Math Olympiad

What is Parity in Mathematics ? 🧐

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’.

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Parity from 2020 AMC 10A Problem – 18

What is Parity?

In mathematics, parity is the property of an integer’s inclusion in one of two categories: even or odd. An integer is even if it is divisible by two and odd if it is not even .

Try the problem

Let $( \textbf a, \textbf b, \textbf c, \textbf d)$ be an ordered quadruple of not necessarily distinct integers, each one of them in the set ${0,1,2,3}$ For how many such quadruples is it true that $ \textbf a\cdot \textbf d – \textbf b\cdot \textbf c$ is odd?

$\textbf{(A) } 48 \qquad \textbf{(B) } 64 \qquad \textbf{(C) } 96 \qquad \textbf{(D) } 128 \qquad \textbf{(E) } 192$

2020 AMC 10A Problem-18


4 out of 10

Mathematics Circle

Knowledge Graph

Parity-Knowledge Graph

Use some hints

We need exactly one term to be odd, one term to be even. Because of symmetry,let us set $\textbf a \textbf d$ to be odd and $\textbf b \textbf c$ to be even,then multiple by $2$.

Now can you complete the sum using odd and even property?

See If  $\textbf a \textbf d$ is odd, then both $\textbf a$ and $\textbf  d$ must be odd, therefore there are $2$.$2$=$4$ possibilities for $\textbf a \textbf d$.

now consider $\textbf b \textbf c$, we can say that $\textbf b \textbf c$ is even,then there are $2$.$4$=$8$ possibilities for $\textbf b \textbf c$ . However, $\textbf b$ can be that case $2$.$2$=$4$ more possibilities for $\textbf b \textbf c$. Thus there are $8$+$4$=$12$ ways for us to choose $\textbf b \textbf c$ and also $4$ ways are there to choose $\textbf a \textbf d$.

Considering symmetry, to $\textbf a \textbf d $- $\textbf b \textbf c$ be odd,there are $12$.$4$.$2$ = $96$ quadruples .So, the answer is $96$.

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Understand the Problem

For each \( n \in \mathbb{N} \) let \( d_n \) denote the G.C.D. of n and (2019 – n). Find the value of \( d_1 + d_2 + … + d_{2019} \).

First, try these problems.

  1. Show that G.C.D. of k and 0 is k for any positive integer k.
  2. Show rigorously that G.C.D. (a, b) = G.C.D. (a, a+b) for any non-negative integers a and b
  3. Can you find and prove a similar result with a negative sign?

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