Grades 11-12 Video Solutions 2023-2023_11-12

Video Transcription

Problem number 27. What is the greatest common divisor of all numbers of the form n cubed times n plus one  cubed times n plus two cubed times n plus three cubed times n plus four cubed, where n is a nonzero natural number? Okay, so forget about the cubed.  These are five consecutive integers, so there will be at least two multiples of two, one of which is going to be a multiple of four, there's going to be at least one multiple  of three, they don't have to be two because we don't have six integers, and at least one multiple of five.  Then this product will be at least always divisible by two times four, so two cubed, times three, times five.  And if you cube this, you're just going to cube each of these exponents, and cubing a number is the same as multiplying its exponent by three, so our final answer will be divisible  by two to the power of nine, times three cubed, times five cubed, and thus that is the greatest common divisor, which is choice E, and just like that, we're done.

Video Summary

The problem involves finding the greatest common divisor (GCD) of numbers of the form \( n^3(n+1)^3(n+2)^3(n+3)^3(n+4)^3 \) for nonzero natural numbers \( n \). Each term represents five consecutive integers, ensuring divisibility by at least \( 2^3 \) (due to two numbers being multiples of two, including one multiple of four), \( 3 \), and \( 5 \). Cubing each calculates the final GCD as \( 2^9 \times 3^3 \times 5^3 \). Thus, the greatest common divisor of these numbers is the product of these prime powers, which corresponds to choice E.

Keywords

greatest common divisor

GCD

consecutive integers

prime powers

natural numbers