How to Solve: Remainders (Basics and Advanced)

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How to Solve: Remainders (Basics and Advanced)

Hi All,

I have recently uploaded two Videos on YouTube to discuss Remainders Basics and Advanced problems on remainder:

Remainders Basics
Remainders by 2,3,5,9,10 and Binomial Theorem

Following is covered in the video:

Video 1

• Definition: What is "Remainder"?
• Remainder Notation
• How to find Remainders : Long Division Method
• Remainder Range
• PT1: Basic Problems on Finding Remainders
• PT2: Find divisor when remainder is given
• PT3: Find dividend when one divisor and remainder are given
• PT4: Find dividend when two divisors and remainders are given
• Remainder of sum of two numbers by a number
• Remainder of difference of two numbers by a number
• Remainder of product of two numbers by a number

Video 2

• Remainder of numbers when divided by 2 , 3, 5, 9, 10
• Remainder of numbers when divided by other numbers
• Binomial Theorem
• Application of Binomial Theorem in finding remainders


Remainders Basics

• Definition: What is "Remainder"?

Remainder is the integer which is left over in a division, when the divisor cannot evenly divide the dividend
Dividend = Divisor * Quotient + Remainder
Ex: 13 when divided by 3 gives 1 remainder. (13 = 3*4 + 1)
13 -> Dividend
3 -> Divisor
4 -> Quotient
1 -> Remainder

• Remainder Notation

1. A number when divided by 5 gives 3 as remainder.
Let the number be n
using Dividend = Divisor * Quotient + Remainder
n = 5k + 3 [where k is quotient and is an integer]

2. A number when divided by 13 gives 5 as remainder.
Let the number be n
using Dividend = Divisor * Quotient + Remainder
n = 13k + 5 [where k is quotient and is an integer]

3. A number when divided by 16 gives 3 as remainder.
Let the number be n
using Dividend = Divisor * Quotient + Remainder
n = 16k + 3 [where k is quotient and is an integer]


• How to find Remainders : Long Division Method -> Check the video link

• Remainder Range
A number when divided by a number k can give remainder from 0 to k-1
Ex: If we are trying to divide a number by 6 then possible values of remainder are from 0-5

Q1. x when divided by "a" gives 3 as remainder.
y when divided by "b" gives 4 as remainder.
Find min(a+b)

Sol: As a and b are positive numbers so min(a+b) = min(a) + min(b)
x when divided by "a" gives 3 as remainder -> Since a is giving 3 remainder that means that a >= 4. So, min a = 4
y when divided by "b" gives 4 as remainder -> Since b is giving 4 remainder that means that b >= 5. So, min b = 5
=> min(a+b) = min(a) + min(b) = 4+ 5 =9

• PT1: Basic Problems on Finding Remainders -> Check out the video above for examples on long division method

Q1. Find the remainder when 200 is divided by 3.
Q2. Find the remainder when 80 is divided by 7.
Q3. Find the remainder when 100 is divided by 9.

Ans:
Q1. 2, Q2 3, Q3. 1

• PT2: Find divisor when remainder is given

Q1. 20 when divided by which number will give 4 as remainder?

Sol: 20 = nk + 4
=> nk= 20-4 = 16
=> n = 16/k
Now, n is giving 4 remainder that means that n>=5
Let's start putting values of k and get values of n
k=1 => n=16
k=2 => n=16/2 = 8
k=3 => n not integer
k=4 => n= 16/4 not possible as n>=5
So, possible values of the number (n) are 8 and 16

Q2. 50 when divided by which number will give 7 as remainder?

Sol: 50 = nk + 7
=> nk = 50-7 = 43 [note 43 is prime so it has only two factors 1 and 43 itself]
=> n = 43/k
k= 1 => n = 43
k = 43 not possible as n becomes 1 [ but we know that n>=8 as n is giving 7 remainder]
So, Answer is 43

• PT3: Find dividend when one divisor and remainder are given

Q1. n when divided by 7 gives 4 as remainder. Find the possible values of the number.

Sol: n when divided by 7 gives 4 as remainder.
n = 7k + 4

We will start taking values of k starting from k=0 and find values of n correspondingly
k=0 n=7*0 + 4 = 4 k=5 n=7*5 + 4 = 39
k=1 n=7*1 + 4 = 11 k=6 n=7*6 + 4 = 46
k=2 n=7*2 + 4 = 18 k=7 n=7*7 + 4 = 53
k=3 n=7*3 + 4 = 25 k=8 n=7*8 + 4 = 60
k=4 n=7*4 + 4 = 32 k=9 n=7*9 + 4 = 67

Q2. n when divided by 5 gives 2 as remainder. Find the possible values of n.

n when divided by 5 gives 2 as remainder.
n = 5k + 2

We will start taking values of k starting from k=0 and find values of n correspondingly
k=0 n=5*0 + 2 = 2 k=5 n=5*5 + 2 = 27
k=1 n=5*1 + 2 = 7 k=6 n=5*6 + 2 = 32
k=2 n=5*2 + 2 = 12 k=7 n=5*7 + 2 = 37
k=3 n=5*3 + 2 = 17 k=8 n=5*8 + 2 = 42
k=4 n=5*4 + 2 = 22 k=9 n=5*9 + 2 = 47

• PT4: Find dividend when two divisors and remainders are given

Q1. n when divided by 7 gives 3 remainder and when divided by 5 gives 3 remainder. Find first 2 non-negative values of n

