Solution to Project Euler Problem 5 in C

Even I got stuck at one point while solving this problem but finally got it right. So here's what the fifth problem said on Project Euler:

2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder. What is the smallest positive number that is evenly divisible by all of the numbers from 1 to 20?

And here's the 22 line solution to this:


#include <stdio.h>
main()
{
    long int i,j,a;
    for(i=21; i<999999999; i++)
    {
        a=0;
        for(j=2; j<21; j++)
        {
            if(i%j != 0)
            {
                a=1;
                break;
            }
        }
        if(a==0)
        {
            printf("Number is %ld", i);
            break;
        }
    }
}



It took just 2.007 seconds for the program to execute giving the final answer 232792560.

Read More

Solution To Project Euler Problem 3 in C

For those who are unable to solve the third problem of Project Euler which says:

The prime factors of 13195 are 5, 7, 13 and 29.

What is the largest prime factor of the number 600851475143 ?

Here's the simple solution to in C language which merely took  25 lines.

#include <stdio.h>
main()
{
    long long int i,j,b,a=0;
    for(i=3; i<=600851475143/2; i++)
    {
        if(600851475143%i == 0)
        {
            for(j=2; j<=i/2; j++)
            {
                if(i%j == 0)
                {
                    a=1;
                    break;
                }            
            }
            if(a==0)
            {
                b=i;
            }
        }
      
    }
    printf("The sum of all multiples is %lld", b);
}


Although since this program involves large numbers, it too around 1 hour to execute.

The final answer which will come is  6857.

Hope it helps you guys.

Read More

Solution to Project Euler Problem 1 in C

Here's the simple solution of the first problem of Project Euler which said:

If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.

Find the sum of all the multiples of 3 or 5 below 1000.

The code is as follows and took just 13 lines:

#include<stdio.h>
main()
{
    int i,j=0;
    for(i=3; i<1000; i++)
    {
        if(i%3==0 || i%5==0)
        {
            j=j+i;
        }
    }
    printf("The sum of all multiples is %d", j);
}

Execution time was  0.01055 seconds

The answer to this question is 233168.

For any queries, you are free to comment.

Read More