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# When the positive integer x is divided by 4, is the

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When the positive integer x is divided by 4, is the [#permalink]  25 May 2008, 18:39
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When the positive integer x is divided by 4, is the remainder equal to 3?

(1) When x/3 is divided by 2, the remainder is 1.

(2) x is divisible by 5.
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Re: DS: remainder [#permalink]  25 May 2008, 18:52
chineseburned wrote:
When the positive integer x is divided by 4, is the remainder equal to 3?

(1) When x/3 is divided by 2, the remainder is 1.

(2) x is divisible by 5.

(1) when x is divided by 6 remainder is 1, so x is 7(OK) , 13(Not OK) ,so Not suff.
(2) x can be 5(Not good), 10(Not good), 15(OK), so insuff

together x can be 25, 65, which should always(?) give remainder of 1, so suff.

Thus C
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Re: DS: remainder [#permalink]  25 May 2008, 20:42
65 doesn't satisfy condition 1/. It is rather 55 and 55 does work, which means that when 55 is divided by 4, the remainder is 3. Since this is not the case for 25, there is not enough information.
Otherwise you did a very good job, fellow!

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Last edited by TheGMATDoctor on 13 Jul 2010, 12:09, edited 2 times in total.
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Re: DS: remainder [#permalink]  25 May 2008, 21:28
Expert's post
I still think that C is correct.

1) When x/3 is divided by 2, the remainder is 1.
It means that x is divisible by 3 because x/3 has to be an integer.
So,
x=3, r=3
x=9, r=1
insufficient

2) x is divisible by 5.

So,
x=5, r=1
x=15, r=3
insufficient

1)&2)
write out x that satisfy 1st condition and find two the closest numbers that satisfy 2nd condition also:

x=3,9,15,21,27,43,49,55

So,
x=15, r=3
x=55, r=3
sufficient

TheGMATDoctor wrote:

Give two examples that satisfy both conditions and give different remainders.
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Re: DS: remainder [#permalink]  25 May 2008, 21:45
E

you have 25 and 55, they meet both conditions

and 25/4 ---> remainder is 1

55/4 ---> remainder is 3
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Re: DS: remainder [#permalink]  25 May 2008, 22:07
Expert's post
gmatnub wrote:
E

you have 25 and 55, they meet both conditions

and 25/4 ---> remainder is 1

55/4 ---> remainder is 3

25 is not divisible by 3 and therefore first condition is not satisfied or maybe you know what is remainder when 8.333333 is divided by 2
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Re: DS: remainder [#permalink]  25 May 2008, 22:12
walker wrote:
gmatnub wrote:
E

you have 25 and 55, they meet both conditions

and 25/4 ---> remainder is 1

55/4 ---> remainder is 3

25 is not divisible by 3 and therefore first condition is not satisfied or maybe you know what is remainder when 8.333333 is divided by 2

hey walker, statement 1 does not say that x has to be divisible by 3, it say x/3 divide by 2 ---> x/6. This is another typical question by chineseburn, where the wording is unnecessary complicated and designed to play "gotcha".
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Re: DS: remainder [#permalink]  25 May 2008, 22:37
Expert's post
(1) When 25/3 is divided by 2, the remainder is 1.

it has no sense, as remainder is applicable only for integers.

I disagree that 1st question simply means x/6 and can say that in real GMAT the condition would state as:

(1) X is divisible by 3 and when x/3 is divided by 2, the remainder is 1.
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Re: DS: remainder [#permalink]  25 May 2008, 23:32
walker wrote:
(1) When 25/3 is divided by 2, the remainder is 1.

it has no sense, as remainder is applicable only for integers.

I disagree that 1st question simply means x/6 and can say that in real GMAT the condition would state as:

(1) X is divisible by 3 and when x/3 is divided by 2, the remainder is 1.

But if this is the case, x can not equal to 55 either.
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Re: DS: remainder [#permalink]  26 May 2008, 00:14
Expert's post
gmatnub wrote:
But if this is the case, x can not equal to 55 either.

Thanks! Silly mistake! should be:

x=3,9,15,21,27,33,39,45

So,
x=15, r=3
x=45, r=1

insufficient.

E
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Re: DS: remainder [#permalink]  26 May 2008, 02:39
I think C is correct.

I think the examples of 25, 55 and 65 do not satisfy (1), i.e. 25/3, 55/3 and 65/3 divided by 2 do not lead to remainder 1.

Agreed with walker.
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Re: DS: remainder [#permalink]  26 May 2008, 09:07
SHIT I got C too..

anyways..i realized my mistake...

stem x=4N+3 i.e x=3, 7, 11, 15 etc

1) x=6N+1 x=1, 7, 13, 19, 25...

clearly Insuff

2) x is divible by 5.., x=5 remainder 1..x=10 remainder 2...x=15 remainder 3...

insuff..

together x can be 15, 55...

HERE is what i did wrong..i accidently ended up dividing by 3..the question clearly asks x/4...GRRRRRR

I agree with OA=e..remainder can be 1 or 3..
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Re: DS: remainder [#permalink]  26 May 2008, 15:27
In my earlier post above,
I assumed that the integer x is being divided by 6. (Otherwise you must assume that x/3 is an integer). Indeed, the division algorithm (actually a theorem that relates to all integers positive, zero and negative with the constraint that d ≠ 0 ) states that for two integers a and d, with d ≠ 0, there are unique integers q and r such that a = qd + r and 0 ≤ r < |d| ,
The integer
-q is called the quotient
- r is called the remainder
-d is called the divisor
-a is called the dividend
This question needs to be more clear as it is the case in the GMAT.
The set of "qualified" integers you get will depend on your choice of "x/3" or "x/6" since an integer can be divisible by 3 without being divisible by 6. If you choose to go with "x/6" then you'll have the "qualified" integer 55. 55 is divisible by 5 and when you divide 55 by 6 the remainder is 1. So the two conditions are satisfied. Finally when you divide 55 by 4 the remainder is 3. The "qualified'' integer 25 will give a remainder of 1 when divided by 4.
Yet, no matter what option you go with, the answer will still be E.
Very nice discussion overall!

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Re: DS: remainder   [#permalink] 26 May 2008, 15:27
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