If a and b are nonnegative integers and 4a+3b=32...

swerve wrote:If a and b are non-negative integers and 4a+3b=32, how many values of b are there?

A. 1
B. 2
C. 3
D. 4
E. 5

If it weren't for the restriction that a and b are non-negative integers, then the equation 4a + 3b = 32 would have infinitely many solutions.
To determine the number of non-negative integer solutions, it's easiest to just list them

Before we start testing possible a and b values, we should recognize that 32 is EVEN, and 4a will be EVEN for all integer values of a
So, we need 3b to be EVEN as well, which means b must be EVEN.

So, let's start with b = 0 and try all even integer values
b = 0 and a = 8 is a solution to the equation 4a + 3b = 32 (SOLUTION #1)

If b = 2, we get 4a + 3(2) = 32, which means b = 6.5. Since 6.5 is NOT an integer, we have NO SOLUTION

b = 4 and a = 5 is a solution to the equation 4a + 3b = 32 (SOLUTION #2)

If b =6, we get 4a + 3(6) = 32, which means b = 3.5. Since 3.5 is NOT an integer, we have NO SOLUTION

b = 8 and a = 2 is a solution to the equation 4a + 3b = 32 (SOLUTION #3)

If b =10, we get 4a + 3(10) = 32, which means b = 0.5. Since 0.5 is NOT an integer, we have NO SOLUTION

If b =12, we get 4a + 3(12) = 32, which means b = -1. Since -1 is NEGATIVE, we have NO SOLUTION

We can safely stop here.

There are 3 solutions on total.

Answer: C

Cheers,
Brent

swerve wrote:If a and b are nonnegative integers and 4a+3b=32, how many values of b are there?

A. 1
B. 2
C. 3
D. 4
E. 5

The OA is C.

Please, can any expert explain this PS question for me? I have many difficulties to understand why that is the correct answer. Thanks.

Simplifying, we have:

3b = 32 - 4a

3b = 4(8 - a)

b = 4(8 - a)/3

Note that b must be an integer, so (8 - a) must be evenly divisible by 3. Since a can be 2, 5, or 8, b has three possible values.

Answer: C