Ages that are each other's digits reversed

Question

A father and his son both have two-digit ages, and the father's age is the son's age with its digits reversed. Their ages add up to 66. The father is more than twice but less than four times as old as the son. How old are they? Show why each clue is needed.

Accepted Answer

The father is 51 and the son is 15. The sum. Let the son's age be 10a + b and the father's 10b + a. Their sum is 11a + 11b = 11(a + b) = 66, so a + b = 6. The possible digit pairs (a, b) with the father older (b greater than a) are (0, 6), (1, 5) and (2, 4): ages 6 and 60, 15 and 51, or 24 and 42. (The pair (3, 3) gives equal ages.) The ratio clue. 60 is ten times 6: more than four times, excluded. 42 is 1.75 times 24: less than twice, excluded. 51 is 3.4 times 15: between twice and four times. So 51 and 15. Why each clue matters. Without the sum, any reversed pair works. Without the ratio, t…

Ages that are each other's digits reversed

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