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Number properties question

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by arvn » Sun Aug 31, 2008 2:35 pm
Can someone please help with the following number properties ds problem:

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.
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Source: — Data Sufficiency |

Re: Number properties question

by mehravikas » Sun Aug 31, 2008 4:44 pm
I would pick numbers for this problem -

Statement 1 - Let's assume x = 16 and y = 6. Therefore, x / 2y the remainder is 4 and when x / y the remainder is again 4.

x = 24, y = 10.
X / 2y = Remainder is again 4
x / y = Remainder is again 4.

Statement 1 is sufficient.

Statement 2 says (x + y) / y gives a remainder 4.
You can break the above equation as - x/y + y/y

y/y will never leave a remainder. Therefore we can say that when x is divided by y the remainder will be 4.

Statement 2 is also sufficient.

Answer should be 'D'
arvn wrote:Can someone please help with the following number properties ds problem:

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.
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by Ian Stewart » Sun Aug 31, 2008 5:32 pm
There's a bit of a trick in this problem, and it illustrates the danger of picking numbers. First, statement 2 is sufficient:

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

Notice (this will be important when we look at Statement 1) that when you divide by y, the remainder must be less than y, by definition. So if Statement 2 is true, y must be larger than 4. Writing the information in the statement using the standard remainder equation (n = qd + r):

x + y = qy + 4
x = (q-1)y + 4

so x is 4 larger than a multiple of y, and 4 is the remainder when x is divided by y.

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

Notice here that the remainder must be less than 2y- that is, 4 < 2y, or y > 2. Writing this statement using the standard remainder equation:

x = 2y*q + 4
x = (2q)*y + 4

So x is 4 larger than a multiple of y. As long as y > 4, then 4 will certainly be the remainder when x is divided by y, and if you test numbers here, and only choose values of y that are larger than 4, you will likely begin to think that 1) is sufficient. It isn't sufficient; if y is equal to 4, for example, then x could be 12. Then the remainder is 4 when x is divided by 2y, but is zero when x is divided by y. Insufficient.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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by mehravikas » Sun Aug 31, 2008 9:59 pm
Oh yeah, I missed out setting the correct value for statement 1.

Thanks Ian for your help.
Vikas
Ian Stewart wrote:There's a bit of a trick in this problem, and it illustrates the danger of picking numbers. First, statement 2 is sufficient:

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

Notice (this will be important when we look at Statement 1) that when you divide by y, the remainder must be less than y, by definition. So if Statement 2 is true, y must be larger than 4. Writing the information in the statement using the standard remainder equation (n = qd + r):

x + y = qy + 4
x = (q-1)y + 4

so x is 4 larger than a multiple of y, and 4 is the remainder when x is divided by y.

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

Notice here that the remainder must be less than 2y- that is, 4 < 2y, or y > 2. Writing this statement using the standard remainder equation:

x = 2y*q + 4
x = (2q)*y + 4

So x is 4 larger than a multiple of y. As long as y > 4, then 4 will certainly be the remainder when x is divided by y, and if you test numbers here, and only choose values of y that are larger than 4, you will likely begin to think that 1) is sufficient. It isn't sufficient; if y is equal to 4, for example, then x could be 12. Then the remainder is 4 when x is divided by 2y, but is zero when x is divided by y. Insufficient.
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by pepeprepa » Mon Sep 01, 2008 7:14 am
Thank you, nice question
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by arvn » Mon Sep 01, 2008 11:38 am
Thanks Ian!
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