Grades 11-12 Video Solutions 2013-Levels 11&12 Video Solutions 2013 problem12

Video Transcription

Question number 12, how many positive integers n exist such that both n divided by 3 and 3n are 3-digit integers?  Let's first use the information that both numbers are integers.  So let k be equal to n over 3 and we suppose that this is an integer.  Then 3k is equal to n. Since clearly both are positive integers, n over 3 must be less than 3 times n. And substituting  for n, we have that k is less than 9 times k. Now,  let's use the information that both are 3-digit integers to put upper bounds on 9k and lower bounds on k. So the smallest number here we can use for k is 100 and 9k is at most 999.  So that tells us that k is an integer between 100 and 111.  And so there are 12 such integers.

Video Summary

The problem asks for the number of positive integers \( n \) such that both \( n/3 \) and \( 3n \) are three-digit integers. By letting \( k = n/3 \), this makes \( n = 3k \). Both \( k \) and \( 3k \) need to be integers. For \( n/3 \) to be a three-digit integer, \( k \) must be between 100 and 111, inclusive. This gives 12 possible values for \( k \), making for 12 positive integers \( n \) that satisfy the conditions.

Keywords

positive integers

three-digit

divisibility

range

conditions