# 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

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

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

2. 7*5=35 so obviously take the remainder of big number (here 7 ) 4 is remainder for 7.

3. Sir