E IMO

To prove: y =2

Stmt 1: y ≠ 1. y can take any value for different x for which x mod y = 0 will be satisfied. INSUFFICIENT.

Stmt 2: (x+2) mod y = 0.
x mod y + 2 mod y = 0
0 + 2 mod y = 0 (Given x mod y = 0)
2 mod y = 0
That means y can be either 1 or 2 for which 2 mod y = 0 will be satisfied. INSUFFICIENT.

Stmt 1 and Stmt 2 combined:
From Stmt 1, y ≠ 1
From Stmt 2, y = 1 or y = 2

Hence together, y = 2.

Ans: C.


What is the OA?

here the question is about factors and multiples and you ultimately need to know whether y always = 2.

first statement does nothing to clarify - x coud be 6 and y=3 or x could be 4 and y = 2

Insufficient

Second statement brings up the idea of factoring with addition. if x is a multiple of y then that means that y divides evenly into x. however if x + 2 is also divisible by y then that means that y is a factor of 2 because if x divides evenly and x+2 divides evenly then that means that 2 must be a multiple of y. HOwever this still doen't answer our quesiton because y could equal 1 or 2.

When you put both pieces of information together you find that y must be 1 or 2 and it can't be 1. Therefore y must equal 2. the answer is C. (This answer might be different if they didn't define y as an integer)

GmatKiss wrote:If x and y are positive integers and x is a multiple of y, is y = 2?

(1) y ≠ 1
(2) x + 2 is a multiple of y.

IMO: E

Answer should be C and Not E.

Solution: --

stmt 1: Y NOT = 1 [ Not sufficient]

stmt 2: X+2 is a multiple of Y

now we already know that X is a multiple of y.
and from statment 2 we also know that x+2 is a multiple of y.

Let's put some values

Let Y =2,3,4,5... etc

for Y =2.. X can be anything from 2.... to infinity (all even)
Let X =2 then x+2= 4 which is also a multiple of y.
if x =4 then x+2 = 6 which is also a multiple of y.

This condition will not hold true if we choose any other value of y, say for example y =3
then X can be 3, 6, 9 ..

but then x+2 will no longer be a factor

LOGIC: -- x+2 means a difference of 2.. (table of 2 -- every consecutive number has a common difference of 2)
similarly: table of 3 every consecutive number has a common difference of 3

But for Y =1 this condition will always remain trueHence, stmt 2 is also Not sufficient.

Combine both the statments and we can eliminate Y =1

Therefore only one value of Y will remain and i.e. 2

OPTION C

C