# Intro. question

If a number divided by 56, leaves remainder 29. If the same number is divided by 8, then what will be the remainder? This type of questions is quite common in SSC exam.

Suppose we divide 9 by 4, then what will be the situation? We know that 4 x 2=8
And 8 +1=9. Therefore,
(4 x 2) + 1 =9. It means “1” is the remainder.
Similarly, if unknown number “N” is divided by 58 and we are getting 29 as remainder, we can write this as

(56 x q) + 29 = N

Distribute these things into green and red buckets in following manner.

 Green Red Total (Green + Red) 56 x q 29 N

Now forget the last column (total N).
Just concentrate on green and red bucket. Assume they’re filled with chocolates.

There is only one rule: Whatever number of chocolates can be perfectly divided by 8, must stay in green bucket. Everything else goes in red bucket.
First check the green bucket itself.
Divide 56 x q by 8.
We know that 56 = 8 x 7. Therefore 56 x q will always be divisible by 8.

 Green Red Total (Green + Red) 8 x 7 x q 29 N Perfectly divisible by 8

How about red bucket: 29 chocolates? Divide them with 8 you get
29 = (8 x 3) + Remainder 5
So shift (8 x 3) number of chocolates from red bucket to green bucket.

 Green Red Total (Green + Red) 8 x 7 x q + 8 x 3 5 N Perfectly divisible by 8 This is our remainder!

We are left with only 5 chocolates in red bucket. Therefore final answer= 5.

# DemoQ: 13s

Q. A number when divided by 65 gives the remainder of 43. If this same number is divided by 13, what’ll be the remainder?

Ans. if an unknown number “N” is divided by 65 and we are getting 43 as remainder, we can write it as (65 x q)+ 43=N
again, Visualize there are two buckets: green bucket and red bucket.

 Green Red Total (Green + Red) 65xq 43 N

There is only one rule: Whatever number of chocolates are perfectly divisible by 8, must stay in green bucket. Everything else goes in red bucket.
But I know that 65=13 x 5

 Green Red total 13 x 5 x q 43 N Perfectly Divisible by 13 Yet to check

Now check the red bucket. Divide it with 13
but 43= (13 x3)+4= (if you’re unsure, just divide 43 by 13 and get the remainder).
it means 13×3 number of chocolates can be shifted from red to green bucket.

 Green Red total (13 x 5 x q) + (13 x 3) 4 N Divisible by 13 Not divisible by 13.

That’s it. Red bucket is our remainder = “4”.
Let’s try one more

# Demo Q: 19s

If a number divided by 114, leaves remainder 21. If the same number is divided by 19, then the remainder will be
(114 x q) + 21 = unknown number N

 Green Red Total (Green + Red) 114 x q 21 N

Divide these buckets with 19. Whatever number of chocolates are perfectly divisible by 19, must stay in green bucket.
First the green bucket itself.
114 = 19 x 6

 Green Red Total (Green + Red) 19 x 6 x q 21 N Perfectly divisible by 19

How about red bucket: 21 chocolates? Divide them with 19 you get
21 = (19 x 1) + Remainder 2
So shift (19 x 1) number of chocolates from red bucket to green bucket.

 Green Red Total (Green + Red) 19 x 6 x q + 19 x 1 2 N Perfectly divisible by 19 This is our remainder!

So what’s the “theme/trend/moral” of these type of questions?
You’ve three numbers A > B > C
If A is perfectly divisible by C
Then just divide “B” by C and whether remainder you get, is your answer.

# DemoQ: 31s (with shortcut)

A Number divided by 899 leaves remainder of 65. When this number is divided by 31, what will be the remainder?
Arrange A> B> C
899 > 65 > 31
Check: is 899 perfectly divisible by 31? Yes. 31 x 29 =899. (if you’re not comfortable with such two digit division, go through previous article click me)
Now divide 65 (B) by 31 and whatever remainder you get, is your answer.
65 = (31 x 2) + 3
Therefore, final answer (remainder is 3)
Let’s complicate the situation

# Complex situation: higher number

A number when divided by 31, leaves a remainder of 29. Find the remainder when same number is divided by 62

1. 29
2. 60
3. Either a or b
4. None of above.

We’ll have go back to red bucket, green bucket concept.
Question says A number when divided by 31 gives 29 remainder
Therefore: (31 x q) + 29 = N

 Green Red Total (Green + Red) 31 x q 29 N

Now divide both buckets with 62. Whatever if perfectly divisible by 62, must be shifted to green bucket.
First check the green bucket, is it perfectly divisible by 62?

 Green Red Total (Green + Red) 31 x q 29 N Can’t say unless we know the value of “q” Not divisible

Two things can happen: either q=1 or q=2,3,…..or some big number.
If q=1

## Situation 1: q=1

 Green Red Total (Green + Red) 31 x 1 29 N But this is not divisible by 62! Not divisible

Since 31 is not divisible by 62, it cannot stay in green bucket.
It must be shifted to red bucket.

