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Collection of remainder problems in GMAT

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Collection of remainder problems in GMAT  [#permalink]

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New post 13 Jan 2009, 11:57
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6. Remainders




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.

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.

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
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New post 12 Aug 2009, 03:56
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.
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New post 11 Nov 2009, 14:56
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.
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New post 04 Dec 2010, 08:36
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)
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New post 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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New post 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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New post 14 Dec 2010, 01:30
spyguy wrote:
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.


Remainders:
Theory: compilation-of-tips-and-tricks-to-deal-with-remainders-86714.html

PS questions:
remainder-101074.html
remainder-problem-92629.html
number-properties-question-from-qr-2nd-edition-ps-96030.html
remainder-when-k-96127.html
ps-0-to-50-inclusive-remainder-76984.html
good-problem-90442.html
remainder-of-89470.html
number-system-60282.html
remainder-problem-88102.html

All PS questions on remainders: search.php?search_id=tag&tag_id=199


DS questions:
remainder-problem-101740.html
remainder-101663.html
ds-gcd-of-numbers-101360.html
data-sufficiency-with-remainder-98529.html
sum-of-remainders-99943.html
ds8-93971.html
need-solution-98567.html
gmat-prep-ds-remainder-96366.html
gmat-prep-ds-93364.html
ds-from-gmatprep-96712.html
remainder-problem-divisible-by-86839.html
gmat-prep-2-remainder-86155.html
remainder-94472.html
remainder-problem-84967.html

All DS questions on remainders: search.php?search_id=tag&tag_id=198

Hope it helps.
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New post 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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New post 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 :wink:

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

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New post 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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New post 01 Jul 2017, 04:53
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,
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New post 28 Oct 2017, 08:40
when the positive integer x is divided by 9 the remainder is 5 . what is the remainder when 3x is divided by 9?
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New post 28 Oct 2017, 08:42
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

Please follow the rules when posting a question: https://gmatclub.com/forum/rules-for-po ... 33935.html
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New post 29 Apr 2018, 10:06

6. Remainders



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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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