This problem can be solved by listing different values of y, and quickly eliminating choices.

y = 3x + 2. In other words, y = (any multiple of 3) + 2. This gives us the values 5 (which is 3*1 + 2), 8, 11, 14, 17, 20 and so on.

Looking at this list we can eliminate the answers 4 (a divisor of 8), 5 (divisor of 20), 7 (divisor of 14), and 8 (divisor of 8). This leaves us with 6, C as the answer. This answer makes sense because 6 is the only answer that's a multiple of 3, and if you add only 2 to any multiple of 3, the resulting number (5, 8, 11, ...) will never be a multiple of three, or of 6.

y=3x+2


3x is a multiple of 3
3x+2 is a multiple of 3 that has been increased by two units.
Therefore 3x+2 is NOT a multiple of 3

Remember the rule for 6: An integer is divisible by 6 if and only if it is divisible by both 2 and 3.
Therefore y is not a multiple of 6 since it is not divisible by 3.

Final Answer: C

Given that x is an integer and y = 3x + 2 and we need to find which of the following CANNOT be a divisor of y

(A) 4
x = 2 we get, y = 3*2 + 2 = 6 + 2 = 8 => 4 IS A DIVISOR of y

(B) 5
x = 6 we get, y = 3*6 + 2 = 18 + 2 = 20 => 5 IS A DIVISOR of y

(C) 6
This wont be possible as 3 is a factor of 6 so 3x + 2 can never be a multiple of 3 or cannot be any multiple of a multiple of 3 => 6 IS NOT A DIVISOR of y

In exam, we don't need to solve further but I am solving to complete the solution.

(D) 7
x = 4 we get, y = 3*4 + 2 = 12 + 2 = 14 => 7 IS A DIVISOR of y

(E) 8
x = 2 we get, y = 3*2 + 2 = 6 + 2 = 8 => 8 IS A DIVISOR of y

So, Answer will be C
Hope it helps!

Watch the following video to learn the Basics of Divisibility Rules

Hi Brent GreenlightTestPrep
Can we write y = 3x + 2 as x = (y-2)/3
From this we can tell that 6 is a multiple of 3 therefore 6 cannot be divisor of 3 here?
Is this way of reasoning correct? Thanks Brent

yes

\(x=\frac{y}{3} -\frac{2}{3}=\frac{y-2}{3}\)

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