how do i check if a number is a prime number, a square, and or a factor of 6?

12 次查看（过去 30 天）
Vjon Joson
Vjon Joson 2020-4-1
编辑： DGM 2023-1-5
here is the current idea for the code so far
integer = 4;
if % ** check if integer is a prime **
Prime = integer % => If condition is met
elseif % ** check if the number is square **
Square = integer % => Else condition is met
elseif % chekc if divisible by 6
Factorof6=integer
end
  3 个评论
显示 1更早的评论
Walter Roberson
Walter Roberson 2023-1-5
https://www.mathworks.com/matlabcentral/answers/514501-how-do-i-check-if-a-number-is-a-prime-number-a-square-and-or-a-factor-of-6#comment_819187
DGM
DGM 2023-1-5
编辑：DGM 2023-1-5
Ran in:
Since the core answers are all basically already on this page, I'll just add that logical indexing, implicit casting, and implicit expansion can make this tidy.
x = 100:999; % a set of numbers as a row vector
issquare = ~mod(sqrt(x),1); % implicit binarization
ismultiple = ~mod(x,[2; 3; 5]); % implicit expansion
validnums = issquare & ~any(ismultiple,1); % logical combination of masks
x(validnums) % logical indexing
ans = 1×7
121 169 289 361 529 841 961
Now you have to make a function out of it.

请先登录，再进行评论。

请先登录，再回答此问题。

采纳的回答

Birdman
Birdman 2020-4-1
n=4;
%check if number is prime
isprime(n) %returns logical value
%check if number is square
mod(numel(factor(n)),2)==0 %returns logical value
%check if number is divisible by 6
mod(n,6)==0 %returns logical value
  7 个评论
显示 5更早的评论
Torsten
Torsten 2022-10-15
编辑：Torsten 2022-10-15
2*3 = 6, and 6 is not a square.
Yes, the answer given is incorrect.
mod(numel(factor(a)),2)==0 is not sufficient for "a" to be a square.
Walter Roberson
Walter Roberson 2022-10-16
https://www.mathworks.com/matlabcentral/answers/514501-how-do-i-check-if-a-number-is-a-prime-number-a-square-and-or-a-factor-of-6#comment_819105 shows a revised test (which I did not examine)

请先登录，再进行评论。

更多回答（0 个）

类别

在 Help Center 和 File Exchange 中查找有关 Function Creation 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by