Solving for Y: Consecutive Squares Sum Problem

The problem requires finding the value of digit $Y$ based on the sum of squares of two consecutive two-digit numbers, related to digits $P, Q, R, S, X, Y$.

Understanding the Conditions

• The digits $P, Q, R, S, X, Y$ must be distinct single digits (0 through 9).
• $PQ$ represents a two-digit number $10P + Q$.
• $RS$ represents a two-digit number $10R + S$.
• $PQ$ and $RS$ are consecutive integers.
• The sum of their squares equals a three-digit number $XYP$, represented as $100X + 10Y + P$.
• A key constraint is that the digit $P$ used as the tens digit of $PQ$ must be identical to the digit $P$ used as the units digit of the sum $XYP$.

Finding Possible Consecutive Numbers

Let the two consecutive numbers be $n$ and $n+1$. We are given $(PQ)^2 + (RS)^2 = XYP$. Since $XYP$ is a three-digit number, $100 \le n^2 + (n+1)^2 \le 999$.

This inequality restricts the possible values for $n$. By testing values:

• The smallest sum is $10^2 + 11^2 = 100 + 121 = 221$.
• The largest sum yielding a three-digit number is $21^2 + 22^2 = 441 + 484 = 925$.
• The next pair, $22^2 + 23^2 = 484 + 529 = 1013$, results in a four-digit number.

Therefore, the consecutive numbers $(n, n+1)$ must be one of the pairs from $(10, 11)$ up to $(21, 22)$.

Testing Consecutive Pairs Systematically

We examine potential pairs $(n, n+1)$ and check against the problem's constraints:

1. Calculate the sum $S = n^2 + (n+1)^2$. Let $S = XYP$.
2. Consider two possibilities for assigning $n$ and $n+1$ to $PQ$ and $RS$.
3. Ensure all digits $P, Q, R, S, X, Y$ are distinct.
4. Verify that the tens digit of $PQ$ equals the units digit $P$ of $XYP$.

Analyzing the Pair (19, 20)

Let $n=19$ and $n+1=20$. The sum is $19^2 + 20^2 = 361 + 400 = 761$. Thus, $XYP = 761$.

• Scenario 1: $PQ=19, RS=20$.
  • From $PQ=19$, we have $P=1$ (tens) and $Q=9$ (units).
  • From $RS=20$, we have $R=2$ (tens) and $S=0$ (units).
  • From $XYP=761$, we have $X=7$, $Y=6$, and $P=1$ (units).
  Checking Constraints:
  • Distinct Digits: The set of digits $\{P, Q, R, S, X, Y\} = \{1, 9, 2, 0, 7, 6\}$. These are all distinct single digits.
  • $P$ Match: The tens digit of $PQ$ is $P=1$. The units digit of $XYP$ is $P=1$. They match.
  • This scenario satisfies all conditions. The value of $Y$ is $6$.
• Scenario 2: $PQ=20, RS=19$.
  • From $PQ=20$, we have $P=2$ (tens) and $Q=0$ (units).
  • From $RS=19$, we have $R=1$ (tens) and $S=9$ (units).
  • From $XYP=761$, we have $X=7$, $Y=6$, and $P=1$ (units).
  Checking Constraints:
  • Distinct Digits: The set of digits $\{P, Q, R, S, X, Y\} = \{2, 0, 1, 9, 7, 6\}$. These are all distinct single digits.
  • $P$ Match: The tens digit of $PQ$ is $P=2$. The units digit of $XYP$ is $P=1$. They do not match.
  • This scenario is invalid due to the mismatch in the value of $P$.

Why Other Pairs Fail

Systematic checking of other consecutive pairs (e.g., $10, 11$; $11, 12$; ... $21, 22$) shows they fail at least one condition:

• Non-distinct digits: Many pairs lead to repeated digits. For instance, $10^2 + 11^2 = 221$. If $PQ=10$, $P=1$. If $RS=11$, $R=1, S=1$. The digits $P$ and $R$ are the same ($1$), and $S$ is also $1$, violating distinctness.
• $P$ Mismatch: For example, with $12^2 + 13^2 = 313$. If $PQ=12$, $P=1$. However, $XYP=313$ implies $P=3$. This conflict means $PQ$ cannot be $12$.

Final Conclusion

The only pair of consecutive numbers satisfying all problem conditions is $19$ and $20$. Assigning $PQ=19$ and $RS=20$ leads to:

• $(19)^2 + (20)^2 = 761$.
• $XYP = 761$, giving $X=7, Y=6, P=1$.
• The digits $\{1, 9, 2, 0, 7, 6\}$ are distinct.
• The tens digit $P=1$ from $PQ$ matches the units digit $P=1$ from $XYP$.

Therefore, the value of $Y$ is $6$.