Even and Odd Numbers: Problems and Solutions

This page presents clear definitions, key properties, and worked problems on even and odd numbers. Additional practice questions and their detailed solutions are also included.

Definitions

An even number is any integer that is divisible by 2.

Examples: \(\ldots, -4, -2, 0, 2, 4, \ldots\)

Any even number can be written in the form

[ 2n ]

where \(n\) is an integer.

An odd number is any integer that is not divisible by 2.

Examples: \(\ldots, -5, -3, -1, 1, 3, \ldots\)

Any odd number can be written in the form

[ 2n + 1 ]

where \(n\) is an integer.

Problems on Even and Odd Numbers

Problem 1

List all even numbers greater than \(-4\) and smaller than \(20\).

Solution

\(-2, 0, 2, 4, 6, 8, 10, 12, 14, 16, 18\)

Problem 2

List all odd numbers greater than \(3\) and smaller than \(30\).

Solution

\(5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29\)

Problem 3

Show that the sum of two even numbers is even.

Solution

Let \(2n\) and \(2k\) be two even numbers. Their sum is

[ 2n + 2k = 2(n + k) ]

Let \(N = n + k\). Then

[ 2n + 2k = 2N ]

Since the sum is a multiple of 2, it is even.

Problem 4

Show that the sum of an even number and an odd number is odd.

Solution

Let \(2n\) be an even number and \(2k + 1\) be an odd number. Their sum is

[ 2n + (2k + 1) = 2(n + k) + 1 ]

Let \(N = n + k\). Then the sum becomes

[ 2N + 1 ]

This is an odd number.

Problem 5

Show that the sum of two odd numbers is even.

Solution

Let \(2n + 1\) and \(2k + 1\) be two odd numbers. Their sum is

[ (2n + 1) + (2k + 1) = 2(n + k + 1) ]

Let \(N = n + k + 1\). Then the sum is \(2N\), which is even.

Problem 6

Show that the sum of three odd numbers is odd.

Solution

Let \(2m + 1\), \(2n + 1\), and \(2k + 1\) be three odd numbers. Their sum is

[ (2m + 1) + (2n + 1) + (2k + 1) = 2(m + n + k + 1) + 1 ]

Let \(N = m + n + k + 1\). The sum is \(2N + 1\), which is odd.

Problem 7

Show that the square of an odd number is odd.

Solution

Let \(2n + 1\) be an odd number. Then

[ (2n + 1)^2 = 4n^2 + 4n + 1 = 2(2n^2 + 2n) + 1 ]

Let \(N = 2n^2 + 2n\). The square can be written as \(2N + 1\), which is odd.

Problem 8

Show that the product of an odd number and an even number is even.

Solution

Let \(2m + 1\) be an odd number and \(2n\) be an even number. Their product is

[ (2m + 1)(2n) = 2(2mn + n) ]

Since the product is a multiple of 2, it is even.

Questions

Complete each sentence using even or odd.

  1. The product of two odd numbers is ______.
  2. The product of two even numbers is ______.
  3. The difference of two even numbers is ______.
  4. The sum of two even numbers and one odd number is ______.
  5. The square of an even number is ______.

Solutions

  1. Odd
    \[(2n + 1)(2p + 1) = 2(2np + n + p) + 1\]
  2. Even
    \[(2n)(2p) = 2(2np)\]
  3. Even
    \[2n - 2p = 2(n - p)\]
  4. Odd
    \[2n + 2p + (2m + 1) = 2(n + p + m) + 1\]
  5. Even
    \[(2n)^2 = 4n^2 = 2(2n^2)\]

More References