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Collection of remainder problems in GMAT [#permalink]

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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]

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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]

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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]

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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]

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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]

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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.

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

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21 Sep 2009, 03:47

Quote:

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

p is the sum of the squares of two positive integers

since p is odd we can take the two integers to be 2k+1,2k+3 So p=(2k+1)^2+(2k+3)^2 which wen expanded will give remainder 2 views?
_________________

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
_________________

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

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21 Sep 2009, 15:16

I have a question on the first problem posted in this thread:

Quote:

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

Based on the information in statement 1, why cannot the remainder be 7, 3 or 9. For instance, (24-3)/21=1. In this example, the remainder is 3.

Also, why is it that:

Quote:

The possible reminders can be 1,2,3,4,5 and 6

Why can't we assume that the remainder is 0 (meaning the number is a multiple of 7).

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

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11 Nov 2009, 13:56

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Remember the following:

over2u wrote:

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.

First we should note the following:

1) The square of an even number is always even. The sum of two even numbers is always even. Therefore, the sum of two even squares is always even.

2) The sum of an odd number is always odd. The sum two odd numbers is always even. Therefore, the sum of two odd squares is always even.

3) The sum of an even number and odd number is always odd. Therefore the sum of an even square and odd square is always odd.

Since we are told that P is odd, if it has to be a sum of two squares then it will have to be case 3. ie. sum of even square and odd square.

Now, P/4 = (sum of even square + sum of odd square)/4

All even numbers are divisible by 2. therefore remainder for first part is 0

Now, in order to proceed to the next step it is imp. to understand the following:

remainder of (x*y)/n = remainder of [(remainder of x/n)*(remainder of y/n)]/n

for eg, remainder of 20*27/25 = remainder of 20*2/25 = remainder of 40/25 = 15

or remainder of 225/13 = remainder of 15*15/13 = remainder of 2*2/13 = remainder of 4/13 = 4.

Now we know that an odd number when divided by 4 will leave remainder of either 1 or 3.

in our case we have (odd number)*(odd number)/4

since it is a square, both the odd numbers will be the same.

thus it can be simplified into either of the two cases: 1) when remainder for both is 1 ---> remainder of 1*1/4 = remainder of 1/4 = 1 2) when remainder for both is 3 ---> remainder of 3*3/4 = remainder of 9/4 = 1

Thus we can see that the square of an odd number when divided by 4 will always leave remainder 1.

As a result, correct answer for this question is (D).

Hope this helps.

Cheers.
_________________

Click below to check out some great tips and tricks to help you deal with problems on Remainders! http://gmatclub.com/forum/compilation-of-tips-and-tricks-to-deal-with-remainders-86714.html#p651942

Word Problems Made Easy! 1) Translating the English to Math : http://gmatclub.com/forum/word-problems-made-easy-87346.html 2) 'Work' Problems Made Easy : http://gmatclub.com/forum/work-word-problems-made-easy-87357.html 3) 'Distance/Speed/Time' Word Problems Made Easy : http://gmatclub.com/forum/distance-speed-time-word-problems-made-easy-87481.html

gmatclubot

Re: Collection of remainder problems in GMAT
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