## Remainder Theorem Aptitude Examples with Answers | Remainder Theorem Tutorial

Contents

**Remainder theorem** basic rules were given in the following link. Here provides some examples with shortcut methods on* remainder theorem aptitude.*

Remainder Theorem for Number System Basic rules

**Application of the remainder theorem:**

Finding the last digit of an expression purpose simply find the remainder of that expression divided by 10. In the same way for finding the last two digits of an expression purpose find the remainder of that expression divided by 100.

#### Remainder Theorem Sums

**Example – 1** : Find the last two digits of the expression of 120 x 2587 x 247 x 952 x 854

**Solution:** Now the above expression divided by 100 and find the remainder then it will equal to last two digits of given sum.

For reducing the calculation purpose cancelled the both numerator and denominator by **20** then

Now find the remainders of the each number

Remainder of the above expression is 2

Now reminder of the initial expression is 40 ( Multiplying the reminder with canceled number i.e )

**Example – 2 :** Find the remainder when 1! + 2! + 3! +4! + 5! + ————– 1000! is divided by 8

**Solution:** Here remember about the **factorial function**

Note : 2! is usually pronounced “2 factorial“

The factorial function says to multiply descending natural numbers series. For example 4! = 4 x 3 x 2 x 1 = 24

Now calculate remainder for each number in the given series

The remainder of the remaining terms is also zero. So

**Example – 3 : **Find the remainder when 1! + 2! + 3! +4! + 5! + ————– 1000! is divided by 14

**Solution:**

Now calculate remainder for each number in the given series

5! / 14 8

The remainder of the remaining terms is also zero. NOW

= 1 + 2 +6 – 4 + 8 + 6 /14

= 19 / 14 5

Remainder of the given sum is 5

**Example – 4 :** Find the remainder when 51^{203} is divided by 7

**Solution:** Find Remainder of the expression 51 / 7

51 / 7 2 So

**Example – 5 :** Find the remainder when 21^{875} is divided by 17

**Solution:** Find Remainder of the expression 21 / 7

Here our aim is obtained number as 2^{4 }= 16 ( 16 = 17 – 1) so according to that rewrite the equation as follow as

Final remainder is 17 – 4 = 13

**Example – 6 :** Find the remainder when 27^{87} x 23^{45} x 19^{92} is divided by 23

**Solution:** While observing the above question middle term is exactly divisible by 23, So the hole expression is also exactly divisible by 23

The final remainder is Zero

**Example – 7 :** Find the remainder when 3^{41} +7^{82 } is divided by 52

**Solution:**

Hint : If a^{n} +b^{n} and n = odd number then (a +b) is exactly divisible of that number a^{n} +b^{n}

The given expression can be written as

3^{41} +7^{82} = 3^{41} +49^{41}

From the above information the given expression is exactly divisible by 52

So remainder is Zero

**Example – 8 :** Find the remainder when 5^{3} + 17^{3} +18^{3} +19^{3 } is divided by 70

**Solution:**

Hint: If A^{n} +B^{n} + C^{n} + D^{n} and n = odd number then (A+B+C+D) is exactly divisible of that number

From the above information n = 3 is odd number and 16 + 17 + 18+ 19 = 70 so

70 is exactly divisible by 16^{3} + 17^{3} +18^{3} +19^{3}

Remainder of the sum is zero

**Example – 9 :** Find the last two digits of the expression of 12899 x 96 x 997

**Solution:** Now the above expression divided by 100 and find the remainder then it will equal to last two digits of given example

Final remainder = -12 +100 = 88

**Example – 10 :** Find the remainder when 1 2 3 4 5 – – – – – – 41 digits is divided by 4

**Solution:** Here first identifying that 1 to 9 numbers having 1 digit after that each number having 2 digits. So

41 digits means – 41 – 9 = 32 / 2 = 16 then last number is 9+ 16 = 25 and given number is

1 2 3 4 5 – – – – – – – – – – 2 4 2 5

Now according to divisibility rules we can find easily ( i. e find remainder while divided last two digits of that number)

= 25/4 1

**Example – 11 :** Find the remainder when 8 8 8 8 8 8 8 8 – – – – – 32 times is divided by 37

**Solution:**

Hint: If any 3-digit which is formed by repeating a digit 3-times then this number is divisible by 3 & 37.

The above expression can be written as 10 pairs of 8 8 8 30 times remaining number is 88

So 88 / 37 14

**Example – 12 :** Find the remainder when 7 7 7 7 7 7 7 – – – – – 43 times is divided by 13

**Solution :**

Hint : If any 6-digit which is formed by repeating a digit 6-times then this number is divisible by 3, 7, 11, 13 and 37.

The above expression can be written as 7 pairs of 7 7 7 7 7 7 7 42 times remaining number is 7

So 7 / 13 7

**Example – 13 :** Find the remainder when 10^{1} + 10^{2} +10^{3} + 10^{4 } + – – – – – – + 10^{100 }is divided by 6

**Solution:** Here find remainder of the each number in the given expression

10^{1}/ 6 +4

10^{2}/ 6 +4

10^{3}/ 6 +4

10^{4}/ 6 +4

So final remainder = 100 x 4 / 6 +4

Hint : The above sum simple we identifying that every three terms the remainder is zero ( 4 + 4+ 4 = 12 / 6 0 ) so upto 99 terms the remainder is zero.

