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Puzzle Details

I have three children.

One is the same age as the first number in my age, another is the same age as the second number in my age, and the third is the same age as the sum of the two numbers in my age.

None of the children are the same age and the total of our ages is 45. How old am I?

[Ref: ZJXS] © Kevin Stone

The children are 9, 7 and 2.

As my children are different ages, the lowest they could be is 1, 2, 3, and as our ages add to 45 this would make me 39. Working backwards the only answer to make the sum correct is 27 + 2 + 7 + 9 = 45.

Alternatively, we could use algebra and say that I am 10A + B years of age. My children are A, B and A+B years of age. Our ages add to 45, so:

10A + B + A + B + A+B = 45

Collecting like terms together gives:

12A + 3B = 45

Dividing throughout by 3 gives:

4A + B = 15

If A = 1 then B would be 11, which makes no sense for an age. If A = 2 then B = 7 and as all values that work give us a correct answer. So I am 27.

Note how A = 3 would make B = 3, which isn't allowed. And A > 3 will take us past 15, so there can be no other solutions.

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