Concept of Remainder

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:

Problems:

1. Find remainder of (32+28+21)/6
   (32+28+21)/6 à (2+4+3)/6 à9/6 à 3
   Remainder is 3.
2. Find remainder of (51+61+71)/4
   (51+61+71)/4 à (3+1+3)/4 à7/4 à 3
   Remainder is 3.

Important Rules to find the remainder

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

1. Remainder of aⁿ/(a+1) is ‘a‘ i.e.
   aⁿ/(a+1) à a, if ‘n’ is odd.
2. Remainder of aⁿ/(a+1) is ‘1‘ i.e.
   aⁿ/(a+1) à 1, if ‘n’ is even.

Problems:

1. Remainder of 2¹⁰⁰/3 = 1.
2. Remainder of 3⁹⁹/4 = 3.
3. Remainder of 65²⁰⁵/66 = 65.
4. Find remainder of (65²⁰⁶+1)/66
   (65²⁰⁶+1)/66 à (1+1)/66 à2

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

1. Remainder of (ax+b)ⁿ/a is remainder of bⁿ/a
   (ax+b)ⁿ/a àremainder of bⁿ/a
2. Remainder of (ax+1)ⁿ/a is 1
   (ax+1)ⁿ/a àremainder of bⁿ/a

Problems:

1. Find the remainder of 51²⁰³/7
   Solution: 51²⁰³/7 = (7×7+2)²⁰³/7 à 2²⁰³/7 = ((2³)⁶⁷×2²)/7 à (8⁶⁷×4)/7 à ((7×1+1)⁶⁷×4)/7 à (1×4)/7 à 4

Click here to get sample questions.