 Green Red Total (Green + Red) 0 29+31=60 N Not divisible by 62

Thus in first situation (when q=1) we get remainder 60.

## Situation#2: q=2

 Green Red Total (Green + Red) 31 x 2=62 29 N This is perfectly divisible by 62. Not divisible by 62

In this case, you get remainder =29.
Similarly, in any other situation like q=3 or q=5, 7, 9…you get remainder 29. Note: you’ll get same remainder 29, if q=0.
Therefore, two remainders are possible: Either 60 or 29

# Mock questions

1. If a number divided by 27, leaves remainder 23. If the same number is divided by 9, then what will be the remainder? This type of questions is quite common in SSC exam.
2. If a number divided by 30, leaves remainder 17. If the same number is divided by 15, then what will be the remainder? This type of questions is quite common in SSC exam.
3. A number when divided by 551, leaves a remainder of 31. Find the remainder when same number is divided by 29
4. A number when divided by 9, leaves a remainder of 7. Find the remainder when the same number is divided by 18
5. A number when divided by 11, leaves a remainder of 5. Find the remainder when the same number is divided by 33
6. A number when divided by 13, leaves a remainder of 5. Find the remainder when the same number is divided by 52
7. A number when divided by 28, leaves a remainder of 7. Find the remainder when the same number is divided by 35.

 Q.No. Answer 1 5 2 2 3 2 4 7 or 16 5 5, 16 or 27 6 5, 18, 31 or 44 7 0, 7, 14, 21 or 28

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## So far 50 Comments posted

thnx mrunal paji,u hav made d concept really easy…(by d way it contains sum mistakes in figures,correct them if possible to remove any ambiguity)

Mrunal ji,i would like to draw your attention to DemoQ: 13s
wherein you have mentioned 43=13×4+4,it should be 43=13×3+4
Answer would be same i.e. 4

Again bothering you,somehow i missed it earlier,
in Demo Q: 19s, it should be 114=19×6 instead of 19×4

3. anand s

Sir,
i am commerce student so pls tell me commerce book list for ias mains.

4. satyendra

sir, nice efforts!!!
i have doubts in (complex situation :higher number)
for given example, if no is 91, then divided by 62 then reminder will be 29(91%62 =29) not 60. so how do you correct this problem that reminder will be 60 or 29 ???

5. nav

Mrunal sir,
please also explain HOW TO FIND THE LAST DIGIT OF A GIVEN POWER WITH SOME OPERATION (+,*) etc WITH TWO OR MORE OPERATOR as it is asked many times in SSC CGL

Q the remainder when 9^19+6 is divided by 8 (SSC 2012)
option—– 5,7,3,2 ????

1. Abhimanyu

7 is the ans .. Use BINOMIAL THEOREM

6. Sujay Shrivastava

Love you Sir…. (Please m not a **y)

7. gaurav

Sir if possible please explain the 7th mock ques.

8. Rohit

Mrunal sir please give me some tips regarding reading comprehension…lately I am practicing lot of passages but still cant improve on this topic…Please help in how to find Conclusions,Inferences and Assumptions and Theme of the passage…

9. k

thnx a lot mrunal sir!

10. Vinay Kumar

Hi Mrunal,

Please come up with some English Section questions as well. Even if you list out some questions from comprehension section here, this would indeed be useful as it will give some practice session. from exam point of view it would be very important because they cover high weight-age and consume much time in solving.
Hope you will understand our concerns.

Thanks and best regards,
Vinay Kumar.

1. Mrunal

I’m working on an article on how to approach comprehension for CSAT. it’ll be published soon.

11. praveen kumar

demo 13s attention plz
There is only one rule: Whatever number of chocolates are perfectly divisible by 8, must stay in green bucket. Everything else goes in red bucket.
But I know that 65=13 x 5
(its must be divisible by 13 not 8 typing mistake :P)

12. SATYA

i want to know abt capital market..plz suggest me way?

13. Bhushan Khedkar

Hi sir,

Thanks for such a nice explanation of each and every aspect of UPSC exam.

I have doubt regarding Q.7 mock test can you explain how answer 0, 7, 14, 21 or 28 is derived.

Regards,
Bhushan

1. Mrunal

7, 28 and 35 are all in the table of 7 so when you keep testing with q=0,1,2,3… you’ll see the remainder keeps getting “cycle or pattern”.
Basically it’ll be from zero to a number below 35 in 7’s table.

The questions like what will be the remainder when (594)^798 is divided by 7 (or any other number) = are solved based on this principle (which we will see in future article.)

1. MOUMITA

sir u mentioned about a article to be uploaded soon…solving what will b the reminder when (594)^798 is divided by 7…if yes plz give the link

2. SMK

Mrunal,

If we apply A>B>C then we can arrange 35>28>7 here C=7
and if I divide the A/7 ,So A is perfectly divisible by 7 and then if I divide B/7 i will get 0 as the remainder so the Answer must have been only 0 and not 0,7,14…
“So what’s the “theme/trend/moral” of these type of questions?
You’ve three numbers A > B > C
If A is perfectly divisible by C
Then just divide “B” by C and whether remainder you get, is your answer”

As per this concept we must arrange the given numbers as A>B>C and A is the largest of the 3.—(I feel i have understood this concept incorrectly.Plz correct me.)

