IAS 2025 — CSAT Maths

Correct Option (B)

The problem describes a 5-digit number PQRST where all digits (P, Q, R, S, T) are distinct. The given conditions are:

• T ≠ 0
• P = 3T
• S = Q + 4
• Q = R + 3

From the relations S = Q + 4 and Q = R + 3, we can derive S = (R + 3) + 4, which simplifies to S = R + 7.

All digits must be distinct and fall within the range {0, 1, 2, ..., 9}. Additionally, P cannot be 0 as it is the leading digit of a 5-digit number.

We proceed by systematically evaluating possible values for T and P:

• Case 1: T = 1
  • P = 3T = 3(1) = 3. Thus, (P, T) = (3, 1).
  • Digits already used: {3, 1}.
  • Now, we find R, Q, S such that Q = R + 3 and S = R + 7, ensuring distinctness from {3, 1} and each other.
    • If R = 0, then Q = 3. This is not allowed as P = 3.
    • If R = 1, this is not allowed as T = 1.
    • If R = 2, then Q = 5 and S = 9. The set of digits {P, Q, R, S, T} becomes {3, 5, 2, 9, 1}. All are distinct. This yields one valid number: 35291.
    • If R ≥ 3, then S = R + 7 would be ≥ 10, which is not a single digit.
• Case 2: T = 2
  • P = 3T = 3(2) = 6. Thus, (P, T) = (6, 2).
  • Digits already used: {6, 2}.
  • We find R, Q, S:
    • If R = 0, then Q = 3 and S = 7. The set of digits {P, Q, R, S, T} becomes {6, 3, 0, 7, 2}. All are distinct. This yields one valid number: 63072.
    • If R = 1, then Q = 4 and S = 8. The set of digits {P, Q, R, S, T} becomes {6, 4, 1, 8, 2}. All are distinct. This yields one valid number: 64182.
    • If R = 2, this is not allowed as T = 2.
    • If R = 3, then Q = 6. This is not allowed as P = 6.
    • If R ≥ 4, then S = R + 7 would be ≥ 11, which is not a single digit.
• Case 3: T = 3
  • P = 3T = 3(3) = 9. Thus, (P, T) = (9, 3).
  • Digits already used: {9, 3}.
  • We find R, Q, S:
    • If R = 0, then Q = 3. This is not allowed as T = 3.
    • If R = 1, then Q = 4 and S = 8. The set of digits {P, Q, R, S, T} becomes {9, 4, 1, 8, 3}. All are distinct. This yields one valid number: 94183.
    • If R = 2, then S = 9. This is not allowed as P = 9.
    • If R = 3, this is not allowed as T = 3.
    • If R ≥ 4, then S = R + 7 would be ≥ 11, which is not a single digit.

If T ≥ 4, then P = 3T would be ≥ 12, which is not a single digit. Therefore, no further cases for (P, T) are possible.

Combining the results from all valid cases, the total number of such 5-digit numbers possible is 1 (from Case 1) + 2 (from Case 2) + 1 (from Case 3) = 4.

The four possible numbers are 35291, 63072, 64182, and 94183.

Incorrect Options:

Options (A), (C), and (D) are incorrect. The detailed and systematic derivation, considering all constraints on distinct digits and their relationships, conclusively demonstrates that exactly four such 5-digit numbers can be formed, thereby invalidating counts of 3, 5, or 6.