If the positions of the digits of a two-digit number are interchanged, the…

Concept: A two-digit number with tens digit a and units digit b equals 10a + b. Interchanging the digits gives 10b + a. Subtracting the two forms shows that the difference between a two-digit number and its digit-reversal is always 9 times the difference of its digits: (10a + b) − (10b + a) = 9(a − b).

Applying to this problem: Let the tens digit be a and the units digit be b, so the number is 10a + b.

1. The reversed number 10b + a is smaller than the original by 27, so (10a + b) − (10b + a) = 27.

2. Using the identity above, this simplifies to 9(a − b) = 27, i.e. a − b = 3 — the tens digit exceeds the units digit by 3.

3. The digits are in the ratio 1:2. Since a > b, the larger digit must be the tens digit, so a = 2k and b = k for some digit k.

4. Substituting into a − b = 3 gives 2k − k = 3, so k = 3. Hence b = 3 and a = 2 × 3 = 6.

5. The original number is 10a + b = 10(6) + 3 = 63.

Cross-check: Reversing 63 gives 36; 63 − 36 = 27, matching the given difference, and the digits 6 and 3 are in the ratio 2:1 (i.e. 1:2), matching the given ratio. Both conditions hold together only for this number.

Answer: 63