In 1941 George's children were both between the ages of 10 and 20. The sum of the cube of one child's age and the square of the other child's age gives the year in which George's wife was born. How old was his wife in 1941?
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In 1941 George's children were both between the ages of 10 and 20.
The sum of the cube of one child's age and the square of the other
child's age gives the year in which George's wife was born.
How old was his wife in 1941?

Start by looking for the possible cubed age:
10^3 = 1000
11^3 = 1331
12^3 = 1728
13^3 = 2197

We can stop there. Obviously the child's age that is cubed can't
be 13; 2197 is nearly 200 years in the future. Now, look for
the age that is squared. Start at the top, 20.

20^2 = 400
Add 400 to 10^3: we get 1400, too far in the past.
Add 400 to 11^3: we get 1731, also too far in the past.
Add 400 to 12^3: we get 2128, at least a hundred years in the future.

We have narrowed the cubed age to 12. Now we need a squared age that
gives a reasonable sum for the mother's birth year.

10^2 = 100: add to 12^3: we get 1828, too far in the past.
11^2 = 121: add to 12^3: we get 1849, still too far in the past.
12^2 = 144: add to 12^3: we get 1872, too far back.
13^2 = 169: add to 12^3: we get 1897; age in 1941 = 44; possible.
14^2 = 196: add to 12^3: we get 1924; age in 1941 = 17; impossible.

George's wife was born in 1897, and she was 44 in 1941.
One child was 12, the other was 13.
by Level 11 User (78.4k points)

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