Cryptarithm PQR - PS = PPT
Question
Let PQR be a 3-digit number, PPT be a 3-digit number and PS be a 2-digit number, where P, Q, R, S, T are distinct non-zero digits. Further, PQR - PS = PPT. If Q = 3 and T = 6, then what is the number of possible values of (R, S)?
Options
2
3
4
More than 4
Explanation
Substitute the known values into the equation: P3R - PS = PP6. Convert this into expanded form: 100P + 30 + R - (10P + S) = 110P + 6. Simplify the algebraic equation: 90P + 30 + R - S = 110P + 6 20P = 24 + R - S Since R and S are single digits, their difference (R - S) is strictly bounded between -9 and +9. Therefore, (24 + R - S) must be a value between 15 and 33. The only multiple of 20 in this restricted range is 20, forcing ⟨MATH⟩P = 1⟨/MATH⟩. Substitute P=1: 20(1) = 24 + R - S \implies S - R = 4. The remaining available digits are \{2, 4, 5, 7, 8, 9\}. Pairs (S, R) with a difference of 4 are (9, 5) and (8, 4). Thus, there are 2 possible pairs.
Answer: (a).
Question details
Year
2025
Paper
CSAT
Question
Q39
Section
Logical & Analytical Reasoning
Sub-topic
Cryptarithmetic
Type
Coding-decoding
Difficulty
Hard
Source hint
Cryptarithmetic puzzle
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