**Remainder Theorem & its application**

We have all learnt the Remainder Theorem in class 10 (now i am in 11) that when you divide a polynomial f(x) by x-c the remainder r will be f(c). Now let’s see how we can use this theorem in other situations.

**Example #1**

Let’s consider the following Product: 65 x 32.

We want to find out what is the remainder when it is divided by a number say 7.

To solve such questions we just need find the individual remainders when the numbers are divided by the divisor.

In this case 65 gives remainder 2 (65 -63) and 32 gives remainder 4 (32 – 28) when divide by 7. Multiplying the remainders we get 2*4=8

Since this number is greater than divisor, divide it again by the divisor again, i.e. 8/7 gives remainder 1.

Thus, when 65*32 is divided by 7 it gives remainder of 1. Isn’t it amazing! We save time and effort of multiplying large numbers and doing complex divisions.

### Example #2

Let’s see another example to find the remainder when 1421 * 1423 * 1425 is divided by 12

By this method 1421 * 1423 * 1425

1st step remainders = 5 * 7* 9 = 35*9

2nd step remainders = 11*9

3rd step remainder = 99/12 = 3

So the monstrous product gives a remainder of 3 when divided by 12.

### Example #3

Let’s suppose we want to find the last two digits of the product

22 * 31 * 44 * 27 * 37 * 43

For such problems we just need to find the remainder when it is divided by 100

(22 * 31) * (44 * 27) * (37 * 43)

1st step remainders = 82*88*91

2nd step remainder = 2 * 28

THATS IT!! The last two digits of the lengthy product is found within seconds and as you see it is 56

This is a guest post by one of the regular QuickerMaths.com follower Debasis Basak. On behalf of all the readers, I thank him for his contribution.