Trick to find Remainder - Remainder Theorem (Polynomial Theorem) for long division

REMAINDER USING BINOMIAL THEOREM

'Find the Remainder of..' is a very common question in competitive exams. In this post we will learn methods to find remainder of long divisions. Except tricks, it is also important to understand the basics of Remainder Theorem (Polynomial Theorem) because tricks originate from basic. By this theorem we can calculate remainder of numbers with large exponential powers easily and quickly. A Binomial is a polynomial with two terms, so we can use the term Binomial theorem or polynomial theorem interchangeably here.

Remainder means something which is "left over" after some operation. In Mathematics, when we divide any dividend by given divisor which is not a factor of that dividend, we get remainder.
⇒ Dividend = Divisor x Quotient + Remainder

There are two types of Remainder:
 1.  Positive Remainder also known as least positive remainder.
 2.  Negative Remainder also known as least absolute remainder.

For example: 23 divided by 5 can be written as: 4 x 5 + 3
Here 3 is least positive remainder

Also we can write like this: 5 x 5 - 2
Here 2 is least absolute remainder

Polynomial Theorem

This theorem is not only useful to determine remainder of numbers but also helpful in deriving numerous algebraic formulas.

Formula: (x + a)n = (x)n + nC1(x)n-1(a)1 + nC2(x)n-2(a)2 +  ..... + (a)n
When (x + a)n is Divided by x, Remainder will be = [(x)n + nC1(x)n-1(a)1 + nC2(x)n-2(a)2 +  ..... + (a)n]/x
               ⇒ Remainder = (a)n/x
Rest all terms contain multiple of x, so no remainder will be left when such terms divided by x.

Example: Find the Remainder of 999 divided by 8.
Solution:  We can write 999 as = (8 + 1)99/8
According to Remainder theorem, the term which will give remainder is: 199/8 = 1/8
                ⇒ Remainder = 1

Negative Polynomial Theorem

This theorem is useful to shorten the calculation. We will get negative remainders by this but remember Remainder can never be negative. We have to convert it in positive remainder.

Formula: (x - a)n = (x)n + nC1(x)n-1(-a)1 + nC2(x)n-2(-a)2 +  ..... + (-a)n 

When (x - a)n is Divided by x, Remainder will be = [(x)n + nC1(x)n-1(-a)1 + nC2(x)n-2(-a)2 + ..... + (-a)n]/x
                 ⇒ Remainder = (-a)n/x
Rest all terms contain multiple of x, so no remainder will be left when such terms divided by x.

Example: Find the Remainder of 31127 divided by 8.
Solution:  We can write 31127 as = (32 - 1)127/8
According to Remainder theorem, the only term which will give any remainder is: (-1)127/8
                 ⇒ Remainder = -1
But Remainder can't be negative, so = -1 + 8 = 7 is the correct remainder.

Formulas

 1.  If fraction is in the form of [an/(a+ 1)], then:

• Remainder will be 1 if n is even
• Remainder will be a if n is odd

 2.  If fraction is in the form of [an/(a- 1)], then:

• Remainder will be 1 always

 

 3.  If a1 ,a2 ,a3 are divided by d, remainder will be D1 , D2 , D3 But when (a1 + a2 + a3) is divided by d then remainder will be obtained by dividing (D1 + D2 + D3) by d.
⇒ Remainder [a1 + a2 + a3]/d = Rem [a1]/d + Rem [a2]/d + Rem [a3]/d

 4.  If a1 ,a2 ,a3 are divided by d, remainder will be D1 , D2 , D3 But when (a1 + a2 + a3) is divided by d then remainder will be obtained by dividing (D1 x D2 x D3) by d. 
⇒ Remainder [a1 x a2 x a3]/d = Rem [a1]/d x Rem [a2]/d x Rem [a3]/d

 5.  If two numbers are separately divided by a certain divisor leaving remainder r1 and r2 respectively, then remainder after their sum is divided by the same number will be = r1 + r2 - d.

 6.   If m and n are two integers then (m + n)! will be divisible by m!n!.

Example 1: Find the remainder of 216/5
Solution:  We can write it in the form of: (22)8/5 => (4)8/5
Now this fraction is same as given in Formula 1, so remainder will be '1'.

Example 2: Find the remainder of 326/8
Solution:  We can write it in the form of: (32)13/8 => (9)8/8
Now this fraction is same as given in Formula 2, so remainder will be '1'.

