Grades 9-10 Video Solutions 2021-video 2021 9-10/26

Video Transcription

Problem number 26 states, each of the numbers a and b is a square of an integer. The difference a minus b is a prime number. Which of the following could be b?  So the problem tells us that we have these two numbers, a and b, and they are the squares of numbers.  They tell us that the difference of these two numbers is going to be prime, so a minus b is going to be equal to m squared minus n squared.  Since we have a difference of squares here on the right side, we can break that up into m minus n on the left and m plus n on the right. And we do know that a minus b must be prime.  Let's put in some values in for m minus n and m plus n and see if we can figure out some things. If we put in 10 and 14, we get a product of 140, which is not prime.  If we put in 6 and 8, then we get a product of 48, which is also not prime. So what could we do here in order to make the product prime?  Well we can't multiply any two integers because then the product would not be prime because it would be divisible by those two numbers.  In order to resolve this, we could look at the definition of the prime number, which states that a prime number is one whose factors are itself and one, which tells us that one  of the numbers must be one and the other one must be a prime number itself. In this case, we know that m and n are going to be positive, so one will be the difference  and the prime number will be the sum. So we know that b is n squared and let's bring up all of the possible values for b. So these are the values.  So we can calculate n by taking the square root of them, and we would get possible n values of 10, 12, 16, 30, and 100. So now simply let's solve for m.  We can do this by adding n to both sides, and we would get m is equal to n plus 1. Now let's calculate m, which would get us 11, 13, 17, 31, and 101.  Let's add them up, and we get 21, 25, 33, 61, and 201. Of these, only one is prime, 61. 21 is divisible by 3, 25 is divisible by 5, 33 is divisible by 11, and 201 is divisible by 3.  So the answer must be d. So the question asked us, which of the following could be b? The answer is 900, letter d.

Video Summary

The problem involves two numbers, \( a \) and \( b \), which are squares of integers. Their difference is given as a prime number. By expressing this difference as a product of two factors, \( m-n \) and \( m+n \), the solution involves finding values where one factor is 1 and the other is prime. Calculating potential \( b \) values as \( n^2 \), and solving for \( m = n + 1 \), five sums are analyzed. Among them, only 61 is prime, leading to the conclusion that the correct value for \( b \), given these conditions, is 900, corresponding to option d.

Keywords

squares of integers

prime number

difference

factors

value 900