Question

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In the figure below, the seven letters correspond to seven unique digits chosen from 0 to 9. The relation among the digits is such that:

\(P \times Q \times R = X \times Y \times Z = Q \times A \times Y\)

P		X
Q	A	Y
R		Z

The value of A is:

• A. 0
• B. 2
• C. 3
• D. 6

Hint:

Try to express all three products in terms of Q, A and Y, then find integer digit combinations that make every product equal.

Answer 1

The Correct Option is B

This puzzle only pins down A once you find at least one full set of seven digits that makes all three products equal, so the fastest path is to guess a workable common product and reverse-engineer the letters from it.

1. A = 0: makes $Q \times A \times Y = 0$. Then $P \times Q \times R$ must also be $0$, forcing one of $P, Q, R$ to be $0$, and separately $X \times Y \times Z$ must be $0$, forcing one of $X, Y, Z$ to be $0$ as well. Since only one of the seven letters can actually be the digit 0, this cannot happen twice, so $A = 0$ fails.
2. A = 3: then $P \times R = 3Y$ and $X \times Z = 3Q$. Trying small distinct $Q, Y$ values other than 3 never lands on two more pairs of distinct single digits, all seven different from each other, that satisfy both equations at once.
3. A = 6: then $P \times R = 6Y$ and $X \times Z = 6Q$, which quickly needs a factor bigger than 9 in at least one pair once $Q$ or $Y$ is even moderately large, so no valid all-distinct-digit set turns up.
4. A = 2: then $P \times R = 2Y$ and $X \times Z = 2Q$. Choosing $Q = 9$, $Y = 4$ gives $P \times R = 8$, so $\{P,R\}=\{1,8\}$, and $X \times Z = 18$, so $\{X,Z\}=\{3,6\}$. Check: $P \cdot Q \cdot R = 1 \cdot 9 \cdot 8 = 72$, $X \cdot Y \cdot Z = 3 \cdot 4 \cdot 6 = 72$, $Q \cdot A \cdot Y = 9 \cdot 2 \cdot 4 = 72$. All three match, and the digits $1,2,3,4,6,8,9$ are all distinct.

Since only $A = 2$ produces a genuinely consistent set of seven distinct digits with all three products equal, the answer is $A = 2$.