Excerpts from Chapter - Remainder from Pearson … from Chapter - Remainder from Pearson Guide to Number System for CAT (by Nishit Sinha) dueNorth Academics, Dehradun (An IIM Alumni

Excerpts from Chapter - Remainder from Pearson Guide to Number System for CAT (by Nishit Sinha)

dueNorth Academics, Dehradun (An IIM Alumni Body) CAT | GRE | GMAT Page 1

Remainder

A number is said to be divisible by another number if the quotient obtained is an integer and remainder

obtained is zero. This statement provides two necessary conditions for a number to be divisible by other:

i. Quotient obtained should be an integer. This can be a +ve integer, or a –ve integer or zero.

ii. Remainder should be equal to zero.

So when I distribute 30 apples among 6 children, each of the children get equal integral number of apples ( 5

in this case – this is known as quotient), as well as I am left with zero apple (this is known as remainder).

Going through the same logic, if i distribute zero apple among 6 children, each of the children still get equal

integral number of apples (0 in this case) and I am left with zero apple (this is known as remainder). And so

we say that 0

6= 0.

Consider another case: I have now 33 apples, and i want to distribute these apples among 6 kids. I may

divide it in two ways- i. I first distribute 5 apples each, and then distribute the remaining 3 apples by cutting

in halves. So quotient obtained = 5.5

ii. If i distribute 5 apples to each of the kids, i am left with 3 apples. This is the case when we obtain

remainder. Quotient = 5, remainder = 3.

Basic framework of remainder:

i. If N is a number divisible by 7, it can be written as -: 7 K = N, where K is the quotient.

ii. When N is divided by 7, remainder obtained is 3 it can be written as -: 7 K + 3 = N, where K is the

quotient.

iii. When N is divided by 7, remainder obtained is 3 is equivalent of saying remainder obtained is (-4) when

divided by 7. It can be understood that When N is divided by 7, remainder obtained is 3 N is 3 more than

a multiple of 7 So N is 4 short of another multiple of 7. So remainder obtained = -4.

iv. When divided by 8, different remainders obtained can be =0, 1, 2, 3, 4, 5, 6, 7 (8 different remainders)

Similarly when divided by 5, different remainders obtained can be =0, 1, 2, 3, 4 (5 different remainders)

Properties of remainder:

1. Remainders are additive

It means that remainder can be obtained by breaking down a number in two or more than two parts, and

then finally calculating the resultant.

For example –

WO1. What is the remainder obtained when 40 is divided by 7?

Solution – Obviously the answer is 5.

40 = 5 × 7 + 5 [Dividend = Quotient × Divisor + Remainder]



Using the property of remainder being additive, we can break down 40 in two or more than two parts, find

the individual remainder and finally add up those individual remainder to get the final remainder.

40

7=

35 + 5

7=

36 + 4

7=

30 + 10

7=

20 + 12 + 8

7

It can be seen in the above example that in each of the cases remainder obtained is going to be same.

35+5

7=

35

7+

5

7 𝑅𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 𝑜𝑏𝑡𝑎𝑖𝑛𝑒𝑑 𝑖𝑛𝑑𝑖𝑣𝑖𝑑𝑢𝑎𝑙𝑙𝑦 = 0 + 5 = 5

We could have even written down 40 as one number being subtracted from other too.

40

7=

50 − 10

7=

49 − 9

7

Remainder obtained from 50−10

7 =

50

7−

10

7

Remainder obtained from 50

7= 1 𝑎𝑛𝑑 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 𝑜𝑏𝑡𝑎𝑖𝑛𝑒𝑑 𝑓𝑟𝑜𝑚

10

7= 3

So net remainder = 1 – 3 = -2 = 5

Concept Enhancer 1 – Remainder obtained = -2 when divided by 7 means that this number is 2 less than a

multiple of 7 Hence this number is 5 more than the previous multiple of 7.

