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POproject euler exercise 5 approach
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copied!<p>Question: 2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder.</p> <p>What is the smallest positive number that is evenly divisible by all of the numbers from 1 to 20?</p> <p>So, I was trying to do exercise 5 on project euler and I came out with this code:</p> <pre><code>#include <stdio.h> #define TRUE 1 #define FALSE 0 int main () { int n, fnd = FALSE, count, i; for (i = 1; fnd == FALSE; i++) { count = 0; for (n = 1; n <= 20; n++) { count += i % n; } printf ("testing %d, count was: %d\n", i, count); if (count == 0) { fnd = TRUE; printf ("%d\n", i); } } return 0; } </code></pre> <p>I believe my apporach is correct, it will surely find the number which is divisible by 1 to 20. But it's been computing for 5 minutes, and still no result. Is my approach correct? If yes, then is there another way to do it? I can't think on another way to solve this, tips would be very much appreciated. Thank you in advance.</p> <p>EDIT: So, based on the advice I was given by you guys I figured it out, thank you so much! So, it's still brute force, but instead of adding 1 to the last number, it now adds 2520, which is the LCM of 1 to 10. And therefore, calculating if the sum of the remainders of the multiples of 2520 divided from 11 to 20 was 0. Since 2520 is already divisible by 1 to 10, I only needed to divide by 11 to 20.</p> <pre><code>#include <stdio.h> #define TRUE 1 #define FALSE 0 int main () { int n, fnd = FALSE, count, i; for (i = 2520; fnd == FALSE; i = i + 2520) { count = 0; for (n = 11; n <= 20; n++) { count += i % n; } printf ("testing %d, count was: %d\n", i, count); if (count == 0 && i != 0) { fnd = TRUE; printf ("%d\n", i); } } return 0; } </code></pre> <p>Thank you so much, I wouldn't solve it without your help : ) PS: It now computes in less than 10 secs.</p>

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