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I guess I am still not sure why in the original post sandipchowdhury wrote:

sandipchowdhury wrote:

St 1: when n is divided by 21 ( 7 and 3) the remainder is an odd number. But it cannot be 7, 3 or 9 . Hence the possibilities are : 1 and 5.

Why remainder cannot be 7, 3 or 9? (because in case of 28 divided by 21 remainder is 7)

OK. It's just not correct.

When n is divided by 21 the remainder is an odd number --> \(n=21q+odd=7*3q+odd\). Now, this odd number can be ANY odd number from 1 to 19, inclusive.

As for r: If \(n=22\) then \(n\) divided by 21 gives remainder of 1 and \(n\) divded by 7 also gives remainder of 1; If \(n=24\) then \(n\) divided by 21 gives remainder of 3 and \(n\) divded by 7 also gives remainder of 3; If \(n=26\) then \(n\) divided by 21 gives remainder of 5 and \(n\) divded by 7 also gives remainder of 5; If \(n=28\) then \(n\) divided by 21 gives remainder of 7 and \(n\) divded by 7 gives remainder of 0.

Re: Collection of remainder problems in GMAT [#permalink]

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13 Dec 2010, 20:55

Great thanks guys. Helps out a great deal however I am still a bit confused. Anyone have any suggestions on where to get more help with these types of problems? Thanks in advance.

Great thanks guys. Helps out a great deal however I am still a bit confused. Anyone have any suggestions on where to get more help with these types of problems? Thanks in advance.

Re: Collection of remainder problems in GMAT [#permalink]

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19 Dec 2010, 20:12

It was a pleasant surprise to see my one year old post still being discussed. Currently I am studying at Stanford GSB. If anyone has any question about Stanford GSB, you may contact me chowdhury_sandip@gsb.stanford.edu

Re: Collection of remainder problems in GMAT [#permalink]

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31 Dec 2010, 16:56

9.If n is a positive integer and r is the remainder when (n - 1)(n + 1) is divided by 24, what is the value of r ?

(1) n is not divisible by 2.

(2) n is not divisible by 3.

ST 1- if n is not divisible by 2, then n is odd, so both (n - 1) and (n + 1) are even. moreover, since every other even number is a multiple of 4, one of those two factors is a multiple of 4. so the product (n - 1)(n + 1) contains one multiple of 2 and one multiple of 4, so it contains at least 2 x 2 x 2 = three 2's in its prime factorization. But this is not sufficient, because it can be (n-1)*(n+1) can be 2*4 where remainder is 8. it can be 4*6 where the remainder is 0.

ST 2- if n is not divisible by 3, then exactly one of (n - 1) and (n + 1) is divisible by 3, because every third integer is divisible by 3. therefore, the product (n - 1)(n + 1) contains a 3 in its prime factorization. Just like st 1 this is not sufficient

the overall prime factorization of (n - 1)(n + 1) contains three 2's and a 3. therefore, it is a multiple of 24. sufficient

9.If n is a positive integer and r is the remainder when (n - 1)(n + 1) is divided by 24, what is the value of r ?

(1) n is not divisible by 2.

(2) n is not divisible by 3.

ST 1- if n is not divisible by 2, then n is odd, so both (n - 1) and (n + 1) are even. moreover, since every other even number is a multiple of 4, one of those two factors is a multiple of 4. so the product (n - 1)(n + 1) contains one multiple of 2 and one multiple of 4, so it contains at least 2 x 2 x 2 = three 2's in its prime factorization. But this is not sufficient, because it can be (n-1)*(n+1) can be 2*4 where remainder is 8. it can be 4*6 where the remainder is 0.

ST 2- if n is not divisible by 3, then exactly one of (n - 1) and (n + 1) is divisible by 3, because every third integer is divisible by 3. therefore, the product (n - 1)(n + 1) contains a 3 in its prime factorization. Just like st 1 this is not sufficient

the overall prime factorization of (n - 1)(n + 1) contains three 2's and a 3. therefore, it is a multiple of 24. sufficient

Answer C

What if N=1? 1 is not divisible by 2 or 3.

Please discuss specific questions in PS or DS subforums.

As for your question: if n=1 then (n - 1)(n + 1) equals to zero and zero is divisible by every integer (except zero itself) so by 24 too.
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Re: Collection of remainder problems in GMAT [#permalink]

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10 Apr 2011, 11:46

8.If n is a positive integer and r is the remainder when 4 + 7n is divided by 3, what is the value of r ?

(1) n + 1 is divisible by 3.

(2) n > 20

Answer A st1... n+1 divisible by 3..so n=2,5,8,11...... this gives 4+7n=18,39,60....remainder 0 in each case...... st2) insufficient ....n can have any value

Re: Collection of remainder problems in GMAT [#permalink]

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02 Jul 2011, 06:33

awesome set of questions...VERY VERY helpful
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Re: Collection of remainder problems in GMAT [#permalink]

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07 Aug 2011, 19:25

1

This post received KUDOS

Galiya wrote:

Quote:

st1. take multiples of 8....divide them by 4...remainder =1 in each case...

but how can it be so? multiples of 8 are also multiples of 4

hey

because P has a remainder of 5 after dividing by 8 , so P= (multiple of 8) + 5 take numbers ( 40 + 5 ) / 8 = 5 + ( 5/8 ) divide 45 by 4 the remainder is 1

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Re: Collection of remainder problems in GMAT [#permalink]

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07 Aug 2011, 22:15

Galiya wrote:

HI,Silver!) yep, I missed it. thanks a lot? +1

you are welcome
_________________

How can i lose my faith in life's fairness when i know that the dreams of those who sleep on the feathers are not more beautiful than the dreams of those who sleep on the ground? - Jubran Khaleel Jubran

Re: Collection of remainder problems in GMAT [#permalink]

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07 May 2012, 16:28

[quote="sandipchowdhury"]I have collected these problems on remainder. This type of problem is frequently asked in DS. Answers are also given.

3.If p is a positive odd integer, what is the remainder when p is divided by 4 ?

(1) When p is divided by 8, the remainder is 5.

(2) p is the sum of the squares of two positive integers.

st1. take multiples of 8....divide them by 4...remainder =1 in each case...

st2. p is odd ,since p is square of 2 integers...one will be even and other odd....now when we divide any even square by 4 v ll gt 0 remainder..and when divide odd square vll get 1 as remainder......so intoatal remainder=1 Ans : D

i don t aggree in your approach it is clearly said that P is not a multiple of 8 Since it has a remainder of 5 So your ST 1 ?

FOR sT2 Why do you consider that the sum of 2 square one is even and one odd do you considered that it is the sum of the Following square I am very confused

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