Question

\(P, Q, R, S, X,\) and \(Y\) are distinct single-digit whole numbers taking values from 0 to 9.
\(PQ\) is a two-digit number with \(Q\) in the units place and \(P\) in the tens place. Similarly, \(RS\) is a two-digit number.
It is known that \(PQ\) and \(RS\) are consecutive numbers and \((PQ)^2 + (RS)^2 = XYP\), with \(XYP\) being a three-digit number.
The value of \(Y\) is __________.

• A. 4
• B. 5
• C. 6
• D. 7

Hint:

Bound the two numbers using \((PQ)^2+(RS)^2 < 1000\), then notice only one consecutive pair in that range has different tens digits.

Answer 1

The Correct Option is C

Step 1: Set up what the numbers can look like.
Write \(PQ = 10P + Q\) and \(RS = 10R + S\). Both are two digit numbers, so each lies between 10 and 99. We are told \(PQ\) and \(RS\) are consecutive, so they differ by exactly 1, and their squares add up to a three digit number \(XYP\), whose own units digit is the same \(P\) that is the tens digit of \(PQ\).

Step 2: Bound how big the two numbers can be.
Since \(XYP\) has only three digits, \((PQ)^2 + (RS)^2 \leq 999\). If we let the smaller number be \(a\) and the larger be \(a+1\), then \(a^2 + (a+1)^2 \leq 999\), which simplifies to \(2a^2 + 2a + 1 \leq 999\), so \(a^2 + a \leq 499\). Checking values, \(a = 21\) gives \(21^2+21 = 462\) (fits), while \(a = 22\) gives \(22^2+22 = 506\) (too big). So \(a\) can only run from 10 up to 21.

Step 3: Use the distinctness condition to narrow it down.
If both \(a\) and \(a+1\) sit in the same tens decade (like 12 and 13, both starting with tens digit 1), then \(P\) (tens digit of \(a\)) and \(R\) (tens digit of \(a+1\)) become equal, which breaks the rule that all six letters must be distinct digits. The only pair in our range 10 to 21 where the tens digit actually changes between \(a\) and \(a+1\) is \(a = 19\) (tens digit 1) and \(a+1 = 20\) (tens digit 2). Every other consecutive pair in this range keeps the same tens digit on both numbers, so this is the only candidate worth testing.

Step 4: Check the candidate \(PQ = 19\), \(RS = 20\).
Here \(P=1, Q=9, R=2, S=0\). Compute \(19^2 + 20^2 = 361 + 400 = 761\). This is indeed a three digit number, and its units digit is 1, which matches \(P = 1\) exactly, as required. So \(XYP = 761\) gives \(X = 7\), \(Y = 6\), \(P = 1\).

Step 5: Confirm all six digits are distinct.
We now have \(P=1, Q=9, R=2, S=0, X=7, Y=6\). These six values, 1, 9, 2, 0, 7, 6, are all different from each other, so every condition in the question is satisfied. Reversing the order, that is trying \(PQ=20, RS=19\), fails because then \(P\) would be 2, but the units digit of \(761\) is 1, so that assignment is rejected.

Final Answer:
The value of \(Y\) is 6, which is option (C). \[ \boxed{Y = 6} \]