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  • \$\begingroup\$ Great idea using primes. \$\endgroup\$ Commented Mar 8, 2011 at 16:06
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    \$\begingroup\$ You can shave a couple of characters off the dictionary initialisation using the sieve. a=[2];for(p=3,j=0;j<26;)if(a[j]){if(p%a[j++]==0){p++;j=0}}else{a[j]=p;j=0} \$\endgroup\$ Commented Mar 8, 2011 at 17:06
  • \$\begingroup\$ I feel like I read about this trick on some blog talking about interview questions... \$\endgroup\$ Commented Mar 9, 2011 at 3:06
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    \$\begingroup\$ @Tom, depends on how well optimised the sorting routines are, given that you've limited inputs to 8 characters :P \$\endgroup\$ Commented Mar 9, 2011 at 11:13
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    \$\begingroup\$ 125 characters. Recursion and ternaries FTW: for(a=[j=p=2];j<123;)a[j]?p%a[++j]<1&&p++&&(j=0):(a[j]=p,j=0);function b(c,i){return c[i=i||0]?a[c.charCodeAt(i)]*b(c,++i):1} \$\endgroup\$ Commented Mar 9, 2011 at 14:03