How do you find the primes from 1 to 100 using loops in C#?

4 Answers

  • 1 decade ago
    Favorite Answer

    I wrote and tested the below code in C#. I might point out if you want to make it more efficient, on the line of the j loop, instead of iterating to i-1, you could iterate to i/2 as it is impossible for any number to be divisible by a number more than half of it. However saying that since you are only looking for 1- 100 , this is fast enough... you should get the results in around 1 second.

    int objStartVal = 1;

    int objEndVal = 100;

    int i, j;

    for ( i=objStartVal; i<=objEndVal; i++ )


    Boolean isPrime = true;

    for (j=1;j<=i-1;j++)


    if (j!=1)


    if ((i % j) == 0)


    isPrime = false;




    if (isPrime)


    MessageBox.Show(i + " is a prime number.");



  • 1 decade ago

    Well, not that it is that difficult, but it is more efficient to just use a hash table if you only need such a small group of prime numbers. There are only 26 numbers less than 100 that are prime (including 1).

    If you're looking for an algorithm that will work for you, I suggest looking online or through just about any C programming book. Prime number finders are pretty common. My buddy writes them in different assembly languages he learns just because it's a simple little problem to solve.

    But, if you just want the answer, here is a simple algorithm, and below is the link to learn more abou it:

    ArrayList primeNumbers = new ArrayList();

    for(int i = 2; i < 100; i++)


    bool divisible = false;

    foreach(int number in primeNumbers)

    if(i % number == 0)

    divisible = true;

    if(divisible == false)



  • 1 decade ago

    I won't write the code for you as it sounds like a homework question, but here's the basic algorythm (skip 1, 2, 3 since you know they're prime):

    (inside the loop) :

    Divide the number by every number between 1 and itself, looking for a remainder each time. If once comes back with no remainder, it's not prime

  • 1 decade ago

    i dunno in c#, but here it is in c++, convert it or something :p

    #include <iostream>

    #include <iomanip>

    using namespace std;

    int main()


    int number, variable, display_num;

    for (number = 3; number < 100; number++)

    for (variable = 2; (number%variable > 0) && variable < (number); variable++)

    for (display_num = 0; ((number - variable) == 1) && (display_num == 0); display_num++)

    cout << setw(5) << number << setw(5);

    cout << endl;


    return 0;


    sorry i don't know c#, i did this as an assignment a long time ago for c++, should be similar to what you are trying to get too. obviously this was a beginning course, i kinda of laughed at the code now that i see it, but it worked.

Still have questions? Get your answers by asking now.