If x and y are positive integer, what is the remainder

[Gmat_mission]

If x and y are positive integers, what is the remainder when x is divided by y ?

(1) When x is divided by 2x, the remainder is 4.
(2) When x + y is divided by y, the remainder is 4.

[spoiler]OA=B[/spoiler].

How can I use the second statement to get an answer here? Can anyone help me?

[Jay@ManhattanReview]

If x and y are positive integers, what is the remainder when x is divided by y ?

(1) When x is divided by 2x, the remainder is 4.
(2) When x + y is divided by y, the remainder is 4.

[spoiler]OA=B[/spoiler].

How can I use the second statement to get an answer here? Can anyone help me?

Given: x and y are positive integers

We have to determine the remainder when x is divided by y.

Let's take each statement one by one.

(1) When x is divided by 2x, the remainder is 4.

Since x < 2x, x itself would be the remainder. Thus, x = 4. But we do not have any information about y. Insufficient.

(2) When x + y is divided by y, the remainder is 4.

=> (x + y)/y = x/y + y/y

=> Remainder of (x/y + y/y) = 4

=> Remainder of x/y + 0 = 4; since y divided by y would render a remainder 0.

=> Remainder of x/y = 4. A unique answer. Sufficient.

The correct answer: B

Hope this helps!

-Jay
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[GMATGuruNY]

If x and y are positive integers, what is the remainder when x is divided by y?

(1) When x is divided by 2y, the remainder is 4
(2) When x + y is divided by y, the remainder is 4

When x is divided by y, the remainder is R.
This statement implies the following:
x is R more than a multiple of y.
Translated into math:
x = ky + R, where k is an integer such that k≥0.

Statement 1: When x is divided by 2y, the remainder is 4
Case 1: y=3, implying that 2y=6
Here, when x is divided by 6, the remainder is 4.
In other words, x is 4 more than a multiple of 6:
x = 6k + 4, where k is an integer such that k≥0.
Options for x = 4, 10, 16...

When these values for x are divided by y=3, we get:
x/y = 4/3 = 1 R1.
x/y = 10/3 = 3 R1.
x/y = 16/3 = 5 R1.
Result:
R=1.

Case 2: y=4, implying that 2y=8
Here, when x is divided by 8, the remainder is 4.
In other words, x is 4 more than a multiple of 8:
x = 8k + 4, where k is an integer such that k≥0.
Options for x = 4, 12, 20...

When these values for x are divided by y=4, we get:
x/y = 4/4 = 1 R0.
x/y = 12/4 = 3 R0.
x/y = 20/4 = 5 R0.
Result:
R=0.

Since x/y can yield different remainders, iNSUFFICIENT.

Statement 2: When x + y is divided by y, the remainder is 4
Case 3: y=5
Here, when x+5 is divided by 5, the remainder is 4.
In other words, x+5 is 4 more than a multiple of 5:
x + 5 = 5k + 4
x = 5k - 1, where k is an integer such that k≥1 (since x must be positive).
Options for x = 4, 9, 14...

When these values for x are divided by y=5, we get:
x/y = 4/5 = 0 R4.
x/y = 9/5 = 1 R4.
x/y = 14/5 = 2 R4.
Result:
R=4.

Case 4: y=6
Here, when x+6 is divided by 6, the remainder is 4.
In other words, x+6 is 4 more than a multiple of 6:
x + 6 = 6k + 4
x = 6k - 2, where k is an integer such that k≥1 (since x must be positive).
Options for x = 4, 10, 16...

When these values for x are divided by y=6, we get:
x/y = 4/6 = 1 R4.
x/y = 10/6 = 1 R4.
x/y = 16/6 = 2 R4.
Result:
R=4.

R=4 in both cases.
The implication is that -- in every case -- when x is divided by y, R=4.
SUFFICIENT.

The correct answer is B.

Algebraic proof for statement 2:
When x + y is divided by y, the remainder is 4.
In other words, x+y is 4 more than a multiple of y:
x+y = ky + 4.

Combining like terms, we get:
x = ky - y + 4
x = (k-1)y + 4, where k-1≥0.

Put into words:
x is 4 more than a multiple of y.
In other words:
When x is divided by y, the remainder is 4.