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    <p>The modulus operator is inefficient. A more faster implementation would be something like this:</p> <pre><code>int multiply2_5(int max) { int i, x2 = 0,x5 = 0,x10 = 0; for(i = 2; i &lt; max; i+=2) x2 += i; // Store all multiples of 2 O(max/2) for(i = 5; i &lt; max; i+=5) x5 += i; // Store all multiples of 3 O(max/5) for(i = 10; i &lt; max; i+=10) x10 += i; // Store all multiples 10; O(max/10) return x2+x5-x10; } </code></pre> <p>In this solution I had to take out multiples of 10 because, 2 and 5 have 10 as multiple so on the second loop it will add multiples of 10 that already been added in the first loop; The three loops combine have O(8/10 max).</p> <p>Another even better solution is if you take a mathematical approach.</p> <p>You are trying to sum all numbers like this 2 + 4 + 6 + 8 ... 10000 and 5 + 10 + 15 +20 + ... 10000 this is the same of having 2 * (1 + 2 + 3 + 4 + … + 5000) and 5 * ( 1 + 2 + 3 + 4 + ... + 2000), the sum of 'n' natural number is (n * (n + 1)) (<a href="http://www.9math.com/book/sum-first-n-natural-numbers" rel="nofollow">source</a>) so you can calculate in a constant time, as it follows:</p> <pre><code>int multiply2_5(int max) { // x = 2 + 4 + 6 + ... = 2 * (1 + 2 + 3 +...) // y = 5 + 10 + 15 + ... = 5 * (1 + 2 + 3 +...) // The sun of n natural numbers is sn = (n (n + 1)) / 2 int x2 = max/ 2; // 2 * ( 1 +2 + 3 … max/2) int x5 = max /5; // 5 * ( 1 +2 + 3 … max/5) int x10 = max/ 10; int sn2 = 0.5 * (x2 * (x2+1)); // (n * (n + 1)) / 2 int sn5 = 0.5 * (x5 * (x5+1)); int sn10 = 0.5 * (x10 * (x10+1)); return (2*sn2) + (5 *sn5) - (10*sn10); } </code></pre>
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    1. COThe faster I know for this one is using the theorm of gauss
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    2. CO`int sn15 = 0.5 * (x15 * (x15+1));return (2*sn3) + (5 *sn5) - (15*sn15);` I think you mean `int sn10 = 0.5 * (x10 * (x10+1));return (2*sn2) + (5 *sn5) - (10*sn10);` in your answer above. You can also move all the multiplies by 0.5 to the return statement.
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    3. CO@carleeto Yep, thanks for point that out.
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