Following is the expression of what we just discussed:

7 K – 2 = 7 (K - 1) + 5 [Remainder obtained are in italics and bold with size enlarged]

It can be further understood through the following graphics:

Graphics 1 – When divided by 7

0

No. = P

1 or -6

No.= P+ 1

2 or

-5

No.=P+2

3 or -4

No.=P+3

4 or -3

No.=P+4

5 or -2

No.=P+5

6 or -1

No.=P+6



It shows that when divided by 7, P is divisible. So when P+1 will be divided by 7, remainder obtained will be

either 1 or -6. Similarly when P+2 is divided by 7, remainder obtained will be 2 or -5. And so on.

Graphics 2 – When divided by 5

It shows that when divided by 5, P is divisible. So when P+1 will be divided by 5, remainder obtained will be

either 1 or -4. Similarly when P+2 is divided by 7, remainder obtained will be 2 or -3. And so on.

2. Remainders are multiplicative

It means that remainder can be obtained by breaking down a number in two or more than two parts, and

then finally calculating the resultant.

For example –

WO2. What is the remainder obtained when (10 × 20) is divided by 7?

Solution – Going by Actual method of finding remainder, remainder obtained = 200/7 = 4

Since remainders are multiplicative, we can find out the individual remainder, and then multiply these to

obtain final remainder.

Remainder obtained when 10 is divided by 7 = 3


Remainder obtained 10×20

7= 𝑅𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 𝑜𝑏𝑡𝑎𝑖𝑛𝑒𝑑

3 × 6

7= 𝑅𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 𝑜𝑏𝑡𝑎𝑖𝑛𝑒𝑑

18

7= 4

Let us understand that why this happens:

10 = 7p + R1 and 20 = 7q + R2 where R1 and R2 are the respective remainder obtained when 10 and 20 are

divided by 7. p and q are respective quotients.

0

No.= P

1 or -4

No.=P+1

2 or -3

No.=P+2

3 or -2

No.=P+3

4 or -1

No.=P+4



10 × 20 = (7p + R1)(7q + R2) = (7p × 7q) + (7p × R2) + (7q × R1) + (R1 × R2) ----------------(1)

When (1) is divided by 7, each of the terms (7p × 7q) + (7p × R2) + (7q × R1) is divisible by 7. So the net

remainder is obtained from the last term (R1 × R2). This is exactly what we have done in the last example.

WO3. What is the remainder when (1421 1423 1425) is divided by 12? (based on CAT 2000)

Solution – Remainder of 1421/12 = 5

Remainder of 1423/12 = 7

Remainder of 1425/12 = 9

Remainder (1421 1423 1425)/12 = Remainder (5 7 9)/12 = Remainder (5 63)/12 = Remainder (5

3)/12 = 3

WO4. What is the remainder when 4100 is divided by 7?

Solution –

𝑅𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 𝑜𝑏𝑡𝑎𝑖𝑛𝑒𝑑 𝑓𝑟𝑜𝑚4100

7=

1650

7= 𝑅𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 𝑜𝑏𝑡𝑎𝑖𝑛𝑒𝑑 𝑓𝑟𝑜𝑚

250

7=

3210

7

= 𝑅𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 𝑜𝑏𝑡𝑎𝑖𝑛𝑒𝑑 𝑓𝑟𝑜𝑚410

7=

165


25

7=

32

7= 4

Ideally when we are using this method of find out the remainder, we should be constantly on a lookout for

an expression giving us remainder either +1 or -1. Although it is not essential that we always get remainder

+ 1 or -1 for every expression. We will discuss this once again while discussing the pattern method of finding

remainder.

