Bunuel 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
If we focus solely on the value of a we can see that the GREATEST value of a is 8, since any a-value greater than 8 will make it impossible to satisfy the equation (4a + 3b = 32) without requiring b to be a negative value.
So, we need only check the values of a from 0 to 8.
a =
0: we get 4(
0) + 3b = 32. This means 3b = 32, so b = 32/3 NOT AN INTEGER VALUE
a =
1: we get 4(
1) + 3b = 32. This means 3b = 28, so b = 28/3 NOT AN INTEGER VALUE
a =
2: we get 4(
2) + 3b = 32. This means 3b = 24, so b = 24/3 = 8
AN INTEGER VALUEa =
3: we get 4(
3) + 3b = 32. This means 3b = 20, so b = 20/3 NOT AN INTEGER VALUE
a =
4: we get 4(
4) + 3b = 32. This means 3b = 16, so b = 16/3 NOT AN INTEGER VALUE
a =
5: we get 4(
5) + 3b = 32. This means 3b = 12, so b = 12/3 = 4
AN INTEGER VALUEa =
6: we get 4(
6) + 3b = 32. This means 3b = 8, so b = 8/3 NOT AN INTEGER VALUE
a =
7: we get 4(
7) + 3b = 32. This means 3b = 4, so b = 4/3 NOT AN INTEGER VALUE
a =
8: we get 4(
8) + 3b = 32. This means 3b = 0, so b = 0/3 = 0
AN INTEGER VALUEAnswer: C
Cheers,
Brent
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