Sol: Method -1
3. n when divided by 7 gives 3 remainder
n = 7k + 3
We will start taking values of k starting from k=0 and find values of n correspondingly
k = 0 , 1, 2, 3, 4, 5, 6, 7, 8, 9
n = 3 , 10, 17, 24, 31, 38, 45, 52, 59, 66

n when divided by 5 gives 3 remainder
n = 5t + 3
We will start taking values of t starting from t=0 and find values of n correspondingly
t = 0 , 1, 2, 3, 4, 5, 6, 7, 8, 9
n = 3 , 8, 13, 18, 23, 28, 33, 38, 43, 48

First two common values are 3 and 38

Sol: Method-2
n when divided by 7 gives 3 remainder
n = 7k + 3

n when divided by 5 gives 3 remainder
n = 5t + 3

7k+3 = 5t+3 => t = 7k/5
So, only those values of k will give us common values of n for which t is integer too.
k = 0 => t=7*0/5 = 0
k = 5 => t=7*5/5 = 7
So, k = 0 => n = 7*0 + 3 = 3
k = 5 => n = 7*5 + 3 = 38

Q2. n when divided by 6 gives 4 remainder and when divided by 4 gives 2 remainder. Find first 2 non-negative values of n

Sol: Method-1
n when divided by 6 gives 4 remainder
n = 6k + 4
We will start taking values of k starting from k=0 and find values of n correspondingly
k = 0 , 1, 2, 3, 4, 5, 6, 7, 8, 9
n = 4, 10, 16, 22, 28, 34, 40, 46, 52, 58

n when divided by 4 gives 2 remainder
n = 4t + 2
We will start taking values of t starting from t=0 and find values of n correspondingly
t = 0 , 1, 2, 3, 4, 5, 6, 7, 8, 9
n = 2, 6, 10, 14, 18, 22, 26, 30, 34, 38

First two common values are 10 and 22

Sol: Method-2
n when divided by 6 gives 4 remainder
n = 6k + 4

n when divided by 4 gives 2 remainder
n = 4t + 2

6k+4 = 4t+2 => t = (6k+2)/4 = (3k+1)/2
So, only those values of k will give us common values of n for which t is integer too.
k = 1 => t=(6*1 + 2)/4 = 2
k = 3 => t=(6*3 + 2)/4 = 5
So, k = 1 => n = 6*1 + 4 = 10
k = 3 => n = 6*3 + 4 = 22

Note: When the numerator is smaller than the denominator then the remainder is the numerator itself
Ex: If 2 is divided by 3 then remainder is 2

• Remainder of sum of two numbers by a number
13 when divided by 3 gives us 1 remainder

When we split 13 as 8 and 5 and divide 8 and 5 individually by 3 then we still get the same remainder
13/3 = (8+5)/3 = 8/3 + 5/3
8/3 will give 2 remainder
5/3 will give 2 remainder

Total remainder is 2+2 = 4, but remainder cant be greater than 3 so remainder will be 4-3 = 1 which is same as the remainder for 13/3

Q1. “A” when divided by 12 gives 3 as remainder. What is the remainder when A is divided by 4?

Sol: A = 12k + 3
When A is divided by 4 then 12k will give 0 remainder and 3 will give 3 remainder. Total remainder is 3


Q2. “B” when divided by 15 gives 6 as remainder. What is the remainder when B is divided by 5?

Sol: B = 15k + 6
When B is divided by 5 then 15k will give 0 remainder and 6 will give 1 remainder. Total remainder is 1

• Remainder of difference of two numbers by a number
13 when divided by 3 gives us 1 remainder

When we split 13 as 15 - 2 and divide 15 and 2 individually by 3 then we still get the same remainder
13/3 = (15-2)/3 = 15/3 - 2/3
15/3 will give 0 remainder
2/3 will give 2 remainder
Total remainder = 0-2 = -2
But remainder cannot be negative so we are going to keep on adding 3 to -2 till the time the sum comes in the range of 0 and 2 [3-1]
=> -2 + 3 = 1
So, remainder is 1

• Remainder of product of two numbers by a number
21 when divided by 5 gives us 1 remainder. Now if we break 21 into product of two numbers and find the remainder of these individual numbers by 5 and multiply the remainders then we are going to get the same remainder as we got when we divided 21 by 5

Let's write 21 = 3*7 and divided 3 and 7 by 5
3/5 * 7/5 will give us 3 * 2 remainder respectively = 6
But we divided set of numbers by 5 so remainder cannot be more than 5, so we divide 6 again by 5 to get final remainder as 1

Q1. Find the remainder of 136 * 148 * 298 by 5.

Sol: 136/5 remainder is 1
148/5 remainder is 3
298/5 remainder is 3
Total remainder = 1*3*3 = 9 [>=5]
=> Final remainder = remainder of 9/5 = 4

Q2. Find the remainder of 1205 * 1208 * 2404 by 12.

Sol: 1205/12 remainder is 5
1208/12 remainder is 8
2404/12 remainder is 4
Total remainder = 5*8*4 = 160 [>=12]
=> Final remainder = remainder of 160/12 = 4

In next section I have just listed down the theory. For problems and Binomial theorem please refer the video

Check above video for examples
• Remainder of numbers when divided by 2

When we try to find remainder of a number by 2 then we just need to know if the number is odd or even.
If the number is odd then remainder is 1
If the number is even then the remainder is 0

• Remainder of numbers when divided by 3

Remainder of a number by 3 is same as the remainder of sum of digits of the number by 3

• Remainder of numbers when divided by 5

Remainder of a number by 5 is same as the remainder of the unit's digit of the number by 5

• Remainder of numbers when divided by 9

Remainder of a number by 9 is same as the remainder of sum of digits of the number by 9

• Remainder of numbers when divided by 10

Remainder of a number by 10 is same as the unit's digit of the number

• Remainder of numbers when divided by other numbers
• Binomial Theorem
• Application of Binomial Theorem in finding remainders -> Please check Video for this


Hope it helps!