Collection of remainder problems in GMAT

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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.

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

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

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.

Here is how I solved these problems:

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

Statement 1

Because 7 divides into 21 evenly, when a number is divided by 28, it will have the same remainder as when it is divided by 7.

In this case we are given that the remainder is an odd number, but it can be any odd number: 1,3,5,7,9

In fact, the way it is stated, the remainder can by any odd number less than 21: 11, 13, 15, 17, 19

This does not tell us anything.

INSUFFICIENT

Statement 2

A number divided by 28 will have the same remainder as a number divided by 7. Thus, r in this case is 3.

SUFFICIENT

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.

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

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."

why are we looking at every other even number?

Thanks!

Nice collection. Thank you.

great collection! I was just looking for remainders problems for practice.

Thank you very much.

Good collection. I like!

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)

Good collection. Bookmarked this page

sorry, why would it be different? seems the same to me, as long as m=1

Good collection. Thank you.

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

Also, what does it mean, "cycles of powers of 3"?

Thanks Man..it will help me a lot.

I have been having issues with these types of DS..thanks Man

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.

Thanks for the great post... I consistently have a difficult time on these kinds of problems.

Thanks again!