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0000is also congruent to 0 modulo 3 but again you can't have the first and last letters the same, so, like 15, you don't need to explicitly test for it. \$\endgroup\$!((a>b)-(b>c)+(c>d)-(d>e))? \$\endgroup\$p<c?0:NaNcan be written as0/(p<c), which saves 2 bytes. \$\endgroup\$k?test because of the possibleNaN.) Regarding your alternate version: that should work indeed. \$\endgroup\$