Let us see that how the same question becomes quite easier now:

𝑅𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 𝑜𝑏𝑡𝑎𝑖𝑛𝑒𝑑 𝑓𝑟𝑜𝑚4100


43 33

×41

7=

𝑅𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 𝑜𝑏𝑡𝑎𝑖𝑛𝑒𝑑 𝑓𝑟𝑜𝑚 64 33 ×41

7=𝑅𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 𝑜𝑏𝑡𝑎𝑖𝑛𝑒𝑑 𝑓𝑟𝑜𝑚

133 ×41

7=

𝑅𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 𝑜𝑏𝑡𝑎𝑖𝑛𝑒𝑑 𝑓𝑟𝑜𝑚 4

7= 4

3. Its wrong to calculate the Remainder by reducing the fraction to its lower terms.

For example- Look at the following fractions:

(a) ½ (b) 10/20 (c) 100 / 200

Despite the fact that all the fractions taken here are same, remainder obtained in each of the cases are going

to be different:

Remainder obtained in (a) ½ = 1

Remainder obtained in (b) 10/20 = 10

Remainder obtained in (a) 100/200 = 100

WO5. 532

.1000!5!4!3!2!1!3



Solution - Denominator = 23 × 31 × 5 = 120 = 5!

So the remainder will come only from 1! + 2! + 3! + 4!. Hence net remainder 1! + 2! + 3! + 4! = 33 (since

numerator is less than denominator, numerator itself will be the remainder).

Summarizing the discussion regarding the property of remainder:

Let remainder obtained when ‘a’ is divided by n = b and remainder obtained when ‘c’ is divided by n

= d.

Then remainder obtained for LHS and RHS will be same.

(i) 𝑎+𝑐

𝑛=

𝑏+𝑑

𝑛

(ii) 𝑎−𝑐

𝑛=

𝑏−𝑑

𝑛

(iii) ) 𝑎𝑐

𝑛=

𝑏𝑑

𝑛

Let us now see some standard formats of remainder questions:

WO 6-WO 9 – FORMAT 1

A number when divided by 7 and 9 gives remainder 1 and 1 respectively.

6. Which is the lowest such natural number?

7. What is the remainder when lowest such natural number is divided by 11?

8. For every ......... natural numbers, there is exactly one such number which satisfies the property given

above. (Fill in the blanks)

9. Only natural numbers can satisfy the above given statement. (True / False)

Solution -

6. Lowest such natural number = 1, and not 64.

7. When 1 is divided by 11, remainder obtained = 1.

8. For every 63 numbers (LCM of 9 and 7), there will be exactly one such number satisfying this property.

9. False. There may be negative numbers satisfying this statement.

Concept enhancer 2 - Understand that the family of numbers satisfying this statement is -: 1 ± 63P, where P

is an integer. Some of the numbers satisfying this statement are - : 1, 64, -63 etc.

WO 10 – WO 13: FORMAT 2

A number when divided by 7 and 6 gives remainder 1 and 4 respectively.

10. Which is the lowest such natural number?



11. What is the remainder when this number is divided by 11?

12. For every ......... natural numbers, there is exactly one such number which satisfies the property given

above. (Fill in the blanks)

13. Only natural numbers can satisfy the above given statement. (True / False)

Solution -

Q10. Let us assume N is that natural number.

N is divided by 7, remainder obtained = 1 N = 7 x + 1

N is divided by 6, remainder obtained = 4 N = 6 y + 4

N = 7 x + 1 = 6 y + 4 7 x – 6y = 3

Now we are required to insert the values of x and y to satisfy the equation given above.

We obtain x = 3 and y = 3, and the number obtained = N = 7 × 3 + 1 = 6 × 3 + 1 = 22

Alternatively, it can be seen through making some combinations directly:

Lowest natural number N divided by 7, remainder obtained = 1 is N = 1, however when N is divided by 6,

remainder obtained = 1 and not 4. Hence 1 is not that number.

Next Natural number N divided by 7, remainder obtained = 1 is N = 8, however when 8 is divided by 6,


Next natural number N divided by 7, remainder obtained = 1 is N = 15, however when 15 is divided by 6,


Next natural number N divided by 7, remainder obtained = 1 is N = 22, and when 22 is divided by 6,

remainder obtained = 4. Hence 22 is that number.

