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

Video Transcription

Question number 27, the sum of the first n positive integers is a three-digit number in which all of the digits are the same.  What is the sum of the digits of that number n?  We have a formula for the sum of the first n positive integers. If we add up 1 plus 2 plus 3 all the way up to n, that is given by the formula n times n plus 1 divided by 2.  And that is supposed to be a three-digit number  with all digits the same. So it must be a multiple of 111 where here k is a number of the set from 1 to 9 so that we have indeed a three-digit number.  And so working with that expression we see that on the left-hand side if we multiply through by 2,  we have a product of 2 consecutive positive integers. And on the right-hand side we have 2 times k times the factorization of 111 into prime factors which gives us 3 times 37.  Now 37 we can't really do anything about so 37 is going to be equal to n or to n plus 1.  And the other factor here is 2 times 3 times k. So 6k is equal to 36 or 38.  And since 38 is not divisible by 6, we have that k has to be equal to 6. So then n is equal to 36, n plus 1 is equal to 37.  And the sum of the digits in the number n, which is 36, is 9. So the answer here is b.

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