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Re: Collection of remainder problems in GMAT [#permalink]
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04 Dec 2010, 08:58
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Fijisurf wrote: 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. So r can equal to 1, 3, 5 or 0. Hope it helps.
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Re: Collection of remainder problems in GMAT [#permalink]
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13 Dec 2010, 13:43
Sandipchowdhury, thanks for the collection
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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.



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14 Dec 2010, 01:30



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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



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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 (n1)*(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.



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Re: Collection of remainder problems in GMAT [#permalink]
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01 Jan 2011, 03:21
girishkakkar wrote: 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 (n1)*(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. This question is discussed here: dsfromgmatprep96712.htmlAs 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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03 Jan 2011, 05:44
great collection. Thanks



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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
Another approach: 4 + 7n = 4 + 4n + 3n = 4(n+1) + 3n
From statement 1) we can see that 4(n+1) and 3n are divisible by 3 so the remainder is 0. From statement 2) " insufficient ....n can have any value"



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02 Jul 2011, 06:33
awesome set of questions...VERY VERY helpful
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07 Aug 2011, 11:31
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



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07 Aug 2011, 19:25
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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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07 Aug 2011, 22:06
HI,Silver!) yep, I missed it. thanks a lot? +1



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07 Aug 2011, 22:15
Galiya wrote: HI,Silver!) yep, I missed it. thanks a lot? +1 you are welcome
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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
please help
best regards



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07 Jul 2012, 10:52
good stuff, book making for future reference.
Thank you



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Re: Collection of remainder problems in GMAT [#permalink]
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18 Apr 2014, 03:01
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
My method  1) n+1 is divisible by 3 => 7n + 4 can be written as (4n+3n + 4)
now, check for divisibility [4(n+1) + 3n]/3 => 4(something divisible by 3)/3 + 3n/3
hence, the expression is divisible by 3.
2) n > 20 not sufficient, as different values of r are possible.



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06 Jul 2015, 05:48
girltalk wrote: Nach0 wrote: sandipchowdhury wrote: 2 If n and m are positive integers, what is the remainder when 3^(4n + 2 + m) is divided by 10 ? (1) n = 2 (2) m = 1
The Concept tested here is cycles of powers of 3.
The cycles of powers of 3 are : 3,9,7,1
St I) n = 2. This makes 3^(4*2 +2 + m) = 3^(10+m). we do not know m and hence cannot figure out the unit digit.
St II) m=1 . This makes 3^(4*n +2 + 1). 4n can be 4,8,12,16... 3^(4*n +2 + 1) will be 3^7,3^11, 3^15,3^19 ..... in each case the unit digit will be 7. SUFF Hence B
I got this question on my GMAT Prep and got it wrong. But after looking at the question again, it seems like I got a different equation. Mine is: 3^(4n+2)+m Which makes a big difference!!! I answered C, which I think would be right for THIS equation, not the 3^(4n+2+m) sorry, why would it be different? seems the same to me, as long as m=1 Can someone explain why "3^(4*n +2 + 1) will be 3^7,3^11, 3^15,3^19 ..... in each case the unit digit will be 7." I know this post is long dead, but I'm trying my luck!



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06 Jul 2015, 06:31
GeeWalia wrote:
Can someone explain why "3^(4*n +2 + 1) will be 3^7,3^11, 3^15,3^19 ..... in each case the unit digit will be 7."
I know this post is long dead, but I'm trying my luck! For unit's digit, you need to remember a couple of things: Concept of cyclicity: cyclicity of 3 is 4. This means that unit's digit wrt 3 as the base will repeat after sets of 4 Example, 3^1 =3, 3^2=9, 3^3=27, 3^4=81, 3^5=243 etc . So you see you have unit digits as {3,9,7,1}, {3,9,7,1} etc . So the digits repeat after every set of 4. Thus, 3^15 would have the same unit digit as 3^3 as 3^15 = 3^(12+3)= same digit as 3^3 as 12 is a multiple of 4. 3^7, 3^11... will be tackled in a similar fashion. Hope this helps.




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