# Odd and even numbers

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Ahmed Nasrullah
on 22 Jan 2016

Answered: Athanasios Paraskevopoulos
on 17 Aug 2024

##### 2 Comments

Munni
on 16 Aug 2023

print the values of odd number and even numbers using 2 seperate veriables in erange 10 to 120

Walter Roberson
on 16 Aug 2023

### Accepted Answer

Guillaume
on 18 Apr 2023

Edited: MathWorks Support Team
on 18 Apr 2023

You simply have to go back to the definition of odd and even. An (integer) number is even if it is divisible by 2, odd otherwise. Divisible by 2 means that the remainder when divided by 2 is 0. That is easy to test, the function to get the remainder is (or you can use ). As with many things in matlab you do not need a loop, the functions work on vector / matrices

m = [1 2 3;4 5 6];

iseven = rem(m, 2) == 0;

Mathworks Support Staff Comments

Depending on the usage, you can use "rem" or "mod" to check if a number is divisible by 2. The "r = rem(a,b)" function finds the remainder after division of a by b, where the sign of the remainder r follows the sign of a. Meanwhile, the "r = mod(a,b)" function finds the remainder after division of a by b, where the sign of the remainder r follows the sign of b.

For example, if you have a matrix of integer numbers that are positive only, you can find the even or odd numbers by first finding the matrix indices of these elements using "rem":

m = [1 2 3; 4 5 6]

iseven = rem(m,2) == 0;

m_even = m(iseven)

isodd = rem(m,2) == 1;

m_odd = m(isodd)

If you want to execute some statements based on whether the positive number in m is even or odd, you can make a conditional statement within a “for” loop:

for i = 1:numel(m)

if rem(m(i),2) == 0

% statements to execute for even m(i)

else

% statements to execute for odd m(i)

end

end

Otherwise, if you have a matrix of integer numbers that can be negative in general, use "mod" instead of "rem" in the above example.

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### More Answers (8)

Tally Miller
on 5 Nov 2018

you can also calculate the remainder after the division. b = mod(a,m) returns the remainder after division of a by m, where a is the dividend and m is the divisor. This function is often called the modulo operation, which can be expressed as b = a - m.*floor(a./m). The mod function follows the convention that mod(a,0) returns a.

by dividing by two, the remainder is either 1 or 0. For odd numbers, the remainder will be 1, and for even, it will be 0. Just be sure to use "==" for a true/false response when querying.

m=1

if mod(m,2) == 1

else

end

##### 0 Comments

ElPerroVerde
on 8 Jun 2020

You can divide the numer by 2 and see if there's any decimal. I use this.

if floor(n/2)==n/2

% code for even

else

% code for odd

end

Note that the "floor" code returns the number without decimals. So if there's a even number, the result of n/2 will not have any decimal.

##### 0 Comments

Medical Imaging
on 23 Sep 2020

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.

If the units digit (or ones digit) is 1,3, 5, 7, or 9, then the number is called an odd number, and if the units digit is 0, 2, 4, 6, or 8, then the number is called an even number (for the set of numbers 0-9). Thus, the set of integers can be partitioned into two sets based on parity:

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

**Parity is a fundamental property of integers, and many seemingly difficult problems can be solved by making parity arguments.

The below function is to find the details of even and odd number in a given list

function [sum_even,sum_odd,n_even,n_odd] = even_odd(x)

sum_even = 0; % initialization of sum of even numbers in the given list

sum_odd = 0; % initialization of sum of odd numbers in the given list

n_even = 0; % initialization of total even numbers in the given list

n_odd = 0; % initialization of total odd numbers in the given list

for i=1:1:x % start of the list

if mod(i,2)==0

sum_even=sum_even+i; % sum of the even numbers

n_even=n_even+1; % % increase sum of even number by 1

else

sum_odd=sum_odd+i; % sum of the odd numbers

n_odd=n_odd+1; % increase sum of even number by 1

end

end

Example set for list:1,2,3,4,5,6,7,8,9,10

sum_even = 30, which is 2+4+6+8+10 = 30

sum_odd = 25 , which is 1+3+5+7+9 = 25

n_even = 5, which are 2,4,6,8,10

n_odd = 5, which are 1,3,5,7,9

I hope this helps.

##### 0 Comments

Jan Siegmund
on 7 Apr 2020

Edited: Jan Siegmund
on 7 Apr 2020

Fancy answer:

m = [1 2 3;4 5 6];

isodd = bitget(m,1)

however this is slower as rem

##### 0 Comments

JAISAI KRISHNAN
on 21 Jun 2022

You can use bitwise 'AND' to check if the number is odd (or) even.

vec = 1:1:10;

oddCheck = bitand(vec,1);

##### 0 Comments

Temidayo Daniel
on 16 Oct 2022

Edited: Walter Roberson
on 24 Mar 2024

x = 1:10

Y = zeros(1,10)

for i = 1:10

if mod (x(i), 2) == 0

Y(i) = 1;

else

Y(i) = 0;

end

end

you can also switch it for odd numbers between 1 t0 10 .

##### 1 Comment

DGM
on 24 Mar 2024

Yeah, but why use loops? Why use literal numeric assignments? It can all be done in one line, for any size array, and the output is more useful, since it can directly be used for indexing.

x = 1:10

isodd = logical(mod(x,2)) % true when odd

Athanasios Paraskevopoulos
on 17 Aug 2024

You can solve this problem using a loop and an if-else statement to check whether a number is odd or even.

% Define the range of numbers to check

startNum = 1; % Starting number

endNum = 10; % Ending number

% Loop through each number in the range

for num = startNum:endNum

% Check if the number is even

if mod(num, 2) == 0

fprintf('%d is even\n', num);

else

fprintf('%d is odd\n', num);

end

end

##### 0 Comments

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