**Example -14:** Find the remainder when 2^{469} + 3^{268} ^{ }is divided by 22

**Solution:** Here find remainder of the each number individually.

Final remainder of this term = 3 x 2 /11 6

Final remainder of the give expression = 6 +5 /22 = 11 /22 11

**Example – 15 :** Find the remainder when 7 ^{7} + 7 ^{77} + 7 ^{777} + 7 ^{7777}^{ } + – – – – – – + 7 ^{777777777}^{ }is divided by 6

**Solution:** Here find remainder of the each number individually.

= 7 ^{7} / 6 = (6+1) ^{7} /6 1

So remaining terms remainders are also 1 and total terms in given expression is 9

= 9/ 6 3

**Example – 16** : Find the remainder when 23^{10} – 1024 is divided by 7

**Solution:** Here 1024 can be written as 23^{10}

The expression is 23^{10 } – 2^{10}

Hint: If the given expression like ( a^{n} – b^{n} ) and “n” is even number then (a-b) and (a+b) are exactly divides that expression.

From the above hint factors of the given expressions = 21 and 25

Factors of 21 = 1 , 3 , 7, 21

Factors of 25 = 1 , 5 , 25

So the numbers 1, 3 , 5 , 21 and 25 are exactly divisible by the given expression 23^{10} – 1024

Remainder = 0

**Example – 17 :** Find the remainder when 3^{41} + 7^{82} is divided by 26

**Solution:** Here 7^{82} can be written as 49^{41}

The expression is 3^{41} + 49^{41}

Hint: If the given expression like ( a^{n} + b^{n} ) and “n” is odd number then (a+b) is exactly divides that expression.

From the above hint factors of the given expressions = 52

Factors of 52 = 1 , 2, 4, 13 , 26, 52

So the numbers 1, 2, 4 , 13 , 26 and 52 are exactly divisible by the given expression 3^{41} + 7^{82}

Remainder = 0

**Example – 18 :** Find the remainder when 27^{23} + 19^{23} is divided by 2

**Solution:**

Hint: If the given expression like ( a^{n} – b^{n} ) and “n” is odd number then (a+b) is exactly divides that expression.

From the above hint factors of the given expressions = 27 + 19 = 46

Factors of 46 = 1 , 2, 23 , 46

So the numbers 1, 2, 23 and 46 are exactly divisible by the given expression 27^{23} + 19^{23}

Remainder = 0

**Example – 19 :** Find the remainder when 5^{2}P is divided by 26 where P = (1 !)^{2} + (2 !)^{2} + (3 !)^{2} + – – – – – + (10 !)^{2}

**Solution:** Here 5^{2p} = 25^{P }= (26 – 1)^{P }So the reminder depend upon value of P.

If P = even number then remainder has 1 & If P = odd number then remainder has 25

Now find the last digit of the expression (1 !)^{2} + (2 !)^{2} + (3 !)^{2} + – – – – – + (10 !)^{2}

For that purpose Find the remainder when P is divided by 10

(1 !)^{2}/ 10 1

( 2 !)^{2}/ 10 4

(3 !)^{2}/ 10 6

( 4 !)^{2}/ 10 = 576 / 10 6

( 5 !)^{2}/ 10 = 0

Remainder is 1 + 4 + 6 + 6 = 17 /10 7 So P is odd number and

Our answer is 25

**Example – 20 :** Find the remainder when 2222^{5555} + 5555^{2222} is divided by 7

**Solution:** Here find remainder of the each number individually.

2222^{5555} / 7 = 3^{5555 } = (3)^{( 3 x 1851 + 2 )} / 7 = (27)^{ 1851 } x 9 / 7 5

5555^{2}222 / 7 = 4^{5555 } = (4)^{( 3 x 740 + 2 )} / 7 = (64)^{ 740 } x 16 / 7 2

Final remainder is = 5 +2 /7 0

**Example – 21 :** Find last digit of the expression 2017 ^{2017}

**Solution:** The last digit of an expression purpose simply find the remainder of that expression divided by 10.

So last digit of the number 2017 ^{2017 } is 7

**Related Topics :**

Rules for Divisibility of numbers

Formulas for Sum of n Consecutive numbers

GCD and LCM Problems & Solutions

Topics in Quantitative aptitude math for all types of exams

Shortcut Math Tricks for helpful to improve speed in all calculations

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## 2 thoughts on “Remainder Theorem Tough Questions for Competitive Exams | Aptitude Questions”

## Jitendra kumar dagur

(February 23, 2019 - 7:22 pm)Thank you very much sir for this valuable information on remainder theorem . I really appreciate this effot of yours. I will be very thankful to you if you kindly give me the source of these valuable questions and if you please mail me the question bank on this topic and other topics of mathematics . My E-mail id is intelect.jkd1@gmail.com and my contact number is 9035930724. Thanks sir

## sivaalluri

(March 12, 2019 - 5:52 pm)Thank you Mr.Jitendra kumar dagur