Concept enhancer 3 - Although a larger question arises that till which number we should keep on checking,

and which number should be that limit till which if we don’t get any number satisfying the given statements,

we should be in a position to conclude that we are not going to get any such number on the whole number

line?

Answer to the above question is – Till the LCM of the divisors. In this case, it is LCM of (7 and 6) = 42.

So if we don’t get any such number till 42, we can conclude that no such number exists that will give

remainder 1 and 4 when divided by 7 and 6 respectively. In other words, if at all numbers satisfying this

statement exists, one and exactly one such natural number will be obtained from 1 to 42, exactly one such

number from 43 to 84, and so on for every next 42 numbers.

It can also be inferred that for every 42 numbers, not necessarily starting from 1, will have exactly one such

natural number satisfying this statement. For example, from 100 to 141, there is exactly one number that

will satisfy the statement given in this question.



11. It can be seen that there are more than one natural numbers satisfying the given statement – 22, 64, 106

etc.




So, more than one remainder will be obtained.

12. Using the Concept enhancer 3, for every 42 natural numbers, there is exactly one such number which

satisfies the statements given in the question. Although the term natural number can be replaced by

integers, and this statement will be true even in that case.

13. False. Using the answer to Q8, even negative integers will satisfy. Understand that the family of numbers

satisfying this statement is -: 22 ± 42P, where P is an integer.

WO 14- FORMAT 3

A number when divided by 6 and 9 gives remainder 4 and 6 respectively. Which is the lowest such natural

number?

Solution – Let us go by the discussion that we had in the above question. If we are going to get a number

satisfying the conditions given in this question, we should obtain this number in the first 18 (18 = LCM of 9

and 6) natural numbers. It is quite evident that no such number exist satisfying the above statements. Hence

no such number is possible.

Alternatively, let us assume that N is the natural number.

So, N = 6x + 4 = 9y + 6

6x – 9y = 6 – 4 3 (2x – 3y) = 2

We can see that LHS of this equation is divisible by 3, whereas RHS is not. Hence if a number divided by 9

gives remainder 6, then the same number cannot give remainder 4 when divided by 6 and vice-versa.

WO 16. What is the remainder when 55100 is divided by 110?

Solution – HCF of 55 and 100 = 55, so the remainder obtained has to be divisible by 55. So the remainders

possible are 0 or 55.

Now, 55100 is an odd number and 110 is an even number, so remainder obtained can never be equal to zero

(or in other words, since an odd number can never be divisible by an even number, 55100 can never be

divisible by 110).

Hence 55 is the only possible option left out for the remainder. So, 55 is the remainder.

WO 17- WO 18:-FORMAT 5

An integer when divided by 17 and 9 gives remainder 14 and 6 respectively.



WO 17. Which is the lowest such integer?

WO 18. Which is the lowest such positive integer?

Solution -

Remainder obtained = 14 when divided by 17 is same as remainder being (-3) when divided by 17.

Remainder obtained = 6 when divided by 9 is same as remainder being (-3) when divided by 9.

WO 17. Integer is going to be of the format -: (-3 ± 153K, where K is any integer). Hence the lowest such

integer is going to be (-).

WO 18. Taking a cue from the above solution, lowest positive integer is going to be (-3 + 153×1) = 150.

Summarizing the above discussion:

Problem states that….. Solution

1 Find the Greatest Number that will exactly divide a, b

and c

Required number = HCF of a, b

and c (greatest divisor)

2 Find the Greatest number that will divide x, y and z

leaving remainders a, b and c respectively.

Required number (greatest

divisor) = HCF of (x-a), (y-b) and

(z-c)

3 Find the least number which is exactly divisible by a, b

and c

Required number = LCM of a, b

and c

4 Find the least number which when divided by x, y and z

leaves the remainders a, b and c respectively, and (x-a)

= (y-b) = (z-c) = N

Required number = LCM of (x, y

and z) – N

5 Find the least number which when divided by x, y and z

leaves the same remainder 'r' each case.

