Hello guys! Today we will discuss about concept of remainder and other methods to find remainder of a number.

Firstly tell me what will be the remainder when number 22 is divided by 5. Yes, as you guess it will be 2 as depicted below:

Now what will be the remainder when -22 is divided by 5? Many of you might guess it will be -2, but you are wrong. So let’s see how:

It is 3.

**Note:** Remainder is always non-negative. It cannot be negative.

**Rule: Dividend = Divisor ****× Quotient + Remainder**

## Remainder Theorem

It states that remainder of product of number is such that:

Where a_{r} = remainder when a is divided by n,

b_{r} = remainder when b is divided by n,

c_{r} = remainder when c is divided by n.

**Problems:**

**1. **Find remainder of (32×28×21)/6

(32×28×21)/6 à (2×4×3)/6 à24/6 à 0

Remainder is 0.

**2. **Find remainder of (51×61×71)/4

(51×61×71)/4 à (3×1×3)/4 à9/4 à 1

Remainder is 1.

Similarly for addition, the method for calculating the remainder is same as given below:

### Important Rules to find the remainder

**Rule 1:** For ‘a’ and ‘n’ be any position integer

- Remainder of a
^{n}/(a+1) is ‘a‘ i.e.

a^{n}/(a+1) à a, if ‘n’ is odd.
- Remainder of a
^{n}/(a+1) is ‘1‘ i.e.

a^{n}/(a+1) à 1, if ‘n’ is even.

**Problems:**

- Remainder of 2
^{100}/3 = 1.
- Remainder of 3
^{99}/4 = 3.
- Remainder of 65
^{205}/66 = 65.
- Find remainder of (65
^{206}+1)/66

(65^{206}+1)/66 à (1+1)/66 à2

**Rule 2: **For ‘a’ and ‘n’ be any position integer

- Remainder of (ax+b)
^{n}/a is remainder of b^{n}/a

(ax+b)^{n}/a àremainder of b^{n}/a
- Remainder of (ax+1)
^{n}/a is 1

(ax+1)^{n}/a àremainder of b^{n}/a

**Problems:**

- Find the remainder of 51
^{203}/7

**Solution: **51^{203}/7 = (7×7+2)^{203}/7 à 2^{203}/7 = ((2^{3})^{67}×2^{2})/7 à (8^{67}×4)/7 à ((7×1+1)^{67}×4)/7 à (1×4)/7 à 4