# Even and Odd Numbers

An **even number** is a number which has a remainder of \(0\) upon division by \(2,\) while an **odd number** is a number which has a remainder of \(1\) upon division by \(2.\)

Thus, the set of integers can be partitioned into two sets based on parity:

- the set of
*even*(or parity 0) integers, and - the set of
*odd*(or parity 1) integers.

## Definition

An even number has parity \(0\) because the remainder upon division by \(2\) is \(0\), while an odd number has parity \(1\) because the remainder upon division by \(2\) is \(1\). For example, \(0,2,4,10,-6\) are all even numbers because they leave a remainder of 0 upon division by \(2\). The integers \(1,3,5,11,-7\) are all odd numbers because they leave a remainder of 1 upon division by \(2\).

Every integer is either even or odd, and no integer is both even and odd. This includes 0, which is even.

Figure out whether 1729 is an odd or even number.

Since the remainder obtained on dividing 1729 by 2 is 1, 1729 is an odd number.

\[\text{OR}\]

The number 1729 ends with the digit "9." Thus it is an odd number. \(_\square\)

Figure out whether 1000 is an odd or even number.

Since the remainder obtained on dividing 1000 by 2 is 0, 1000 is an even number.

\[\text{OR}\]

The number 1000 ends with the digit "0." Thus it is an even number. \(_\square\)

Is the number 2222452122 odd or even?

The last digit is 2, and 2 is an even number. So, 2222452122 is an even number. \(_\square\)

## Even and Odd (Parity) Properties

The following are the parity properties of even and odd numbers:

- even \( \pm\) even = even
- odd \( \pm\) odd=even
- even \( \pm\) odd= odd
- even \( \times\) even= even
- even \(\times\) odd= even
- odd \( \times\) odd= odd

These properties are often useful for testing whether an equality is false by using the parity rules of arithmetic to see whether both sides have the same parity. The application of these rules becomes clear through the following examples and problems:

## If \(n\) is an integer, what is the parity of \(2n+2\)?

Since \(n\) is an integer, \(n+1\) is also an integer. Then, \(2n+2 = 2(n+1) + 0\) shows that the parity of \(2n+2\) is \(0,\) which implies \(2n+2\) is always an even number. \(_\square\)

Is the number \(\left(47630750675+453407032\right) \times 549068453\) even or odd?

To answer this question, it would be unwise to actually multiply out these numbers. Instead, we can apply the properties of even and odd numbers.

Since \(47630750675\) ends in a 5, it is odd. On the other hand, since \(453407032\) ends in a 2, it is even. By property 3, even \( \pm\) odd= odd, so \(47630750675+453407032\) is odd. Since that sum is being multiplied by \(549068453,\) which is odd, the entire number is odd since property 6 gives odd \( \times\) odd= odd. \(_\square\)

Here are some problems to try.

## Problem Solving

Here are the examples and problems motivated to enhance the problem solving skills based on the parity of odd and even numbers. Go thorough them to achieve the objectives of this section.

## If \(a\) and \(b\) are integers, what is the parity of \(a \times b\)?

We know that an odd number multiplied by an odd number remains odd, an even number multiplied by an odd number is even, and an even number multiplied by an even number is even. This can be summarized as

\[ ( \mbox{Parity of } a ) \times ( \mbox{Parity of } b ) = (\mbox{Parity of } ab). \ _\square\]

## Let \( P \) be the product of the first 100 prime numbers. What is the parity of \( P \)?

We see that the first prime number is 2, which is even. The rest of the 99 prime numbers are all odd. The product of these 99 primes will be an integer, say \( k \). Multiplying an even number by another integer always gives an even number; so we can write \( P \) as \( 2 k \). Dividing \( P \) by 2 does not leave a remainder, and therefore \( P \) is even. \(_\square \)

If \( k \) is an integer, which of the following is always even?

\(\quad \text{A})\) \( 2k + 1 \)

\(\quad \text{B})\) \( k^2 \)

\(\quad \text{C})\) \(4k + 4 \)

\(\quad \text{D})\) \( k^2 -1 \)

A) is always odd for any \( k\).

B) is odd whenever \( k \) is odd.

D) is odd whenever \( k \) is even.C) can be rewritten as \( 4k+4 = 2(2k+2) +0 \), which means the remainder upon division by \(2\) is always \(0.\) Thus, \( 4k+4\) is always even, showing the correct answer is C). \( _\square \)

## If \(k\) is an integer, what is the parity of \( k^2 + k\)?

Observe that \( k^2 + k = k (k+1),\) where \( k\) and \( (k+1)\) have different parity. Then by the arithmetic rules of parity, the parity of \( k(k+1)\) is \( 0\). \(_\square\)

## For integers \(x\) and \(y\), show that \(\frac{{x}^{2}+{y}^{2}}{2}+\frac{x+y}{2}\) is also an integer.

Rewriting the expression as \( \frac{x^2 + x } { 2} + \frac{ y^2 + y}{2},\) we will show that for any integer \(n\), \( \frac{n^2 + n } { 2} \) is also an integer.

This follows because \( n^2 + n = n (n+1) \), which is the product of \(2\) consecutive integers. Since one of these integers is even, the product is even. Thus, when we divide \( n^2 + n\) by \(2,\) we will obtain an integer. \(_\square \)

Try the following problems.

For further applications of parity in combinatorics, see Parity - Intermediate.

## See Also

**Cite as:**Even and Odd Numbers.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/even-and-odd-numbers/