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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

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

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

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Oct 15, 2018

#1
+5038
+1
$$3b=8-2a\\ 2b+12 = \dfrac 23(8-2a)+12=\\ \dfrac{16-4a+36}{3} = \dfrac{52-4a}{3} \\ \text{If I understand the question were are looking for the values of }\\ a \in \{1,2,3,4,5,6\} \text{ such that } \dfrac{52-4a}{3} \in \mathbb{Z}$$
$$\dfrac{52-4a}{3} = 16+\dfrac{4-4a}{3} \text{ so we want values of }a \ni \dfrac{4-4a}{3} \in \mathbb{Z}\\ 4-4a = \{0, -4, -8, -12, -16, -20\}\\ \text{ and the two of those divisible by 3 are 0 and -12} \\ \text{corresponding to }a=1,~a=4\\ \text{ so there are 2 values of a that satisfy the original statement}$$