Required number = (LCM of x, y

and z) + r

[This table presents just a part of the above discussion, and does not present the exhaustive list that we

discussed above.]

Test Your Learning 1

(You may solve some of these questions using options. However, I have removed the options, because I want

you to develop the equation method of solving which may be the only method of solving a particular

question like Q4 as given below)

Q1. A natural number divided by 9 gives remainder 4, and when divided by 7 gives remainder 2. What will be

remainder obtained when lowest such natural is divided by 11?



Q2. A natural number divided by 15 gives remainder 3, and when divided by 6 gives remainder 4. What will

be remainder obtained when lowest such natural is divided by 13? (Not possible)

Q3. What is the smallest five-digit number which when divided by 7, 11 and 21 leaves a remainder of 3 in

each case? (10167)

Q4.What is the sum of digits of the least multiple of 13 which when divided by 6, 8 and 12 leave 5, 7 and 11

as the remainder respectively? (8)

Q5. Find the lowest four digit natural number which when divided by 5 leaves remainder 1 but when divided

by 6 leaves remainder 2? (1016)

Q6. How many three digit natural numbers are there which when divided by 6 gives remainder 2, but when

divided by 7 gives remainder 1?

Fermat’s Remainder Theorem

Let P be a prime no. and N be a no. not divisible by P (in other words N and P are co-prime). Then remainder

obtained when NP-1

is divided by P is 1.

Remainder obtained when 𝑁𝑃−1

𝑃= 1 provided HCF(N, P) = 1, and P is a prime number.

Using Fermat’s theorem we can say that:

Remainder of 66

7 = Remainder of

56

7 = Remainder of

46

7 = Remainder of

36

7 = Remainder of

26

7 = Remainder of

16

7

= 1

WO 19. What is the remainder when 2100

is divided by 101?

Solution – Since it satisfies the Fermat’s theorem format, remainder = 1

Some generalizations:

i. (an + b

n) is divisible by (a + b), if n is odd.

ii. Extension of the above formula - (an + b

n + c

n) is divisible by (a + b + c), if n is odd and a, b

and c are in Arithmetic Progression.

iii. (an + b

n + c

n + d

n) is divisible by (a + b + c + d), if n is odd and a, b, c and d are in Arithmetic

Progression.

Similarly this formula can be extended for any number of terms provided:

Terms as in a, b, c, d etc are in AP, and power n = Odd.

iv. (an - b

n) is divisible by (a + b), if n is even.

v. (an - b

n) is divisible by (a - b), if n is any natural number.

vi. (an - b

n) is divisible by (a

k - b

k), if k is a factor of n.

Direction for Questions 1 to 6: Find the remainder in each of the questions when numerator is divided by

denominator:



2. 9090

/13

1. 2 2.1 3. 3 4. 4

3. 3202

/ 101

1. 8 2. 11 3. 9 4. 7

4. 10200

/ 8

1. 0 2.1 3. 3 4. 4

5. (1010

+10100

+101000

) / 7

1. 5 2.4 3.2 4. 1

Q6. What is the remainder when [225 + 235 + 245 + ..................+875 + 885] is divided by 110?

1. 0 2. 60 3. 110 4. 55

(Answers are given in bold)

Disclaimer:

This is a copyleft material taken from the book “Number System for CAT, published by Pearson

publication and written by Nishit Sinha (Alumnus, IIM Lucknow). You may circulate this freely among

your circle subject to the conditions as mentioned below:

(a) It is circulated for NOT FOR PROFIT or sales.

(b) The footer and header has not been removed.

(c) This disclaimer has not been removed.

Documents

Excerpts from Chapter - Remainder from Pearson … from Chapter - Remainder from Pearson Guide to Number System for CAT (by Nishit Sinha) dueNorth Academics, Dehradun (An IIM Alumni