Tough DS: please help

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Tough DS: please help

by gmat620 » Mon Nov 02, 2009 6:13 am
If x and y are positive integer, 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.

OA: B
Source: — Data Sufficiency |

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by scoobydooby » Mon Nov 02, 2009 6:24 am
1) x=2kx+4
no definite value of x, x and k both are unknown
not sufficient


2)(x+y)/y leaves a remainder of 4
=>x/y+y/y since y/y leaves no remainder, x/y must leave 4 as the remainder.
sufficient

hence, B

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Re: Tough DS: please help

by maihuna » Mon Nov 02, 2009 6:32 am
gmat620 wrote:If x and y are positive integer, 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.

OA: B
1 doesnt say anything about y so not ok

2 saya: x+y = my + 4 or x/y + 1 = m + 4/y since 4 will be less than y 4/y = 4 only so x/y = m + 3 or remainder will be 3.
Charged up again to beat the beast :)

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Re: Tough DS: please help

by gmat620 » Mon Nov 02, 2009 6:47 am
maihuna wrote:
gmat620 wrote:If x and y are positive integer, 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.

OA: B
1 doesnt say anything about y so not ok

2 saya: x+y = my + 4 or x/y + 1 = m + 4/y since 4 will be less than y 4/y = 4 only so x/y = m + 3 or remainder will be 3.
I don't understand the last step of your solution, we need exact remainder but we can't be sure by "m + 3"

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Re: Tough DS: please help

by life is a test » Mon Nov 02, 2009 6:58 am
gmat620 wrote:
maihuna wrote:
gmat620 wrote:If x and y are positive integer, 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.

OA: B
1 doesnt say anything about y so not ok

2 saya: x+y = my + 4 or x/y + 1 = m + 4/y since 4 will be less than y 4/y = 4 only so x/y = m + 3 or remainder will be 3.
I don't understand the last step of your solution, we need exact remainder but we can't be sure by "m + 3"
x+y/y = some int + 4
-> x/y + y/y = some int + 4
-> x/y + 1 = some int + 4
-> x/y = some int + 3

the last step basically says that if the remainder was 4 when 1 was added to the ratio x/y then remainder must be 3 without that 1 being added to the ratio.

hope that helps.

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Re: Tough DS: please help

by gmat620 » Mon Nov 02, 2009 7:28 am
Thanks a lot friends

@life is test

however, does it mean we can not get an exact value ? is it 3 or 4 ?
life is a test wrote:
gmat620 wrote:
maihuna wrote:
gmat620 wrote:If x and y are positive integer, 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.

OA: B
1 doesnt say anything about y so not ok

2 saya: x+y = my + 4 or x/y + 1 = m + 4/y since 4 will be less than y 4/y = 4 only so x/y = m + 3 or remainder will be 3.
I don't understand the last step of your solution, we need exact remainder but we can't be sure by "m + 3"
x+y/y = some int + 4
-> x/y + y/y = some int + 4
-> x/y + 1 = some int + 4
-> x/y = some int + 3

the last step basically says that if the remainder was 4 when 1 was added to the ratio x/y then remainder must be 3 without that 1 being added to the ratio.

hope that helps.

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by chipbmk » Mon Nov 02, 2009 10:17 am
Can someone please give an example of statement number 1 being true?

How can x/2x ever have a remainder of 4? Doesn't it always equal 1/2?

Also, i understand the logic being used above for statement two, but the way I looked at it was like this.

I got remainder is 0, please tell me where I am going wrong?

For example:

4+6/6 = 1r4 (statement 2 is satisfied)
14+10/10 = 2r4 (statement 2 is satisfied)

Now plug these into the original question:
4/6 = r0
14/10 = r0

Thanks!

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Re: Tough DS: please help

by life is a test » Mon Nov 02, 2009 1:25 pm
chipbmk. - seems 1 should read x/2y has remainder 4.

gmat620 I would like to change my explanation :P the remainder is still 4, you subtract the 1 from the integer value; it doesnt impact the remainder:

x+y/y = k + 4 (k being some int) -> x+y=yk + 4 -> x=yk + 4 - y -> x = y(k-1) + 4 -> this means that the remainder of 4 still holds but the 1 is subtracted from the whole multiple of k, e.g. x=4 y=11 gives 4+11/11 has remainder 4. x=4 y=10 gives 4+10/10 which also has 4 as remainder even though y has been decreased by 1.

hope that helps

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by scoobydooby » Tue Nov 03, 2009 3:36 am
one can even see it this way:

remainder of (x+y)/y is equal to the sum of the remainders of x/y and y/y (remainder theorem)

y/y leaves 0 remainder, the sum of the remainders is given as 4=> x/y must leave a remainder of 4

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by chipbmk » Tue Nov 03, 2009 8:51 am
I just want to make sure I am understanding remainders correctly, is the remainder of 1/2 = 0?

or like I said in my previous post, does 4/6= remainder 0?

Thanks!

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by scoobydooby » Tue Nov 03, 2009 9:12 am
remainder of 1/2 is 1 and remainder of 4/6 is 4

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by chipbmk » Tue Nov 03, 2009 9:16 am
OOHHHH right, so 4/6 would equal 0r4! and 1/2=0r1!!

Got it, thanks!