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# help!

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Suppose that (Pb^2)=QPb, where P, Q, and b represent three distinct digits 1-9. If Q=P/2 and P is two less than b, what is the value of the digit P?

(b means base b)

Dec 17, 2018

#1
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Let's try to express $$(Pb^2) = QPb$$ in terms of b. Since Q = P/2 and P = b - 2, $$((b - 2) \cdot b^2) = \dfrac{b-2}{2} (b - 2) b \Rightarrow b^3 - 2b^2 = \dfrac{1}{2}b^3 -2b^2 + 2b$$. Combining like terms, we have $$\dfrac{1}{2}b^3 = 2b \Rightarrow b^3 = 4b$$. Dividing both sides by b, we have $$b^2 = 4 \Rightarrow b = \pm 2$$, which in this case, changes to b = 2. We know that P is two less than b (P = b - 2), so P = 2-2 = 0.

$$\boxed{P = 0}$$.

- PM

Dec 17, 2018
#2
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If you want to check, b = 2, P = 0, and Q = 0.

$$(0\cdot 2^2) = 0\cdot 0\cdot2$$ simplifies to $$0=0$$ because anything multiplied by 0 is 0.

- PM

PartialMathematician  Dec 17, 2018
#3
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Sorry, it's wrong, b means base b, not multiply by b

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MathCuber  Dec 20, 2018