Remainder Theorem Tough Questions for Competitive Exams | Aptitude Questions

Remainder Theorem Aptitude Examples with Answers | Remainder Theorem Tutorial

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.

= \frac{ 120 \ \times \ 2587 \times 247 \times \ 952 \times \ 854 } {100}

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

= \frac{ 6 \ \times \ 2587 \times 247 \times \ 952 \times \ 854 } {5}

Now find the remainders of the each number

= \frac{ 1 \ \times \ 2 \times 2 \times \ 2 \times \ 4 } {5} = \frac{ 32 } {5}

Remainder of the above expression is 2

Now reminder of the initial expression is  40 ( Multiplying the reminder with canceled number i.e   \frac{2 \times \ 20 }{100}  )

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

= \frac{ 1! \ + \ 2! + \ 3! + \ 4! \ + 5! \ +6! \ + \ - \ - \ - \ - \ + 1000!}{8}

Now calculate remainder for each number in the given series

\frac{ 1! }{8} \ = \ \frac{1 }{8 } \ \overset{R}{\rightarrow} \ 1

 \frac{ 2! }{8} \ = \ \frac{2 \times \ 1 }{8 } \ \ \overset{R} {\rightarrow} \ \ \ 2

\frac{ 3! }{8} \ = \ \frac{3 \ \times2 \times \ 1 }{8 } \ \ \overset{R} {\rightarrow} \ \ 6

\frac{ 4! }{8} \ = \ \frac{4 \times \ 3 \ \times2 \times \ 1 }{8} \ \overset{R} {\rightarrow} \ \ 0

The remainder of the remaining terms is also zero. So

= \frac{ 1 \ + \ 2 + \ 6 + \ 0 \ + 0 + \ - \ - \ - \ - \ + 0}{8} = \frac{9}{8 } \ \ \overset{R}{\rightarrow} \ \ 1

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

Solution:  = \ \frac{ 1! \ + \ 2! + \ 3! + \ 4! \ + 5! \ +6! \ + \ - \ - \ - \ - \ + 1000!}{14}

Now calculate remainder for each number in the given series

\frac{ 1! }{14} \ = \ \frac{1 }{14 } \ \overset{R}{\rightarrow} \ 1

\frac{ 2! }{14} \ = \ \frac{2 \times \ 1 }{14 } \ \ \overset{R} {\rightarrow} \ \ \ 2

 \frac{ 3! }{14} \ = \ \frac{3 \ \times2 \times \ 1 }{14 } \ \ \overset{R} {\rightarrow} \ \ 6

 \frac{ 4! }{14} \ = \ \frac{4 \times \ 3 \ \times2 \times \ 1 }{14 } \ \overset{R} {\rightarrow} \ \ - 4

5! / 14  \overset{R}{\rightarrow}  8

\frac{ 6! }{14} \ = \ \frac{6 \times\ 5 \ \times \ 4 \times \ 3 \ \times2 \times \ 1 }{14 } = \frac{ 720 }{14} \ \ \overset{R} {\rightarrow} \ 6

\frac{ 7! }{14} \ = \ \frac{7 \ \times6 \times\ 5 \ \times \ 4 \times \ 3 \ \times2 \times \ 1 }{14 } \ \ \overset{R} {\rightarrow} \ \ 0

The remainder of the remaining terms is also zero.  NOW

= 1 + 2 +6 – 4 + 8 + 6 /14

= 19 / 14 \overset{R}{\rightarrow} 5

Remainder of the given sum is 5

Example – 4 : Find the remainder when 51203 is divided by 7

Solution: Find Remainder of the expression 51 / 7

51 / 7 \overset{R}{\rightarrow}  2  So

\frac{51 ^{203}}{7} \ \ = \frac{2 ^{203}}{7} \ \

= \ \frac{(2^3) ^{67} \times 2^2}{7} \ \ = \frac{8^{67} \times \ 4}{7} \ \= \ \frac{(7 \ +1) ^{67} \times \4}{7} \ \ \overset{R}{\rightarrow} \ \ 4

Example – 5 : Find the remainder when 21875 is divided by 17

Solution: Find Remainder of the expression 21 / 7

\frac{21 ^{875}}{17} \ \ = \frac{4 ^{875}}{17} \ \

Here our aim is obtained  number as   24   = 16 ( 16 = 17 – 1) so according to that rewrite the equation as follow as

= \frac{(2^2) ^{875}}{17} \ \ = \frac{(2^4) ^{437} \times \ 2^2}{17} \ \

= \frac{(17 \ - 1 ) ^{437} \times \ 2^2}{17} \ \ \overset{R}{\rightarrow} \ -1 \times 4 = -4

Final remainder is 17 – 4 = 13

Example – 6 : Find the remainder when 2787 x 2345 x 1992 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 341 +782  is divided by 52

Solution:

Hint : If an +bn  and  n = odd number then (a +b) is exactly divisible of that number an +bn

