Understanding the Distinct Non-Zero Digits Problem

The question asks for the value of the digit Q based on the equation \(PP \times PQ = RRSS\), where P, Q, R, and S are distinct non-zero digits. We are also given the constraints \(P \le 3\) and \(Q \le 4\).

The numbers PP, PQ, and RRSS are represented using place values:

• PP represents a two-digit number where both digits are P. This is \(10P + P = 11P\).
• PQ represents a two-digit number where the tens digit is P and the units digit is Q. This is \(10P + Q\).
• RRSS represents a four-digit number where the thousands and hundreds digits are R, and the tens and units digits are S. This is \(1000R + 100R + 10S + S = 1100R + 11S\).

The given equation is therefore:

\(11P \times (10P + Q) = 1100R + 11S\)

We can divide both sides of the equation by 11:

\(P(10P + Q) = 100R + S\)

We need to find the value of Q using the constraints:

• P, Q, R, S are distinct non-zero digits (from the set \{1, 2, 3, 4, 5, 6, 7, 8, 9\}).
• \(P \le 3\), so P can be 1, 2, or 3.
• \(Q \le 4\), so Q can be 1, 2, 3, or 4.

Analyzing Possible Values for Digit P

Let's test the possible values for P based on the constraint \(P \le 3\).

Case 1: P = 1

If P=1, the equation becomes:

\(1(10 \times 1 + Q) = 100R + S\)

\(10 + Q = 100R + S\)

Given \(Q \le 4\), the possible values for \(10+Q\) are \(10+1=11\), \(10+2=12\), \(10+3=13\), or \(10+4=14\).

The term \(100R + S\) represents a number formed by the non-zero digits R and S. Since R is a non-zero digit, the smallest possible value for R is 1. The smallest possible value for S is also 1 (but R and S must be distinct, so the smallest distinct values are R=1, S=2 or R=2, S=1 etc.). In any case, if R is a non-zero digit (\(R \ge 1\)), \(100R + S \ge 100 \times 1 + 1 = 101\) (assuming S is any non-zero digit).

Comparing the possible values:

• Maximum of \(10+Q\) is 14.
• Minimum of \(100R+S\) (with R non-zero) is 101.

Since 14 is much less than 101, the equation \(10 + Q = 100R + S\) cannot hold for any non-zero digit R. Therefore, P cannot be 1.

Case 2: P = 2

If P=2, the equation becomes:

\(2(10 \times 2 + Q) = 100R + S\)

\(2(20 + Q) = 100R + S\)

\(40 + 2Q = 100R + S\)

P and Q must be distinct non-zero digits. Since P=2, Q cannot be 2. Q can be 1, 3, or 4 (given \(Q \le 4\)). R and S must be non-zero and distinct from 2 and Q.

Let's find the possible values for \(40 + 2Q\):

• If Q=1: \(40 + 2(1) = 42\). Equation: \(100R + S = 42\). For this to be true, R would have to be 0, which is not allowed (R must be a non-zero digit).
• If Q=3: \(40 + 2(3) = 46\). Equation: \(100R + S = 46\). Again, R must be 0, not allowed.
• If Q=4: \(40 + 2(4) = 48\). Equation: \(100R + S = 48\). Again, R must be 0, not allowed.

In all valid cases for Q when P=2, the right side \(100R+S\) would require R=0, which contradicts the condition that R is a non-zero digit. Therefore, P cannot be 2.

Case 3: P = 3

If P=3, the equation becomes:

\(3(10 \times 3 + Q) = 100R + S\)

\(3(30 + Q) = 100R + S\)

\(90 + 3Q = 100R + S\)

P and Q must be distinct non-zero digits. Since P=3, Q cannot be 3. Q can be 1, 2, or 4 (given \(Q \le 4\)). R and S must be non-zero and distinct from 3 and Q.

Let's find the possible values for \(90 + 3Q\):

• If Q=1: \(90 + 3(1) = 93\). Equation: \(100R + S = 93\). R must be 0, not allowed. (Also, digits P=3, Q=1 are valid so far).
• If Q=2: \(90 + 3(2) = 96\). Equation: \(100R + S = 96\). R must be 0, not allowed. (Also, digits P=3, Q=2 are valid so far).
• If Q=4: \(90 + 3(4) = 102\). Equation: \(100R + S = 102\).

  For \(100R + S = 102\) to be true where R and S are single non-zero digits, R must be 1 and S must be 2. Let's check if these digits satisfy all conditions:

  • P=3, Q=4, R=1, S=2.
  • Are they distinct? Yes (3, 4, 1, 2).
  • Are they non-zero? Yes.
  • Is \(P \le 3\)? Yes (P=3).
  • Is \(Q \le 4\)? Yes (Q=4).

  All conditions are met. Let's verify the original equation \(PP \times PQ = RRSS\):

  \(33 \times 34\)

  \(33 \times 34 = 1122\)

  With R=1 and S=2, RRSS = 1122. The equation holds.

  This gives us a unique solution for the digits: P=3, Q=4, R=1, S=2.

Determining the Value of Q

From our analysis, the only set of distinct non-zero digits P, Q, R, S satisfying the given equation and constraints (\(P \le 3\), \(Q \le 4\)) is P=3, Q=4, R=1, S=2.

In this unique solution, the value of Q is 4.

Conclusion on Statement Sufficiency

We were able to determine the value of Q (which is 4) definitively by only using the information provided in the question itself, without needing to refer to Statement I (R=1) or Statement II (S=2).

Therefore, the question can be answered even without using any of the Statements.

Revision Table: Solving the Digit Puzzle

Additional Information on Digit Puzzles

This problem is an example of a cryptarithmetic puzzle involving digits. Key steps in solving such puzzles often include:

• Representing the letter variables as digits.
• Translating the letter-based number representations (like PP or RRSS) into algebraic expressions based on place value.
• Setting up equations based on the given operations (multiplication in this case).
• Using constraints (like distinct non-zero digits, range of values) to limit the possibilities for the variables.
• Systematically testing the limited possibilities to find the digit values that satisfy the equations and constraints.
• Often, there is a unique solution in well-posed problems of this type.

The structure RRSS implies a number is formed by repeating a two-digit block RS, like 1122 or 3399. This can be written as \(100 \times RS + RS = 101 \times RS\). In our case, RRSS is \(11 \times (100R + S)\), which is a different structure than \(101 \times (10R+S)\). Our initial breakdown of RRSS = \(1100R + 11S\) was correct.