Children's puzzle

[gotardz]

Since folks seemed to like the puzzle I posted yesterday, I thought I'd put up another. Comments are screened this time so that's it'll be safe to look at them until tomorrow. Absolutely everyone on my friends list has the math skills necessary to solve this puzzle

Comments

[revjim]

Damn... that sucks...

my first GUESS was 9,2,2... and it worked... and met all the criteria (other than the stupid address clue). So I thought, well, that was too easy. So then I found a few more combinations. That worked. Then I got cranky. So I decided to find ALL of the combinations that worked. There were 8 of them. Two of them had sums that made 13. So... THAT was the reason for the address clue and why we didn't need to know the adress to be helped by that clue. Since the woman STILL didn't know after that clue, it must be one of the combinations adding to 13. So that left 2 combinations: 6, 6, 1 and 9, 2, 2. Of course, since the "oldest" (singular) has red hair, we know that the oldest children cannot be twins. Therefore, it must be 9,2,2.

Yes?

[gotardz]

Right on. And in record time. I'll try to make sure the next puzzle I post wastes more of your time :)

[revjim]

Yeah. This one was fun... required a bit of thought... but... mostly I lucked into it.

The KEY is figuring out why the address is relevant. I happened to think that finding all of the different ways to factor 36 was important. If I hadn't started there, I'd have been quite stuck.

Then again... where else WOULD one start?

[mister_tonberry]

Ok, I'll bite at this one and look like the ignoramus that I am. The way I see it, there's only one piece of relevant information we know (product of x,y, and z = 36). We also know that x+y+z = N and that x = RedHair. I'm sure I'm missing something in between all those lines. That said, I'll just say that the lil buggers are 3,3, and 4 ( we do also know there is one who is definitely the "oldest"). This means that the woman asking after the children lives in unit 10, while her neighbor and the children live in unit 8 or 12. The woman asking the question, unless I am mistaken, appears to have blonde hair, but is actually a brunette. Am I right? :P

[gotardz]

Heh, no. Keep thinking. Once you get it, you'll have an "Aha!" moment. Don't try to quantify hair color :P

[stick_figure]

If there aren't twins in the 3, then 2, 3, and 6.

[gotardz]

Twins are allowed. Make sure you take into account all of the clues. The address and hair color are NOT just there to confuse you.

[ybeayf]

I worked on this for a while, and couldn't find the answer, so I googled it as it was driving me nuts, and you're wrong, I would NEVER in a million years have figured this one out.

[gotardz]

I didn't say it was an easy one. The math isn't the hard part, it's the logic that's tricky. Oh, and the post is entitled "Children's puzzle" because it deals with children, not due to its level of difficulty. And in the future, I can give you a hint before you give up and start googling if you'd like.

I hope you're enjoying these, by the way.

[ybeayf]

No, seriously, it was the math that I was stuck on. I would never have realized (ok, maybe I would have eventually realized, if I worked on it long enough) that if 1, 3, and 12 were factors of 36, then 3, 3, and 4 were also factors, and so on. I listed about 4 groups of factors and that was it -- I honestly didn't realize there were more.

And yeah, they're pretty cool.

[yts88]

It would be entertaining, i'll let you know when one of my neighbors moves.

The ages are assumed to be integral.
By the fundamental theorem of arithmetic: 36=3*3*2*2

The woman does not know after knowing the sum, so the sum of the ages is not unique when you consider all the possible summations.

Possible 3-term Summations that consist of 3,3,2,2:
1+1+36=38
1+2+18=20
1+3+12=16
1+4+9=14
1+6+6=13
2+2+9=13
2+3+6=11
3+3+4=10

Therefore the solution is either 1,6,6 or 2,2,9

And there exists an eldest, so the largest of the 3 numbers is unique.
Therefore the ages of the children are 2, 2, and 9.

[gotardz]

Correct, sir.