Q68. 68. The 5-digit number PQRST (all distinct digits) is such that T is not equal to 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?

We are given a 5-digit number PQRST where P, Q, R, S, T are distinct digits. 1. T ≠ 0 2. P = 3T 3. S = Q + 4 4. Q = R + 3 From (3) and (4), we can express S and Q in terms of R: Q = R + 3 S = (R + 3) + 4 = R + 7 Now let's consider possible values for T, keeping in mind P=3T and P must be a single digit (P ≠ 0 as it's the first digit of a 5-digit number): * If T = 1, then P = 3 × 1 = 3.

Digits used so far: P=3, T=1. (Distinct) Now consider R. R must be a digit distinct from 1 and 3. * If R = 0: Q = 0 + 3 = 3, S = 0 + 7 = 7. Digits: P=3, Q=3, R=0, S=7, T=1. Here Q=P, which violates the 'distinct digits' condition.

(Invalid) * If R = 2: Q = 2 + 3 = 5, S = 2 + 7 = 9. Digits: P=3, Q=5, R=2, S=9, T=1. All are distinct. Number: 35291. (Valid) * If T = 2, then P = 3 × 2 = 6. Digits used so far: P=6, T=2.

(Distinct) Possible values for R (distinct from 2 and 6): * If R = 0: Q = 0 + 3 = 3, S = 0 + 7 = 7. Digits: P=6, Q=3, R=0, S=7, T=2. All are distinct. Number: 63072. (Valid) * If R = 1: Q = 1 + 3 = 4, S = 1 + 7 = 8.

Digits: P=6, Q=4, R=1, S=8, T=2. All are distinct. Number: 64182. (Valid) * If T = 3, then P = 3 × 3 = 9. Digits used so far: P=9, T=3. (Distinct) Possible values for R (distinct from 3 and 9): * If R = 0: Q = 0 + 3 = 3, S = 0 + 7 = 7.

Digits: P=9, Q=3, R=0, S=7, T=3. Here Q=T, which violates the 'distinct digits' condition. (Invalid) * If R = 1: Q = 1 + 3 = 4, S = 1 + 7 = 8. Digits: P=9, Q=4, R=1, S=8, T=3.

All are distinct. Number: 94183. (Valid) * If T = 4, then P = 3 × 4 = 12. P must be a single digit, so T cannot be 4 or greater. The possible 5-digit numbers are: 35291, 63072, 64182, 94183.

There are 4 such numbers. This question tests logical deduction and systematic enumeration of possibilities, common in number puzzles.