I am bit confused.Plz clarify.

Thanks for the great work.

14. emm

in Q13 u ‘ve written…There is only one rule: Whatever number of chocolates are perfectly
divisible by 8,…it should be 13 inplace of 8

15. Amrinder

Thanks for this.

16. lily doley

sir,plz give me some suggestions regarding substance writng and spotting error in sentence.m alwyz confuse abt it…

17. jawahar

there is another simple short cut…
for example take the 7th qus.
let the number be 28+7=35 which is exactly multiplied by 35.so the remainder is 0
then 28*2+7=63 when divided by 35 leaves the remainder 28.
28*3+7=91 divided by 35 leaves 21
28*4+7=119 divided by 35 leaves 14
28*5+7=147 divide by 35 leaves 7
28*6+7=175 divided by 35 leaves 0….so the remainders are 0,28,21,14,7.
this may be much more easier..i think so..

1. Optimus

Just saw yours. Much much easier lol. Thank you.

18. divyasri

thank u mrunal sir……..

19. Parthi

Easy to absorb and very empathetic of you

hats off to your helping nature

there’s a small correction

in Situation#2: q=2

2nd line below the table… q should = 0,2,4…

20. Rohit Jagtap

Even simpler method :
1.If a number divided by 27, leaves remainder 23. If the same number is divided by 9, then what will be the remainder? This type of questions is quite common in SSC exam

Just divide the first remainder by next divisor -> the remainder is your ANSWER !
in this case : 23/9 : remainder = 5 (ANSWER !)

21. kavan

hi mrunal,
i think there is a simpler method. for xample
Q: a number is devided by 56 and give remainder 29. what would be the remainder if the same number is divided by 8?
ans. assume that the number is 29. i.e. minimum such number possible.
divide that number by 8 the rwmainder is 5.
done.

for greater numbers the calculation is similar except that you have to add alternatives.

Sir, I am not able to solve last mock ques i.e. no.7

Kindly help

23. Akanksha

Hey Sir,
I got stuck in one question.
Thnx

Q : A number when divided by 5 gives a number which is 8 more than the remainder obtained on dividing the same number by 34. Such a least possible number would be –

1. HK

number is 95..

number is multiple of 5.. let it be x

x/5 = y(suppose)

(x-(y+8)) = multiple of 34

(x-(y+8))/34 = N next put n = 1 2 3 4 and solve for least value of n (N= 2)

x = 95

95/5 = 19

95/34 remainder = 27 that is 19 + 8 ..

i’d rather go for putting options..

1. Akanksha

thanks HK 🙂

2. ak

x/5=y

y=remainder(r)+8

x/34=a+(r/34)

when we take all 3 above and eliminate r, we get

x=(85a-20)/2

a is an integer and a=1,2,3,4….

so minimum value comes for x = 75 for a=2 as in case of a=1; x is not an integer

make correction if wrong

24. DDDD

hello sir plz provide some english grammer notes like parts of speech….I have many doubts by grammer books..in english grammer books there are lots of different different rules and lotes of confusion so plz give me clear notes in parts of speech..

25. arjun

When a number is divided by 13, the remainder is 6. When the same number is divided by 7, then remainder is 1. What is the number ?

1. Mudasir

Simple dudu ,71

1. abc

How did u do that?

26. Vignesh

Sir I have one doubt….If a number when divided by 5 leaves a remainder 2 and when the same number is divided by 7 leaves a remainder 4.Find the remainder when the same number is divided by 35.Can we Solve it by the above method???????

27. neeharika

can any one tell me best books for aptitude, reasoning and english comprehension and grammar book for csat paper 2 & ssc cgl

1. ashwin

QUANTUM CAT by sarvesh verma. I think this is a good one and it helped me in both CSAT P2 and CGL.

28. Anand Jwala

First go through old question papers.You will get to know from where to pick as many student possess some experience in dealing with questions. For english comprehension go for Arun Sharma TMH,For reasonin rely on BSC .Wren & Martin for Grammar. For aptitude follow CL or Bsc books.

1. neeharika

what abt rs agarwal reasoning and aptitude

in studyplan of csat2014, u have mentioned that dont use R S Agrawal but in your maths practise u have used this source e.g solution of HCF, LCM

30. Amarjit

[Aptitude Q] Mixture and Alligiation: Change Alcohol concentration from 15% to 32% not working….

31. Weknowth . . .

7*5=35 so obviously take the remainder of big number (here 7 ) 4 is remainder for 7.

32. chakrapani

Sir
If 578xy6 is devisivle by 18 then find the maximum value of x
1. 6
2. 7
3. 8
4. 9

1. sam

33. Arindam Paul

Thanks its very helpful for me.Just some confusion remain in higher situation you have explained.If you get time pls make me clear about this in my mail