- Intro. question
- DemoQ: 13s
- Demo Q: 19s
- DemoQ: 31s (with shortcut)
- Complex situation: higher number
- Mock questions

# 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

- 29
- 60
- Either a or b
- 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

- 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.
- 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.
- A number when divided by 551, leaves a remainder of 31. Find the remainder when same number is divided by 29
- A number when divided by 9, leaves a remainder of 7. Find the remainder when the same number is divided by 18
- A number when divided by 11, leaves a remainder of 5. Find the remainder when the same number is divided by 33
- A number when divided by 13, leaves a remainder of 5. Find the remainder when the same number is divided by 52
- A number when divided by 28, leaves a remainder of 7. Find the remainder when the same number is divided by 35.

## Answers

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 |

## 52 Comments on “[Aptitude] Remainder: One number and two divisors (Number Theory)”

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

fixed, thanks

Again bothering you,somehow i missed it earlier,

in Demo Q: 19s, it should be 114=19×6 instead of 19×4

:->

Sir,

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

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 ???

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

please help me to solve this:

Q the remainder when 9^19+6 is divided by 8 (SSC 2012)

option—– 5,7,3,2 ????

7 is the ans .. Use BINOMIAL THEOREM

have to find the unit digit of 9^19,, i,e, 9. add 9+6=15..divde 15 by 8 and the reminder is 7..

Love you Sir…. (Please m not a **y)

Sir if possible please explain the 7th mock ques.

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…

thnx a lot mrunal sir!

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.

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

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)

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

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

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.)

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

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.

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

Thanks for this.

Please help us with difficult topics (for SSC CGL): Algebra, Geometry, Trigonometry , etc

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

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..

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

thank u mrunal sir……..

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…

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 !)

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

Hey Sir,

I got stuck in one question.

Could you please help me out.

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 –

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..

thanks HK :)

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

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..

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 ?

Simple dudu ,71

How did u do that?

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???????

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

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

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.

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