Thread: 51 problem
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Old 2004-01-29, 11:48   #2
Nov 2002
Vienna, Austria

41 Posts

what do you mean with "pow" ? only exponents formed by "4"s (like "4^SQRT(4)"), or any power, like 4^0? Then it'l be 4!*SQRT(4)+4-4^0

If any "Function" is allowed, one could do the following

51 = "LEFT"((4^4*SQRT(4)),SQRT(4))

which is "LEFT"(512,2)
which is 51

but I'm sure, that's illegal ...
koal is offline