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    <p>This reminds me of an old trick from base 10 called "casting out the 9s". This was used for checking the result of large sums performed by hand. In this case <code>123 mod 9 = 1 + 2 + 3 mod 9 = 6</code>. </p> <p>This happens because 9 is one less than the base of the digits (10). (Proof omitted ;) )</p> <p>So considering the number in base 16 (Hex). you should be able to do:</p> <pre><code>0xABCE123 mod 0xF = (0xA + 0xB + 0xC + 0xD + 0xE + 0x1 + 0x2 + 0x3 ) mod 0xF = 0x42 mod 0xF = 0x6 </code></pre> <p>Now you'll still need to do some magic to make the additions disappear. But it gives the right answer.</p> <p>UPDATE:</p> <p>Heres a complete implementation in C++. The <code>f</code> lookup table takes pairs of digits to their sum mod 15. (which is the same as the byte mod 15). We then repack these results and reapply on half as much data each round. </p> <pre class="lang-c++ prettyprint-override"><code>#include &lt;iostream&gt; uint8_t f[256]={ 0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,0, 1,2,3,4,5,6,7,8,9,10,11,12,13,14,0,1, 2,3,4,5,6,7,8,9,10,11,12,13,14,0,1,2, 3,4,5,6,7,8,9,10,11,12,13,14,0,1,2,3, 4,5,6,7,8,9,10,11,12,13,14,0,1,2,3,4, 5,6,7,8,9,10,11,12,13,14,0,1,2,3,4,5, 6,7,8,9,10,11,12,13,14,0,1,2,3,4,5,6, 7,8,9,10,11,12,13,14,0,1,2,3,4,5,6,7, 8,9,10,11,12,13,14,0,1,2,3,4,5,6,7,8, 9,10,11,12,13,14,0,1,2,3,4,5,6,7,8,9, 10,11,12,13,14,0,1,2,3,4,5,6,7,8,9,10, 11,12,13,14,0,1,2,3,4,5,6,7,8,9,10,11, 12,13,14,0,1,2,3,4,5,6,7,8,9,10,11,12, 13,14,0,1,2,3,4,5,6,7,8,9,10,11,12,13, 14,0,1,2,3,4,5,6,7,8,9,10,11,12,13,14, 0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,0}; uint64_t mod15( uint64_t in_v ) { uint8_t * in = (uint8_t*)&amp;in_v; // 12 34 56 78 12 34 56 78 =&gt; aa bb cc dd in[0] = f[in[0]] | (f[in[1]]&lt;&lt;4); in[1] = f[in[2]] | (f[in[3]]&lt;&lt;4); in[2] = f[in[4]] | (f[in[5]]&lt;&lt;4); in[3] = f[in[6]] | (f[in[7]]&lt;&lt;4); // aa bb cc dd =&gt; AA BB in[0] = f[in[0]] | (f[in[1]]&lt;&lt;4); in[1] = f[in[2]] | (f[in[3]]&lt;&lt;4); // AA BB =&gt; DD in[0] = f[in[0]] | (f[in[1]]&lt;&lt;4); // DD =&gt; D return f[in[0]]; } int main() { uint64_t x = 12313231; std::cout&lt;&lt; mod15(x)&lt;&lt;" "&lt;&lt; (x%15)&lt;&lt;std::endl; } </code></pre>
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    1. POHow to find x mod 15 without using any Arithmetic Operations?
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    1. USMichael Anderson
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    1. COTo get rid of the additions, you can use a 16x16 lookup table. :-)
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    2. CO@ShreevatsaR - indeed the algorithm then looks hacky, but can be reduced to not very many operations. So I've added an implementation of it...
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      1. USMichael Anderson
    3. CO(head explodes) Good idea anyway :D
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