NUMBER SYSTEMS EXAMPLE QUESTIONS PART 4

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  + – – – – – – +  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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