ARITHMETICMATHSRemainder & Divisibility

QUESTIONS ON REMAINDERS & DIVISIBILITY (PART-I)

QUESTIONS ON REMAINDERS & DIVISIBILITY (PART-I)

QUERY 1

Finding out smallest number which leaves same remainder with different divisors. Type # 1.

Find Smallest number other than 4, that leaves remainder 4 when divided by 6, 7, 8 or 9.

A) 504
B) 500
C) 508
D) 516

JAYANTCHARAN CHARAN
To solve such questions, take LCM of the divisors and add the remainder to it.
Now, LCM of 6, 7, 8, 9 = 504
Therefore the required number = 504 + 4 = 508 (option ‘C)

QUERY 2

Finding out smallest number when the difference between the divisors and their remainders is same. Type # 2

Find smallest number that leaves remainder 3, 5, 7 when divided by 4, 6, 8 respectively.

A) 23
B) 25
C) 35
D) 37

JAYANTCHARAN CHARAN
Unlike the case mentioned at TYPE # 1, this time the remainder is not the same; but if you see carefully the difference between the divisor and remainder is having a certain trend. i.e.
4 – 3 = 6 – 5 = 8 – 7=1

In such questions, take LCM of divisors and subtract the common difference from it.

Now the LCM of 4, 6, 8 = 24
Therefore the required number here = 24 – 1 = 23 (option ‘A’)

QUERY 3

Finding out smallest number when the difference between the divisors and their remainders is same; and leaves different remainder when divided by another number. Type # 3

Find smallest number that leaves remainder 3, 4, 5 when divided by 5, 6, 7 respectively and leaves remainder 1 when divided by 11.

A) 208
B) 426
C) 600
D) 628

JAYANTCHARAN CHARAN
We have just seen above in TYPE-2 how to tackle the first part of the questionThus the number for the first part would be the [(LCM of 5, 6, 7) – (Common difference of divisors and their remainders)] i.e. 210 – 2 = 208

Here now, we have one more condition to satisfy i.e. remainder 1 when divided by 11

Here we should remember that if LCM of the divisors is added to a number; the corresponding remainders do not change i.e if we keep adding 210 to 208… the first 3 conditions for remainders will continue to be fulfilled.

Therefore now, let 208 + 210k be the number that will satisfy the 4th condition i.e. remainder 1 when (208 + 210k)/11

Now let’s see how
The expression (208 + 210k)/11 = 208/11 + 210k/11
Now the remainder when 208 is divided by 11 = 10
And remainder when 210k is divided by 11 = 1*k = k

Therefore the sum of both the remainders i.e. 10 + k should leave remainder 1 on division of the number by 11
Obviously k = 2
Hence the number = 208 + 210*2 = 628 (option ‘D’)

QUERY 4

Finding out smallest number which leaves same remainder with different divisors; and leaves different remainder when divided by another number. Type # 3…………continued

Find smallest number that leaves remainder 2 when divided by 3, 4, 5; and is fully divisible by 7 (means zero as remainder).

A) 218
B) 282
C) 182
D) 176

JAYANTCHARAN CHARAN
Here the first part of the question is like TYPE # 1
Hence the desired number for that = LCM of 3, 4, 5 + the remainder = 60 + 2 = 62

Now we have one more condition to satisfy i.e. the number is fully divisible by 7 (means zero as remainder); obviously we’ll need to follow the same thing as has been described in TYPE-3 above
Therefore now, let 62 + 60k be the number that will satisfy the 4th condition i.e. remainder zero when (62 + 60k)/7

The expression (62 + 60k)/7 = 62/7 + 60k/7
Now the remainder when 62 is divided by 7 = 6
And remainder when 60k is divided by 7 = 4*k = 4k

Therefore 6 + 4k should leave remainder zero on division of the number by 7 (the number is divisible by 7)
So k must be 2

Hence the number = 62 + 60*2 = 182 (option ‘C’)

QUERY 5

Finding out smallest number when there is no relation in the trends of divisors and remainders. Type # 4

Find smallest number that leaves remainder 1 with 5, 4 with 7, 6 with 11 and 7 with 13.

A) 500
B) 501
C) 478
D) 505

JAYANTCHARAN CHARAN
We can see there’s no relation among these divisors-remainders sets……..neither the remainder nor the difference between divisor and remainder is having a trend.

In such cases take 1 case and target another case…e.g. First take the case 7 with 13……..and target 6 with 11.

We see that when the number is divided by 13 the remainder is 7 (means 7 is the smallest number that we get as remainder. So all numbers of the form 7 + 13k will give 7 as remainder on division by 13.

Now target 6 with 11…………so we will have to divide 7 + 13k by 11
By doing so the remainder = 7 + 2k
Obviously for minimum value of k = 5
Therefore the number = 7 + 13*5 = 72

Now that 2 conditions are fulfilled, lets target the third condition…say 4 with 7.
Now, if we add LCM of 11 and 13 i.e 143 to 72, 2 conditions already satisfied would continue being satisfied…
Hence the number is of the form 72 + 143 k.
Thus the remainder when 72 + 143k is divided by 7 = 2 + 3k
So the minimum value of k = 3
Therefore the number satisfying all 3 condition considered = 72 + 143*3 = 501

Now, if we see carefully…..the 4th condition i.e. ‘remainder 1 with 5’ considering 501 as the number satisfies itself.
Hence 501 is the appropriate answer here. (option ‘B’)

NOTE-I: For ease of calculation, start from biggest divisor and gradually move to smaller ones…you’ll always see that last 1-2 conditions will be satisfied automatically.

