Q48·CSAT · Prelims 2025
5-digit number PQRST with digit constraints
ReasoningNumber PuzzleCoding-decoding● Hard
Question
The 5-digit number PQRST (all distinct digits) is such that T ≠ 0. P is thrice T. S is greater than Q by 4, while Q is greater than R by 3. How many such 5-digit numbers are possible?
Options
a
3
bAnswer
4
c
5
d
6
Explanation
Establish the relationships:
1P = 3T, where T \ne 0. Valid (P, T) pairs are (3,1), (6,2), (9,3).
2S = Q + 4 and Q = R + 3. This links them into a single chain: S = R + 7.
Since S \le 9, R can only be 0, 1, or 2. This gives the (R, Q, S) triplets:
If R=0, (0, 3, 7)
If R=1, (1, 4, 8)
If R=2, (2, 5, 9)
Now, pair the (P, T) sets with the (R, Q, S) sets and enforce the "all distinct digits" rule:
For (3,1): Fails with (0,3,7) (3 repeats). Fails with (1,4,8) (1 repeats). Valid with (2,5,9) \rightarrow 35291.
For (6,2): Valid with (0,3,7) \rightarrow 63072. Valid with (1,4,8) \rightarrow 64182. Fails with (2,5,9) (2 repeats).
For (9,3): Fails with (0,3,7) (3 repeats). Valid with (1,4,8) \rightarrow 94183. Fails with (2,5,9) (9 repeats).
Exactly 4 distinct combinations survive the constraint check.
Break multiple interrelated variable constraints into independent sub-groups (here, P/T and R/Q/S), list all exhaustive possibilities for each, and cross-match them against global rules (like distinct digits).
Answer: (b).
Question details
Year
2025
Paper
CSAT
Question
Q48
Section
Logical & Analytical Reasoning
Sub-topic
Number Puzzle
Type
Coding-decoding
Difficulty
Hard
Source hint
Constraint puzzle
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