\), we determine that both terms need to meet specific conditions, given that their product must be a multiple of 100 and they differ by 18. Analyzing factor pairs of 100, we chose \( 2 \times 50 \). The possible \( x \) values are 41, 59, and 91, where the factor conditions hold. These numbers sum to 200. Thus, the result is 200.

Keywords

positive integers

multiple of 100

factor pairs

sum

solution