Remainder Theorem and its application in advance

Remainder Theorem:

If a and b are natural numbers, with b is not equal to zero 0, then there exist unique integers q and r, such that a = b*q + r and 0 ≤ r < b. 

The number q is called the quotient, while r is called the remainder.

Dividend = Divisor × Quotient + Remainder.

Example(1):-

23 = 4 * 5 + 3, which means 23 when divided by 4 will give 3 as remainder. 

It is simple that remainder is less than 4

Used in advanced 

Example(2):-

Suppose we have to find the remainder When 45*22*67 is divided by 7.

Simple divide all number by 7

             45=7*6+3

             22=7*3+1

             67=7*9+4

Multiply all remainder    3*1*4=12

                      then divide 12 by 7

                      12=7*1+5

Answer is 5

Example (3):-

Find the reminder when 1! + 2! + 3! + . . . . .. . .. 50! is divided by the  7. 

From 7! to 50! the remainder will be zero.since 7! is nothing but product of first 7 natural numbers and all factorial after that will have 7 as one of the factor. so we are concerned only factorials 1! to 6!, i.e, 1! + 2! + 3! + 4! + 5! + 6!

1! + 2! + 3! + 4! + 5! + 6! = 1+2+6+24+120+720 divided by 7

 Remainder is 1+2+6+3+1+1=14

                        14 divided by 7  remainder is 0

Example (4):-

What is the remainder when 50100 is divided by 7

Simple all 50 is divide by 7 remainder is 1

and 1100 is 1 hence remainder is 1

Concept of negative remainder

By the definition remainder can not be negative. But considering negative
 remainder is a very useful exam trick.
Note:-Whenever remainder is negative number  make it positive by 
           adding the divisor to the negative remainder.

Example (5):-

What is the remainder when 413 divided by 5.

Here 4 when divided by 5 gives a remainder of -1

So we are asked to find (-1) * (-1) * … 13 times=-1 divided by 5.

Which is Remainder[-1/5] = -1.

Here required answer is 5+ (-1) = 4.

Example (6):-

Remainder when (21 * 23) is divided by 24

Remainder[(-3) * (-1) / 24 ]( as 21 = 24 * 1 – 3

 and 23 = 24 * 1 – 1)

= Remainder [ -3*-1 /24 ] = 3

Example (7):-

Remainder when (99*98 *97) is divided by 100

Remainder [ 99 *98*97 / 100 ] = Remainder[(-1)*(-2) * (-3) / 100 ]

( as 99 = 100 * 1 – 1 , 98 = 100 * 1 – 2 and 97 = 100 * 1 – 3)

= Remainder [ -6 /100 ] = -6

Here required answer is 100+ (-6) = 94.

Remainder by Binomial Theorem

Binomial Theorem is 

(a+b)ⁿ=aⁿ+C(n,1)a^n-1 b+ C(n,2) a^n-2b²+ C(n,3) a^n-3b³+………+ C(n,n)bⁿ.

Example (8):-

Remainder when 2175 is divided by 17.

2175 = (17+4)²⁵=4⁷⁵ since all terms containing either 17 or power of 17 
except last term by Binomial theorem

Hence remainder is 4⁷⁵=4^2*37+1=16³⁷*4=(17-1) ³⁷*4=(-1) ³⁷*4=-4

Here remainder is 17+ (-4) = 13.

Fermat’s little theorem

      a^{( p – 1 )}  is divided by p remainder is 1 where p is prime.

     a prime number and a and p are co-primes

Example (9):- 17¹⁰ is divided by 11 remainder is 1  

                             17 and 11 are co prime and p is prime.

Wilson’s theorem

If (p-1)! is divided by p then remainder is p-1 (where p is prime)

Example (10):-

   Remainder of [16!/17] = 16.
Note:-If (p-1)! is divided by p then remainder is 0 (zero) (if p is not prime)
            Since P can split into at least two prime factor and both the factor
             present in (p-1)! 
Example (11):-Remainder [15!/16] = 0.