+537 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)
(Please help)
+775 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
+775 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
