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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)
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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

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

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11 Nov 2009, 15:12

1

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Quote:

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).

IMO, you are right in both instances.

It is possible for 24 to have a remainder of any number from 0 to 23. (Unless there is some rule we are not aware of.. )

(can someone please confirm that this is true?)

And, 7 can have remainders of 0,1,2,3,4,5 or 6.
_________________

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

As you can see the remainders repeat every four numbers: [3,9,7,1]

So now when we look at 3^(4n + 2 + m), we can see that 4*n is not relevant to determining what the remainder is. What we really need to know is the value of m.

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

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14 Nov 2009, 04:40

Hi guys,

I'm trying to get together a collection of tips and tricks for remainder problems. Please visit the link below and let me know if you have anything interesting to add.

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

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

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14 Nov 2009, 06:00

sriharimurthy wrote:

Hi guys,

I'm trying to get together a collection of tips and tricks for remainder problems. Please visit the link below and let me know if you have anything interesting to add.

A is twice as good a workman as B and together they finish a set of work in 14 days. The number of days taken by A alone to finish the work is: 1) 11 2)21 3)28 4)30

see my solution below

Ratio of A and B = A:B = 2:1 A+B 's together work = 14 days Divide 1/14 into ratio 2:1 so A's work = 1/14*2/1 = 1/7 means 7 days--- so ans shud be 7... but there is no options given.

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

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20 Jan 2010, 03:53

sandipchowdhury wrote:

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.

Quote:

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

Answer C

For first question, St 1: when n is divided by 21 ( 7 and 3) the remainder is an odd number.

Quote:

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 i agree not sufficient but why we do say so : when n is divided by 21 ( 7 and 3) "the remainder is an odd number.But it cannot be 7, 3 or 9 " Look at an example of 45 , it gives 3 as a reminder.

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

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20 Apr 2010, 10:18

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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

if n is a odd positive integer , why can't it be 1? I see that you assumed n to be 3

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

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02 Dec 2010, 15:33

sriharimurthy wrote:

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).

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

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03 Dec 2010, 14:43

Thank you for posting these problems. (+1 Kudos). I could not follow the explanation for the one below.

sandipchowdhury wrote:

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

My question is why when n is divided by 21 the remainders cannot be 7, 3, 9? What about when 24 is divided by 21? My reasoning was that remainder can be any odd integer between 1 and 19 inclusive... Thanks.

Thank you for posting these problems. (+1 Kudos). I could not follow the explanation for the one below.

sandipchowdhury wrote:

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

My question is why when n is divided by 21 the remainders cannot be 7, 3, 9? What about when 24 is divided by 21? My reasoning was that remainder can be any odd integer between 1 and 19 inclusive... Thanks.

PLEASE DISCUSS SPECIFIC QUESTIONS IN DS OR PS SUBFORUMS (post a problem there and ask a question).

If r is the remainder when the postive integer n is divided by 7, what is the value of r ?

Positive integer \(a\) divided by positive integer \(d\) yields a reminder of \(r\) can always be expressed as \(a=qd+r\), where \(q\) is called a quotient and \(r\) is called a remainder, note here that \(0\leq{r}<d\) (remainder is non-negative integer and always less than divisor).

(1) when n is divided by 21 the remainder is an odd number --> \(n=21q+odd=7*3q+odd\), now as \(21q\) is itself divisible by 7 then if \(odd=1\) then \(n\) divided by 7 will yield the same reminder of 1 BUT if \(odd=3\) then \(n\) divided by 7 will yield the same reminder of 3. Two different answers, hence not sufficient.

Or try two different numbers for \(n\): 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. Two different answers, hence not sufficient.

(2) when n is divided by 28, the remainder is 3 --> \(n=28p+3=7*(4p)+3\), now as \(28p\) is itself divisible by 7, then \(n\) divided by 7 will give remainder of 3. Sufficient.

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

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04 Dec 2010, 08:25

Bunuel wrote:

(1) when n is divided by 21 the remainder is an odd number --> \(n=21q+odd=7*3q+odd\), now as \(21q\) is itself divisible by 7 then if \(odd=1\) then \(n\) divided by 7 will yield the same reminder of 1 BUT if \(odd=3\) then \(n\) divided by 7 will yield the same reminder of 3. Two different answers, hence not sufficient.

Or try two different numbers for \(n\): 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. Two different answers, hence not sufficient.

(1) when n is divided by 21 the remainder is an odd number --> \(n=21q+odd=7*3q+odd\), now as \(21q\) is itself divisible by 7 then if \(odd=1\) then \(n\) divided by 7 will yield the same reminder of 1 BUT if \(odd=3\) then \(n\) divided by 7 will yield the same reminder of 3. Two different answers, hence not sufficient.

Or try two different numbers for \(n\): 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. Two different answers, hence not sufficient.

Thanks. It makes sense now!

That's correct. But what's your point?
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