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    <p>Or you can use a property of vampire numbers described on <a href="http://users.cybercity.dk/~dsl522332/math/vampires/" rel="noreferrer">this page</a> (linked from Wikipedia) :</p> <blockquote> <p>An important theoretical result found by Pete Hartley:</p> <pre><code> If x·y is a vampire number then x·y == x+y (mod 9) </code></pre> <p>Proof: Let mod be the binary modulo operator and d(x) the sum of the decimal digits of x. It is well-known that d(x) mod 9 = x mod 9, for all x. Assume x·y is a vampire. Then it contains the same digits as x and y, and in particular d(x·y) = d(x)+d(y). This leads to: (x·y) mod 9 = d(x·y) mod 9 = (d(x)+d(y)) mod 9 = (d(x) mod 9 + d(y) mod 9) mod 9 = (x mod 9 + y mod 9) mod 9 = (x+y) mod 9</p> <p>The solutions to the <a href="http://en.wikipedia.org/wiki/Congruence_relation" rel="noreferrer">congruence</a> are (x mod 9, y mod 9) in {(0,0), (2,2), (3,6), (5,8), (6,3), (8,5)}</p> </blockquote> <p>So your code could look like this :</p> <pre class="lang-c prettyprint-override"><code>for(int i=18; i&lt;100; i=i+9){ // 18 is the first multiple of 9 greater than 10 for(int j=i; j&lt;100; j=j+9){ // Start at i because as @sh1 said it's useless to check both x*y and y*x checkVampire(i,j); } } for(int i=11; i&lt;100; i=i+9){ // 11 is the first number greater than 10 which is = 2 mod 9 for(int j=i; j&lt;100; j=j+9){ checkVampire(i,j); } } for(int i=12; i&lt;100; i=i+9){ for(int j=i+3; j&lt;100; j=j+9){ checkVampire(i,j); } } for(int i=14; i&lt;100; i=i+9){ for(int j=i+3; j&lt;100; j=j+9){ checkVampire(i,j); } } // We don't do the last 2 loops, again for symmetry reasons </code></pre> <p>Since they are 40 elements in each of the sets like <code>{(x mod 9, y mod 9) = (0,0); 10 &lt;= x &lt;= y &lt;= 100}</code>, you only do <code>4*40 = 160</code> iterations, when a brute-force gives you 10ˆ4 iterations. You can do even less operations if you take into account the <code>&gt;= 1000</code> constraint, for instance you can avoid checking if <code>j &lt; 1000/i</code>.</p> <p>Now you can easily scale up to find vampires with more than 4 digits =)</p>
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    1. POFind all the 4 digit vampire numbers
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    1. USkeelar
    1. USSamuel Peter
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    1. POFind all the 4 digit vampire numbers
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    1. COEdited to avoid the (6,3) and (8,5) loops, which give the same numbers as (3,6) and (5,8)
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      1. USSamuel Peter
    2. COI was considering suggesting a fast-out in my answer (something like `x*y%9 != (x*100+y)%9`) but I realised it was unrefined and that it was going to look stupid if somebody posted something properly thought-out like this.
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    3. CO@SamuelPeter yeah, ok, as it can be made to scale over larger amount of digits I'll agree with you and vote up (otherwise it *would* be too specific... )
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      1. USMaarten Bodewes
 

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