The 5-digit number PQRST (all distin 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 ?
- (a) 3
- (b) 4 ✓ UPSC's answer
- (c) 5
- (d) 6
Why the answer is (b)
• Since P is thrice T and P must be a single digit (1-9), T can only be 1, 2, or 3 (as 4*3=12 is not a single digit). Also, T ≠ 0 is given.
• The condition S = Q + 4 and Q = R + 3 implies S = R + 7. Since S and R are distinct digits (0-9), R can only be 0, 1, or 2 (as R+7 must be ≤ 9).
• We must check combinations of (T, R) such that all digits P, Q, R, S, T are distinct.
• Case 1: T=1. Then P=3. R can be 0, 1, 2. If R=1, R=T (invalid). If R=0, Q=3, S=7. Digits: P=3, Q=3 (P=Q, invalid). If R=2, Q=5, S=9. Digits: P=3, Q=5, R=2, S=9, T=1. All distinct. Valid number: 35291.
• Case 2: T=2. Then P=6. R can be 0, 1, 2. If R=2, R=T (invalid). If R=0, Q=3, S=7. Digits: P=6, Q=3, R=0, S=7, T=2. All distinct. Valid number: 63072. If R=1, Q=4, S=8. Digits: P=6, Q=4, R=1, S=8, T=2. All distinct. Valid number: 64182.
• Case 3: T=3. Then P=9. R can be 0, 1, 2. If R=0, Q=3, S=7. Digits: P=9, Q=3, R=0, S=7, T=3. Q=T (invalid). If R=1, Q=4, S=8. Digits: P=9, Q=4, R=1, S=8, T=3. All distinct. Valid number: 94183. If R=2, Q=5, S=9. Digits: P=9, Q=5, R=2, S=9, T=3. P=S (invalid).
• The valid numbers are 35291, 63072, 64182, and 94183. There are exactly 4 such numbers.
Why the other options are wrong
- (a) 3
- There are 4 valid numbers, not 3, as the combination T=2, R=1 yields the valid number 64182.
- (c) 5
- There are only 4 valid numbers, not 5, as the combination T=3, R=0 is invalid because Q equals T.
- (d) 6
- There are only 4 valid numbers, not 6, as most combinations of T and R result in duplicate digits among P, Q, R, S, or T.
Asked in the CSAT Paper II of the UPSC Civil Services Preliminary Examination 2025, held on 25 May 2025. Question and answer key: Union Public Service Commission. Explanation: UPSC Answer Check.