NOTE-II: There are theorems for solving above type of questions…viz Chinese theorem etc…but I find this approach very practical and easy. And I believe I can tackle any twist in the question devised by Cat Makers through this method.

NOTE-III: At times such a question can be solved by just seeing the answer options.

QUERY 6

Finding out smallest number when there is no relation in the trends of divisors and remainders. Type # 4………..Continued

Find smallest number that must be subtracted from 1000 so that the resultant number leaves remainder 1, 3, 4, 8 with divisors 2, 6, 5, 13 respectively.

A) 121
B) 125
C) 213
D) 135

JAYANTCHARAN CHARAN
From TYPE 4 we can find out the smallest number satisfying all 4 conditions is 99.
Obviously if we add LCM of 2, 6, 5, 13 i.e. 390 to 99 the remainders will remain unchanged.

So we need such value of 99 + 390k so that the resultant number is just below 1000.
Easily we can see that for k = 2, we get that such value which is 99 + 390*2 = 879.

Hence the required number = 1000 – 879 = 121 (option ‘A’)

QUERY 7

Finding out smallest number which leaves specific remainders when SUCCESSIVELY divided by two or three numbers. Type # 5

Find the smallest number that leaves remainders 3, 2, 4 when successively divided by 5, 6, 7 respectively.

A) 425
B) 133
C) 135
D) 335

JAYANTCHARAN CHARAN
For such questions start doing from the rear end.

So, we want 4 as the remainder with 7 as the divisor; of course the smallest such number is 4 itself.

Obviously this 4 must have come as the quotient when the number was divided by 6
Therefore the number must have been 4*6 + 2 = 26 ……….(see 2 is the remainder with 6 as divisor)

It’s clear this 26 was the quotient when the number was divided by 5
Therefore the number must have been 26*5 + 3 = 133 ……….(see 3 is the remainder with 5 as divisor)

Hence the required number is 133 (option ‘B’)

QUERY 8

Finding out smallest number when divided by the LCM of the earlier divisors. Type # 6

A number leaves remainder 3 when divided by 5 and remainder 8 when successively divided by 11. What is the remainder when this number is divided by 55?

A) 53
B) 43
C) 48
D) 50

JAYANTCHARAN CHARAN
If looked carefully we’ll find that 55 is the LCM of earlier divisors 11 and 5.

It would be easier if we start solving such a question from the rear end
Now take the smallest number that leaves remainder 8 with 11 as the divisor. Obviously it’s 8.

Thus according to the question this 8 will be the quotient when the main number is divided by 5. We can see that it also leaves remainder 3 as given in the question.

Hence, the main number = 8*5 + 3 = 43

Now dividing 43 by 55 we see that the remainder is 43 itself.
Hence the answer 43

So 43 is the answer (option ‘B’)

QUERY 9

Finding out the largest number that leaves the same remainder when it divides two different numbers. Type # 7

Find the largest number that leaves same remainder when it divides 3398 and 6578.

A) 3180
B) 3115
C) 2065
D) 2290

JAYANTCHARAN CHARAN
This concept is very much simple to understand. Hence it could easily be understood that the difference between two dividends must be divisible by the divisor.

Therefore 6578 – 3398 = 3180 should be divisible by the divisor to leave same remainders.

You can see the largest number that divides 3180 is 3180 itself.

Hence, the answer is 3180 (option ‘A’)

QUERY 10

Finding out the largest number that leaves the same remainder when it divides more than two different numbers. Type # 7……….Continued

Find the largest number that leaves same remainder when it divides 16009, 9009, 7509 and 14009.

A) 525
B) 480
C) 235
D) 500

JAYANTCHARAN CHARAN
The approach is same as is in the above type of question
Take difference of the numbers in ascending or descending order as below
16009 – 14009 = 2000,
14009 – 9009 = 5000,
9009 – 7509 = 1500.

Now to leave same remainder, each of the intervals must be divisible by the divisor.

Obviously we’ll need the HCF of these numbers 2000, 5000 and 1500, and that is 500

Hence, the answer is 500 (option ‘D’)

QUERY 11

Finding out the remainder when a multiple of a number is divided by a different number. Type # 8

If a number is divided by 15, it leaves the remainder 7, if thrice the number is divided by 5; then what is the remainder?

A) 1
B) 2
C) 3
D) 0

JAYANTCHARAN CHARAN
Such questions are difficult to frame as one has to find a pattern between divisors n remainders; but actually they are easy and can be cracked very comfortably. Why I’m putting it here because I have a very short…practical approach for solving such questions.

Now choose a number that leaves 7 as the remainder with 15 as divisor Let’s take it 7 only.

So thrice of 7 = 21

Now, according to the question we have to divide it by 5 and find the remainder. Of course it’s 1
Hence the answer is 1 (option ‘A)

NOTE: Since the number should be giving same result for all values that give 7 as remainder with 15
as divisor, therefore its always better to take some value and solve it instead of taking an algebraic approach.

Previous post

SSC CGL: QUESTIONS ON PARTNERSHIP (PART-I)

Next post

QUESTIONS ON REMAINDERS & DIVISIBILITY (PART-III)

Maha Gupta

Maha Gupta

Founder of www.examscomp.com and guiding aspirants on SSC exam affairs since 2010 when objective pattern of exams was introduced first in SSC. Also the author of the following books:

1. Maha English Grammar (for Competitive Exams)
2. Maha English Practice Sets (for Competitive Exams)