We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.

+0

# Help!

0
145
1

Suppose that $$a$$ and $$b$$ are integers such that $$3b = 8 - 2a.$$How many of the first six positive integers must be divisors of $$2b + 12$$?

Oct 14, 2018

### Best Answer

#1
+1

$$3b=8-2a\\ 2b+12 = \dfrac 23(8-2a)+12=\\ \dfrac{16-4a+36}{3} = \dfrac{52-4a}{3} \\ \text{If I understand the question were are looking for the values of }\\ a \in \{1,2,3,4,5,6\} \text{ such that } \dfrac{52-4a}{3} \in \mathbb{Z}$$

$$\dfrac{52-4a}{3} = 16+\dfrac{4-4a}{3} \text{ so we want values of }a \ni \dfrac{4-4a}{3} \in \mathbb{Z}\\ 4-4a = \{0, -4, -8, -12, -16, -20\}\\ \text{ and the two of those divisible by 3 are 0 and -12} \\ \text{corresponding to }a=1,~a=4\\ \text{ so there are 2 values of a that satisfy the original statement}$$

.
Oct 15, 2018

### 1+0 Answers

#1
+1
Best Answer

$$3b=8-2a\\ 2b+12 = \dfrac 23(8-2a)+12=\\ \dfrac{16-4a+36}{3} = \dfrac{52-4a}{3} \\ \text{If I understand the question were are looking for the values of }\\ a \in \{1,2,3,4,5,6\} \text{ such that } \dfrac{52-4a}{3} \in \mathbb{Z}$$

$$\dfrac{52-4a}{3} = 16+\dfrac{4-4a}{3} \text{ so we want values of }a \ni \dfrac{4-4a}{3} \in \mathbb{Z}\\ 4-4a = \{0, -4, -8, -12, -16, -20\}\\ \text{ and the two of those divisible by 3 are 0 and -12} \\ \text{corresponding to }a=1,~a=4\\ \text{ so there are 2 values of a that satisfy the original statement}$$

Rom Oct 15, 2018