Age of kids

Here is a problem by J.A.H. Hunter.

Mike looked up from his scratch pad. 'I've got it,' he said. 'She must be forty-four.' 'Just a lucky guess,' Doug laughed. 'I know the way you work.' 'Okay. You're so smart, so I'll give you one,' replied his friend. 'There are three kids and one is as old as the other two together. If you multiply the difference between the ages of the younger two by itself as many times as the number of years old the youngest was a year ago, you get the age of the oldest kid.' Doug could not refure this challenge. 'You've ignored odd months, of course,' he commented, taking up his pen. Can you solve Mike's little teaser?

Would you please give the ages of the kids(one or more solutions) and the method you used to get these solution(s).

Peter Heichelheim

: itself as many times as the number of years old the youngest was a year : ago, you get the age of the oldest kid.'

By inspection, the kids are 3, 5 and 8. The difference between the ages of the younger two is 2; multiply this by itself twice (age of the youngest a year ago) and you get 2 x 2 x 2 = 8.

Once you realise the youngest kid's age is the exponent you must raise the difference between the two youngest to, there are very few options that could result in an oldest child still young enough to be considered a child.

Where do all these J.A.Hunter problems come from anyway?

S P O I L E R

There seems to be some ambiguity in the problem. Does 'multiply the difference ...... by itself as many times as .....' mean that if the youngest child was two last year, you square the difference, or do you cube it, as 'uncle monty' seems to assume. In other words, are we to find x^n or x^(n+1).

If uncle monty's assumption is correct, then his solution is valid (ages 8, 5, 3)

If not, then the solution is 9, 6, 3 (difference between ages is 3, youngest was 2 last year, and 3^2 = 9)

Clearly, the age of the oldest kid (kid < 20?) must be a second or higher power, i.e. 4, 8, 9 or 16. Also, the difference between the ages of the younger two must be between 2 and 4 (because 5^2 > 20), and the present age of the youngest child must be less than 6 (because 2^5 > 20). This narrows the problem down to manageable proportions.

: youngest child was two last year, you square the difference, or do you cube : it, as 'uncle monty' seems to assume. In other words, are we to find x^n : or x^(n+1).

: If uncle monty's assumption is correct, then his solution is valid (ages : 8, 5, 3)

: If not, then the solution is 9, 6, 3 (difference between ages is 3, : youngest was 2 last year, and 3^2 = 9)

I thought about this, and the first solution I saw after translating teh problem into an equation was your {3, 6, 9} one. But then it seemed clear that to 'multiply something by itself once' means to 'multiply something by itself' i.e. to square it, so to 'multiply something by itself twice' must mean n x n x n, etc.

Actually there is another interpretation of 'multiplying something by itself twice' which I believe is more natural than your suggestion of squaring it. If 'multiplying something by itself' means squaring it, then performing this operation *twice* should yield a fourth power. In this case there would be no integer solutions to the problem though.

I read the solutions by uncle monty and Keith Engers, but don't understand how the following part of the puzzle applies:

Doug could not refure this challenge. 'You've ignored odd months, of course,' he commented, taking up his pen.

Can someone explain its significane, and has this been accounted for in the solutions?

Very good. You got J.A.H. Hunter's answer and explained how he got it. I thought the youngest kid a year ago's age was multiplied by the difference rather than the power but now I realize 'as many times' implies a power. I am getting J.A.H. Hunter's problems from old issues of the Toronto Globe and Mail's Fun With Figures. There are more than 8000 problems. He started in 1952 and stopped in 1983 when he became too sick to continue and he died in 1986.

He just meant the ages had to be integers, i.e. age last birthday.

Ah I see, when I first read it I actually thought that he had missed 'odd months', ie the 1st, 3rd, ..., 11th month, so people are really only around about half of their age and etc etc.

Does anyone have a mathematical solution to this?

Ive gotten as far as:

youngest kid = x years old middle kid = y years old oldest kid = (x + y) years old

so the word problem would turn into this equation:

(y-x)^(x-1) = x + y (the exponent could also be interpreted to be just x)

is this solvable?