If x and y are positive integers and 21x + 23y = z, what is the value

Target question: What is the value of y?

Given: x and y are positive integers and 21x + 23y = z

Statement 1: x = 4
This statement doesn't FEEL sufficient, so I'll TEST some values.
There are several values of x, y and z that satisfy statement 1 (and the given information). Here are two:
Case a: x = 4, y = 1 and z = 107, in which case y = 1
Case b: x = 4, y = 2 and z = 130, in which case y = 2
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: z = 130
NOTE: Since we're told that x and y are POSITIVE INTEGERS, this statement is very limiting.
To determine whether or not this helps us determine the value of y, we need to check all possible solutions to the equation 21x + 23y = z where z = 130
Since x must be a positive integer, we'll start looking for solutions ( to the equation 21x + 23y = 130) when x = 1
Since (7)(21) = 147, and since 147 is greater than 130, we can see that x cannot equal 7 and x cannot be greater than 7
So, we'll check all values of x from 1 to 6:
x = 1. We get: 21(1) + 23y = 130. So, 23y = 109. This equation has no INTEGER solution for y, and since y must be a positive integer, we know that x cannot equal 1
x = 2. We get: 21(2) + 23y = 130. So, 23y = 80. This equation has no INTEGER solution for y. So, x cannot equal 2
x = 3. We get: 21(3) + 23y = 130. So, 23y = 67. This equation has no INTEGER solution for y. So, x cannot equal 3
x = 4. We get: 21(4) + 23y = 130. So, 23y = 46, which means y = 2.
x = 5. We get: 21(2) + 23y = 130. So, 23y = 25. This equation has no INTEGER solution for y. So, x cannot equal 5
x = 6. We get: 21(2) + 23y = 130. So, 23y = 4. This equation has no INTEGER solution for y. So, x cannot equal 6

We've checked all possible solutions to the equation, and only one of them yielded an INTEGER value for y
So, it must be the case that y = 2
Since we can answer the target question with certainty, statement 2 is SUFFICIENT


Answer:

Cheers,
Brent