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If (P_b)*(P_b)=31_b, where P and b represent two distinct digits 0-9 and P is one less than b, what is the value of base b?

So I have ((b-1)_b)^2=31_b

how do I do this?!  please explain. :)

Guest Oct 6, 2018
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$$P=b-1\\ (b-1)^2 = 31_b = 3b+1\\ b^2 -2b+1 = 3b+1\\ b^2 = 5b \\ b=5 \text{ (or 0, but 0 is not a valid base)}\\ b=5, ~P=4\\ (4_5)^2 = 16_{10} = 31_5$$

Rom  Oct 6, 2018