Data Sufficiency

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


Ans-B

[email protected] wrote: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


Ans-B

Hi Shibsriz,
From
Statement 1: When x is divided by 2y, the remainder is 4
Let X= 10 so 2Y be 6(2*3),(Remainder will be 4)
So, X/Y = 10/3 => Remainder will be 1

Let X= 20 So 2Y be 16(2*8), (Remainder will be 4)
So, X/Y = 20/8 => Remainder will be 4

Hence Statement 1 is Insufficient.

Statement 2: When x + y is divided by y, the remainder is 4

(X+Y)/Y gives remainder 4

above statement can be written an

(X/Y) + (Y+Y) => (X/Y) + 1

Hence X/Y give remainder 4

Hence Statesmen 2 is Sufficient.

Answer is B

Regards,
Uva.

[email protected] wrote: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.

Hi Brent,
I have a question here,
u wrote,

Factor to get: x = y(k - 1) + 4
As we can see, x is 4 greater than some multiple of y.
So, if we divide x by y, the remainder must be 4
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

which i too agree,
my doubt is
in option A) too we can write x = 2y (K) + 4 = y (2K) + 4
so, in this case too we can see, x is 4 greater than some multiple of y,
then y we are not saying statement 1 is sufficient ?
Kindly help me out..

Maybe an easier way of dealing with this is to recognize that remainders are additive: in other words,

(Remainder of x divided by y) + (Remainder of y divided by y) = Remainder of (x+y) divided by y.

Now consider Statement 2. Since y has a remainder of 0 when divided by y -- as does any positive integer when divided by itself, we really have

(Remainder of x divided by y) + 0 = 4

Piece of cake!

GMATGuruNY wrote: 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.

Hi GMATGuru,

I have put y=2
x+2=2k+4
x=2k+2
options for x= 4,6,10 .

in all cases Reminder is 0

I have put y=4

x+4=4k+4
x=4k
options for x= 4,8,12
In all cases reminder is 0

combined to your cases mentioned above y=5 & y=6. we have two answers R=1 & R=0. So statement 2 is insufficient.

What do I miss?

Thanks in advance

RULE:
If positive integer x is divided by positive integer y, the greatest possible remainder is equal to y-1.

Example:
If y=3, the following remainders are possible:
3/3 = 1 R0.
4/3 = 1 R1.
5/3 = 1 R2.
6/3 = 2 R0.
7/3 = 2 R1.
8/3 = 2 R2.
As the red options illustrate, if y=3, the greatest possible remainder = y-1 = 3-1 = 2.

Mo2men wrote: Statement 2: When x + y is divided by y, the remainder is 4

I have put y=4

If y=4, then the greatest possible remainder = y-1 = 4-1 = 3.
Since statement 2 indicates that the remainder is 4, y=4 is not viable.
To yield a remainder of 4, y must be GREATER THAN 4.