**Parity** in Mathematics is a term which we use to express if a given integer is even or odd. It is basically depend on the remainder when we divide a number by 2.

Parity can be divided into two categories – **1. Even Parity**

**2. Odd Parity **

**Even Parity :** If we divide any number by 2 and the remainder is ‘0’,the parity is even or ‘0’ parity

**Odd Parity :** If we divide any number by 2 and the remainder is ‘1’,the parity is odd or ‘1’ parity. Till this we all know but let’s try to explore this by some of the problems.

## Let’s understand it with the help of a few problems:

**Problem 1:** In how many ways can 10001 be written as the sum of two primes? (AMC 8,2011 Prob.28)

This problem can easily be solved using the ‘Parity’ concept.

The above rules have wide range of utility in Mathematics.

According to the problem the 10001 need to be expressed by the sum of two Prime Numbers.

10001 is odd number and to get the odd sum we need to add one odd number and one even number.This is the basic criteria of Parity.

Again the numbers should be Prime.

The only Even Prime Number is 2 but we cannot consider the other number to be 10001 – 2 = 9999 which is not a Prime Number.

Hence, **We can’t express 10001 as the sum of the two Prime Numbers.**

**Problem 2 :**

Suppose you have written the numbers 1 2 3 4 5 6 7 8 9 10.

You have to plug in ‘+’ or ‘-’ in between these numbers. You have the complete freedom to plug in anywhere.

The question is can the be sum zero? Ever?

**Step 1:** Let’s start by taking some numbers :

1 2 3 4 5 6 7 8 9 10

**Step 2:** Plug in the signs ‘+’ or ‘-‘.

1 + 2 + 3 – 4 + 5 – 6 – 7 + 8 – 9 – 10 = -17 ——– (1)

(You can take any signs its just an example)

**Step 3:** We need to make all the -ve numbers in the LHS of \(eq^n\) (1) into +ve numbers. It’s an easy calculation that if we add double the number (the numbers in negative) we will get the same number in positive, eg. -x + 2x = +x .

1 + 2 + 3 – 4 +8+ 5 – 6+12 – 7 +14+ 8 – 9+18 – 10 +20= -17+8+12+14+18+20

1+2+…………+10 = 55

Basically, (1 + 2 + 3 – 4 + 5 – 6 – 7 + 8 – 9 – 10) + (8+12+14+18+20) = 55

Initial Sum + Bunch of Positive numbers = Odd Number

Initial sum was = -17 again an odd number

Odd Number + Even Numbers(as double of any number is even) = Odd

But if the initial sum is ‘0’ which is an even number then it’s not possible.

As only , odd + even = odd .

Hence, **It is not possible to be the initial sum as ‘0’.**

### Watch the Video – Parity in Mathematics:

### Some Useful Links:

- Triangular Number Sequence – Explanation with application
- How to solve weird equations using Inequality? – Video