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### paxwill says

They could be 2, 6, 14, and 15 years old, since 2*6*14*15 = 2520 and 3*7*15*16 = 5040, which is 2 times 2520.

Or they could be 2, 8, 9, and 15 years old, since 2*8*9*15 = 2160 and 3*9*10*16 = 4320, which is 2 times 2160.

That leaves one more solution I guess. For fun I tried to find some solutions that included twins or triplets. I couldn't find any where all three ages were the same, but I did find these two that have twins: {4, 4, 5, 15} and {4, 5, 5, 9}.

I found the last one: 3, 5, 8, 9! 3*5*8*9 = 1080, *2 = 2160 = 4*6*9*10. I made a list of all we found, 4 starting with 2, 3 starting with 3, and one other. and figured the last one had to start with 3, and hunted it down.

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### Sid Kemp says

I've got this one part way. One set of ages is 4, 5, 6, 7. And I'm kind of seeing the pattern in prime factors. But I haven't found another solution, or a way to find solutions, yet.

I found one more set, 2, 7, 12, 13, with the help below from bubba-math and tussin. Thanks to all three of you (Calculus-Geometry included) for enjoyable brain-strain this Thanksgiving weekend.

### Jim Bob says

I found 9 solutions, which I believe are all of them. Not to spoil the fun for others I will just give two: {3, 6, 7, 8} and {2, 9, 10, 11}. When finding solutions, I noticed several restrictions that helped me narrow down the field.

1) The set of current ages can't contain 0 (for obvious resaons), 1 (because the product of next year's ages will be more than double since it necessarily contains 2), 16 (because 17 is prime, and the set of current ages will need a multiple of 17), or 17 (because 18 is out of range)

2) The set of current ages can't contain both 2 and 3, because then the set of next year's ages will contain 3 and 4, and then the product will be more than double.

3) The only solution set that doesn't contain either 2 or 3 is Sid's solution, {4, 5, 6, 7}. If you take any other set of four numbers greater than 3, the product you get from adding 1 to each element will be less than twice the current product.

4) If the current set contains 11 or 13, then it must also contain 10 and 12 respectively. That way the set of next year's ages will also contain 11 or 13 and the factors will cancel out.

### I'M BANNED Y'ALL!!!!!!!!!! says

{3, 4, 10, 11} works since 3*4*10*11 = 1320 and 4*5*11*12 = 2640 = 2*1320.

{3, 4, 8, 15} also works since 3*4*8*15 = 1440 and 4*5*9*16 = 2880 = 2*1440

Regarding (1) in Bubba Math's answer, there's no problem in and of itself with having 16 and 17 in the set of current ages. The set of next year's ages would contain 17 and 18, which takes care of the 17. And there's no violation if 18 is in the set of next year's ages because the problem states that their *current* ages are all less than 18; someone could become an adult the following year.

The problem arises in trying to find distinct ages x and y such that (x*y*16*17)/((x+1)*(y+1)*17*18) = 1/2. There's no solution (among the integers) if x and y are distinct. However, if twins were permitted, then the set {3, 3, 16, 17} would be a valid solution to the problem. 3*3*16*17 = 2448 and 4*4*17*18 = 4896 = 2*2448.

You're right, thanks for correcting that.