Grades 9-10 Video Solutions 2013-Level 9&10 Video Solutions 2013 problem22

Level 9&10 Video Solutions 2013 problem22

Video Transcription

Video Summary

The problem involves finding how many positive integers are multiples of 2013 and have exactly 2013 divisors. By identifying the prime factorization of 2013 as \(3, 11, 61\), you derive that the product of their exponents plus one has to match the 2013 divisors condition. Since each prime factor must appear at least once, permutations of assigning exponents to 3, 11, and 61 determine the solution. The solution shows there are 6 unique ways to assign these exponents, leading to 6 positive integers that meet the conditions.

Keywords

positive integers

multiples of 2013

prime factorization

divisors condition

unique solutions