The given expression can be written as

341 +782  =  341 +4941

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

So remainder is Zero

Example – 8 : Find the remainder when 53 + 173 +183 +19 is divided by 70

Solution:

Hint: If An +Bn + Cn  + Dn 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 163 + 173 +183 +193

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

= \ \ \frac{12899 \ \times \ 96 \ \times \ 997}{100}

= \ \ \frac{ -1 \ \times \ -4 \ \times \ -3}{100} = \ \ \frac{ -12}{100}

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 \overset{R}{\rightarrow} 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 \Rightarrow 30 times remaining number is 88

So 88 / 37 \overset{R}{\rightarrow}  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 \Rightarrow 42 times remaining number is 7

So 7 / 13 \overset{R}{\rightarrow}  7

Example – 13 : Find the remainder when 101 + 102 +103 + 10 + – – – – – – +  10100   is divided by 6

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

101/ 6   \overset{R}{\rightarrow} +4

102/ 6   \overset{R}{\rightarrow} +4

103/ 6   \overset{R}{\rightarrow} +4

104/ 6   \overset{R}{\rightarrow} +4

So final remainder = 100 x 4 / 6 \overset{R}{\rightarrow} +4

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

Example -14: Find the remainder when 2469 + 3268   is divided by 22

Solution: Here find remainder of the each number individually.

\frac{2^{469}}{22} \ = \ \frac{2^{468}}{11}

 = \ \frac{2^{5 \times93 \ + 3 }}{11} \ \ = \ \frac{32^{93} \times2^3}{11} \ \

 = \ \frac{(33 - 1)^{93} \times 8 }{11} \ \ = \frac {-1 \times 8 }{11} \ \ \overset{R}{\rightarrow} 3

Final remainder of this term = 3 x 2 /11 \overset{R}{\rightarrow} 6

\frac{3^{268}}{22} \  = \ \frac{3^{5 \times53 \ + 3 }}{22} \ \ = \ \frac{243^{53} \times3^3}{22} \ \

 = \ \frac{(242+1)^{53} \times 27 }{22} \ \ = \frac {27 }{22} \ \ \overset{R}{\rightarrow} 5

Final remainder of the give expression = 6 +5 /22 = 11 /22  \overset{R}{\rightarrow} 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    \overset{R}{\rightarrow}  1

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

= 9/ 6    \overset{R}{\rightarrow}  3

Example – 16 : Find the remainder when 2310 – 1024 is divided by  7

Solution: Here 1024 can be written as 2310

The expression is 2310 – 210

Hint: If the given expression like ( an – bn ) 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 2310 – 1024 

Remainder = 0

Example – 17 : Find the remainder when 341 + 782 is divided by  26

Solution: Here   782 can be written as 4941

The expression is 341 + 4941

Hint: If the given expression like ( an + bn ) 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 341 + 782

Remainder = 0

Example – 18 : Find the remainder when 2723 + 1923 is divided by  2

Solution:

Hint: If the given expression like ( an – bn ) 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 2723 + 1923

Remainder = 0

Example – 19 : Find the remainder when 52P  is divided by  26 where P = (1 !)2 + (2 !)2 + (3 !)2 + – – – – –  + (10 !)2

Solution: Here  52p = 25P  = (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 \overset{R}{\rightarrow} 1

( 2 !)2/ 10 \overset{R}{\rightarrow} 4

(3 !)2/ 10 \overset{R}{\rightarrow} 6

( 4 !)2/ 10 = 576 / 10 \overset{R}{\rightarrow} 6

( 5 !)2/ 10 =  \overset{R}{\rightarrow} 0

Remainder is 1 + 4 + 6 + 6 = 17 /10 \overset{R}{\rightarrow} 7  So P is odd number and

Our answer is 25

Example – 20 : Find the remainder when 22225555  + 55552222  is divided by 7

Solution: Here find remainder of the each number individually.

22225555 / 7 = 35555  = (3)( 3 x 1851 + 2 ) / 7 = (27) 1851 x 9 / 7  \overset{R}{\rightarrow}  5

55552222 / 7 =  45555  = (4)( 3 x 740 + 2 ) / 7 = (64) 740 x 16 / 7  \overset{R}{\rightarrow}  2

Final remainder is = 5 +2 /7  \overset{R}{\rightarrow} 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.

\frac{2017 ^{2017}}{10 } \ = \frac{7 ^{2017}}{10 } \

= \ \frac{7 ^{2 \ \times \ 1008 \ + 1}}{10 } \ = \ \frac{49 ^{ \ 1008 \ } \ \times7}{10 } \

 = \ \frac{(50 -1) ^{ \ 1008 \ } \ \times7}{10 } \ = \ \frac{( -1) ^{ \ 1008 \ } \ \times7}{10 } \

 = \frac{7}{10} \ \overset{R}{\rightarrow} 7

So last digit of the number  2017 2017  is  7

 

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