Example 3: Find the remainder when 15 x 17 x 19 is divided by 7.
Solution:  When 15, 17, and 19 are divided by 7 individually, we get 1, 3 & 5 as remainder respectively.
⇒ (15 x 17 x 19)/7 = (1 x 3 x 5)/7 = 15/7 => (14 + 1)/7
⇒ Remainder will be '1'.

More Theorems

 1.  Number of numbers which are less than N = Pa x Qb x Rc and co prime to each other are:
⇒ Ø(N) = N x (1 - 1/a)(1 - 1/b)(1 - 1/c)

If M and N are co prime then: Remainder [MØ(N)/N] = 1

Example: Find the remainder of 750/90
Solution:  90 = 2 x 32 x 5
First find number of numbers which are less than N and co prime to each other.
⇒ Ø(N) = 90 x (1 - 1/2)(1 - 1/3)(1 - 1/5) = 24
Since 7 & 90 are co prime to each other, therefore:
Rem [724/90] = Rem [748/90] = 1
⇒ Rem [750/90] = (72 x 748)/90 = 49 x 1 = 49

 2.  If M and N are co prime numbers and N is prime number then:

• Remainder [MN/N] = M

• Remainder [MN-1/N] = 1

Example: Find the remainder of 629/29
Solution:  Rem [629/29] = 6

 3.  If N is a prime number, then:

• Remainder [(N-1)!/N] = N - 1
  

• Remainder [(N - 2)!/N] = 1

Example: Find the remainder of 28!/29
Solution:  Rem [28!/29] = 28

Transformation

In any question, try to cancel out parts of the numerator and denominator since it directly reduces the calculations required. But remember to multiply that common part in initial remainder to get the correct remainder.

 CASE 1:  When the Dividend (A) and Divisor (B) have a factor in common, then: A = ka and B = kb
⇒ Rem [A/B] = Rem [ka/kb] = k x Rem [a/b]

Example 1: Find the remainder of 14/4?
Solution:  We know that 14/4 will give a remainder '2'.

Lets transform 14/4 = 7/2
When we divide 7 by 2 we will get Remainder = '1'
But this is incorrect answer.

In order to take care of this problem, we need to reverse the effect of this transformation. We divided the numerator and denominator by '2', so now we will multiply the answer by '2'.
So Remainder will be = 1 x 2 = '2'

Example 2: Find the last two digits of expression: 22 x 31 x 44 x 27 x 37 x 43
Solution:  To find last two digits, we need to divide it by 100.
         ⇒ (22 x 31 x 44 x 27 x 37 x 43)/100

Since 4 is a common factor, so divide numerator and denominator by 4.
         ⇒ (22 x 31 x 11 x 27 x 37 x 43)/25

Now by using polynomial theorem we will get remainder:
         ⇒ [(-3) x 6 x 11 x 2 x 12 x (-7)]/25
         ⇒ (18 x 77 x 24)/25
         ⇒ [(-7) x 2 x (-1)]/25
         ⇒ 14/25
Hence Remainder is '14'.

But remember we divided numerator and denominator by 4, so new remainder will be = 14 x 4 = 56
Hence, the last two digits of the product are = '56'.

 CASE 2:  When divisor can be broken into co-prime factors.
Let Dividend = A and Divisor = B which is a product of a*b
   ⇒ Remainder [A/B] = Rem [A/a*b]

Let Rem [A/a] = R1 and Rem [A/b] = R2

   ⇒ Remainder [A/B] = [axR2 + byR1]
But condition is: ax + by = 1

Example: Find the remainder of 290/91
Solution: Here divisor 91 = 7 x 13
   ⇒ R1 = Rem [290/7] = 1
   ⇒ R2 = Rem [290/13] = 12
Now we know that ax + by should be equal to 1
   ⇒ 7x + 13y = 1
For x = 2 and y = -1, this condition is satisfied.
   ⇒ Remainder [290/91] = [7 x 2 x 12 + 13 x (-1) x 1] = 165 which is more than 91 so remainder will be = 165 - 91 = 64

These are few tricks helpful in calculating remainder for long division easily. It is suggested to transform fractions to find remainder of long division quickly. Overall Remainder theorem is a great tool to calculate remainder. No method needs calculator or pen-paper, just practice. Good Luck 👍