When you reverse the digits of the father's age, you get the age of the son.…

Concept: In a digit-reversal age problem, if a two-digit age has tens digit x and units digit y, the age itself is 10x + y, and reversing the digits gives 10y + x. A past-age condition of the form "k years ago, A's age was m times B's age" translates to the equation (A's present age − k) = m × (B's present age − k), which is then solved for the digits x and y (each between 0 and 9, with the leading digit at least 1).

1. Let the father's present age be 10x + y and the son's present age be 10y + x, where x and y are single digits and x is at least 1 (the father's age has two digits).

2. One year ago, the father's age was twice the son's age: (10x + y − 1) = 2 × [(10y + x) − 1].

3. Expanding: 10x + y − 1 = 20y + 2x − 2.

4. Rearranging: 8x + 1 = 19y, that is, 19y − 8x = 1.

5. Testing digit values of y from 0 to 9 for an integer x from 1 to 9: only y = 3 gives an integer x, namely 8x = 56, so x = 7.

6. So the father's present age is 10(7) + 3 = 73 years and the son's present age is 10(3) + 7 = 37 years.

Cross-check: one year ago the father was 72 and the son was 36, and 72 is exactly twice 36, confirming the condition. So the present age of the son is 37 years.