🚀 go-pugleaf

in the paper "A Fast Technique for Constant Divisors" CACM V19 nr 2
all such division operations are considered for this class of problems.

let N be the given. 
	M = N * 5; /* 2^2+1 = 5*/
	M = M * 17; /* 2^4 +1 = 17 */
	M = M * 257; /* 2^8 +1 = 257 */
/* at this point our intermediate quantity has overflowed our
   original 11 bit register: for larger register sizes we may have
   to continue with 2^16+1 and 2^32+1 */

	ANS = -M    /* VOILA*/

Note that if the N is not evenly divisible by 3 the result ANS 
appears larger than N. HOWEVER, ANS * 3 results in N (within 11 bits)