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Collection of remainder problems in GMAT [#permalink]
13 Jan 2009, 10:57

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I have collected these problems on remainder. This type of problem is frequently asked in DS. Answers are also given. Please dont mind any typo error.

1.If r is the remainder when the positive integer n is divided by 7, what is the value of r

1. when n is divided by 21, the remainder is an odd number 2. when n is divided by 28, the remainder is 3

The possible reminders can be 1,2,3,4,5 and 6. We have the pinpoint the exact remainder from this 6 numbers.

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. Hence there can be two remainders ,1 and 5, when divided by 7. NOT SUFFICIENT

St 2: when n is divided by 28 the remainder is 3. As 7 is a factor of 28, the remainder when divided by 7 will be 3 SUFFICIENT

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

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

4.If p and n are positive integers and p > n, what is the remainder when p^2 - n^2 is divided by 15 ?

(1) The remainder when p + n is divided by 5 is 1.

(2) The remainder when p - n is divided by 3 is 1.

multiply these two to get p^2-n^2.....multiplying any ttwo values from the above results in different remainder......

also can be done thru equation....p+n=5a+1..and so on

5.What is the remainder when the positive integer x is divided by 3 ?

(1) When x is divided by 6, the remainder is 2.

(2) When x is divided by 15, the remainder is 2.

Easy one , answer D

st 1...multiple of 6 will also be multiple of 3 so remainder wil be same as 2.

st2) multiple of 15 will also be multiple of 3....so the no.that gives remaindr 2 when divided by 15 also gives 2 as the remainder when divided by 3...

6.What is the remainder when the positive integer n is divided by 6 ?

(1) n is a multiple of 5.

(2) n is a multiple of 12.

Easy one. Answer B

st 1) multiples of 5=5,10,15....all gives differnt remainders with 6

st2)n is divided by 12...so it will be divided by 6...remainder=0

7If x, y, and z are positive integers, what is the remainder when 100x + 10y + z is divided by 7 ?

(1) y = 6

(2) z = 3

We need to know all the variables. We cannot get that from both the statements. Hence the answer is E.

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

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

Re: Collection of remainder problems in GMAT [#permalink]
04 Dec 2010, 07:58

2

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Expert's post

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.

multiply these two to get p^2-n^2.....multiplying any ttwo values from the above results in different remainder......

also can be done thru equation....p+n=5a+1..and so on

wen we take the two stmts together, (5k+1)(3k+1)/15 = 15k+8k+1/15 so we have to check whether 8k+1 is divisible putting values,we get 9,17,25,33,41...therefore not divisible.(C) IMO

we have been asked to find the value of the remainder. not whether it is divisible or not. Hence answer is (E) _________________

Re: Collection of remainder problems in GMAT [#permalink]
07 Aug 2011, 18:25

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

_________________

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]
07 Feb 2009, 19:29

good stuff there sandip..helps a lot.

I have follow up question for the last problem: I couldn't quite follow explanation of statement 1.. starting on.. "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."

Re: Collection of remainder problems in GMAT [#permalink]
16 Apr 2009, 18:15

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)

Re: Collection of remainder problems in GMAT [#permalink]
27 Apr 2009, 09:25

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

Re: Collection of remainder problems in GMAT [#permalink]
28 Apr 2009, 20:18

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

I also got a problem with the solution for this one. I get C here. Could someone pls explain?

Re: Collection of remainder problems in GMAT [#permalink]
12 Aug 2009, 02:56

Quote:

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 disagree with step 2 Even number divided by 4 could have a remainder 2 or 0. You can check it if you plug 3 and 11 as two positive integers, than you will have 3^2+11^2=130 and 130 divided by 4 will have remainder 2. However the first part is correct, so in my choise A will be a good answer. By the way could anyone proove that a^2+b^2 divided by 4 will never have 3 as remainder. I now that it is true but I can't prove it.