GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 20 Nov 2018, 11:52

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

## Events & Promotions

###### Events & Promotions in November
PrevNext
SuMoTuWeThFrSa
28293031123
45678910
11121314151617
18192021222324
2526272829301
Open Detailed Calendar
• ### All GMAT Club Tests are Free and open on November 22nd in celebration of Thanksgiving Day!

November 22, 2018

November 22, 2018

10:00 PM PST

11:00 PM PST

Mark your calendars - All GMAT Club Tests are free and open November 22nd to celebrate Thanksgiving Day! Access will be available from 0:01 AM to 11:59 PM, Pacific Time (USA)
• ### Free lesson on number properties

November 23, 2018

November 23, 2018

10:00 PM PST

11:00 PM PST

Practice the one most important Quant section - Integer properties, and rapidly improve your skills.

# Collection of remainder problems in GMAT

Author Message
Manager
Status: Stanford GSB
Joined: 02 Jun 2008
Posts: 88
Collection of remainder problems in GMAT  [#permalink]

### Show Tags

13 Jan 2009, 10:57
56
137

6. Remainders

I have collected these problems on remainder. This type of problem is frequently asked in DS.

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.

Ans: E

st1) p+n=6,11,16....insuff.
st2) p-n=4,7,10....insuff...

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.

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.

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

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

Intern
Joined: 09 Aug 2009
Posts: 19
Re: Collection of remainder problems in GMAT  [#permalink]

### Show Tags

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.
Manager
Joined: 29 Oct 2009
Posts: 197
GMAT 1: 750 Q50 V42
Re: Collection of remainder problems in GMAT  [#permalink]

### Show Tags

11 Nov 2009, 13:56
2
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

= { [(even number)*(even number)]/(2*2) } + { [(odd number)*(odd number)]/(2*2) }

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

1) Translating the English to Math : http://gmatclub.com/forum/word-problems-made-easy-87346.html

Manager
Joined: 10 Sep 2010
Posts: 119
Re: Collection of remainder problems in GMAT  [#permalink]

### Show Tags

04 Dec 2010, 07:36
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)
Math Expert
Joined: 02 Sep 2009
Posts: 50711
Re: Collection of remainder problems in GMAT  [#permalink]

### Show Tags

04 Dec 2010, 07:58
2
1
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.
_________________
Intern
Joined: 05 Nov 2010
Posts: 43
Re: Collection of remainder problems in GMAT  [#permalink]

### Show Tags

13 Dec 2010, 19:55
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.
Math Expert
Joined: 02 Sep 2009
Posts: 50711
Re: Collection of remainder problems in GMAT  [#permalink]

### Show Tags

14 Dec 2010, 00:30
Manager
Joined: 15 Jan 2011
Posts: 101
Re: Collection of remainder problems in GMAT  [#permalink]

### Show Tags

07 Aug 2011, 10:31
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
Manager
Status: Trying to survive
Joined: 29 Jun 2011
Posts: 166
GMAt Status: Quant section
Concentration: Finance, Real Estate
Schools: WBS (D)
GMAT Date: 12-30-2011
GPA: 3.2
Re: Collection of remainder problems in GMAT  [#permalink]

### Show Tags

07 Aug 2011, 18:25
1
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

Intern
Joined: 25 Mar 2015
Posts: 7
Re: Collection of remainder problems in GMAT  [#permalink]

### Show Tags

06 Jul 2015, 04:48
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!
CEO
Joined: 20 Mar 2014
Posts: 2635
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
Re: Collection of remainder problems in GMAT  [#permalink]

### Show Tags

06 Jul 2015, 05:31
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.
Intern
Joined: 01 May 2017
Posts: 30
Re: Collection of remainder problems in GMAT  [#permalink]

### Show Tags

01 Jul 2017, 03:53
peraspera

if you see the "power cycle of 3" i.e 3^1, 3^2, 3^3, 3^4, 3^5, 3^6, 3^7, 3^8, 3^9, 3^10,...... the units digit is 3,9,7,1,3,9,7,1,3,9 and so on. SO you see the 3,9,7,1 pattern continues.

In the given question when m=1, we get 3^(4n+2+1)=3^(4n+3) . From the "power cycle of 3" we can see that every 4th power of 3 will end in 1 and 3 places ahead on the power cycle would have a unit digit of 7.

Remainder when a number is divided by 10 would be the units digit of that number. For the question above as we have solved the units digit would be 7 and hence remainder would be 7.

Hope this helps.

Regards,
Shreya
Intern
Joined: 25 Aug 2017
Posts: 1
Re: Collection of remainder problems in GMAT  [#permalink]

### Show Tags

28 Oct 2017, 07:40
when the positive integer x is divided by 9 the remainder is 5 . what is the remainder when 3x is divided by 9?
Math Expert
Joined: 02 Sep 2009
Posts: 50711
Re: Collection of remainder problems in GMAT  [#permalink]

### Show Tags

28 Oct 2017, 07:42
SHATHY wrote:
when the positive integer x is divided by 9 the remainder is 5 . what is the remainder when 3x is divided by 9?

Discussed here: https://gmatclub.com/forum/when-the-pos ... 56032.html

_________________
Math Expert
Joined: 02 Sep 2009
Posts: 50711
Re: Collection of remainder problems in GMAT  [#permalink]

### Show Tags

29 Apr 2018, 09:06
Re: Collection of remainder problems in GMAT &nbs [#permalink] 29 Apr 2018, 09:06
Display posts from previous: Sort by