NUMBER SYSTEMS EXAMPLE QUESTIONS PART 4

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 +19³ 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 + 10⁴ + – – – – – – + 10¹⁰⁰is 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 ^777777777is 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 5^2p = 25^P= (26 – 1)^PSo 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)¹⁸⁵¹x 9 / 7 $\overset{R}{\rightarrow}$ 5

5555²222 / 7 = 4⁵⁵⁵⁵= (4)^{( 3 x 740 + 2 )} / 7 = (64)⁷⁴⁰x 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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NUMBER SYSTEMS EXAMPLE QUESTIONS PART 4

Remainder Theorem Sums

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