| http://www.w3.org/ns/prov#value | - 27 Jun 2011, 06:06 Let me share how we can do this in ~2 mins.St1: 3a=f=6y => f must be a multiple of 2*3 but if we multiply 2*3 with any number other than 1, value of f will be > 9, which is not allowed. so, f=6; a=2 and y=1.now, from the given condition c+f=z => (9-0) + 6 = z, from the set for c we can rule out values of 0, 1, 2 (as 1 and 2 are already taken and 0 will make z=6, which is not all
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