Puzzle of not yet 10yr old boy

[lostdummy]

I initially posted this as reply to 1020yr old puzzle ( http://www.greylabyrinth.com/discussion/viewtopic.php?t=14408 ), but decided to repost as separate , since it is completely different puzzle (and one not so easy at that):

My nephew is little boy in his first decade, but his age is NOT "Twenty plus eight times twelve divided by four minus two times six plus ten times eight divided by four plus twenty" years.

How old is he?

[MNOWAX]

to make the obvious first post....

[Suspence]

I get 72, but 72 is ruled out by the fact that the boy is not yet 10.
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[Scurra]

The only bit I could do straight off was that the last step must be "divided by 24" as otherwise you'd never be able to get a value <11.

However, the wording of the initial bit of the question is slightly weird too - all it says is that his age is NOT whatever the answer to the sum is. Well, if the answer to the sum is indeed 82 then that would be a perfectly true statement. So there must be some other trick going on here.
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[Zag]

Assuming that the speaker might have had parens in mind as he spoke, let's try to make every number from 0 to 10, just by adding parens. I'll bet that there is exactly one that you can't make.


(20 + 8 * 12 / 4 - 2 * 6 + 10) * 8 / (4 + 20) = 14 oops, too big

20 + 8 * 12 / 4 - 2 * (6 + 10) * 8 / 4 + 20 = 0

(20 + 8 * (12 / 4 - 2) * 6 + 10 * 8) / (4 + 20) = 6 and 1/6

Edit: I'm starting to despair of my idea. I don't think you can make every number (but one), even if you drop partial fractions.

[ChickenMarengo]

He's 7 assuming that you use two's complement arithmetic:

20 + 8 * 12 / 4 - (2 * 6 + 10 * 8 /4 + 20) = -8
bitwise-NOT -8 = 7

[lostdummy]

Zag, you are on right track ;p

[ralphmerridew]

He's 7.
0 -> (((20 + ((8 * 12) / 4)) - (((2 * (6 + 10)) * 8) / 4)) + 20)
1 -> ((20 + ((8 * 12) / 4)) - ((((2 * 6) + (10 * 8)) / 4) + 20))
2 -> ((20 + ((((8 * 12) / 4) - (((2 * 6) + 10) * 8)) / 4)) + 20)
3 -> ((20 + ((((8 * 12) / 4) - (2 * (6 + (10 * 8)))) / 4)) + 20)
4 -> ((((20 + ((8 * 12) / 4)) - (2 * (6 + 10))) * 8) / (4 + 20))
5 -> ((((20 + (8 * 12)) / 4) - ((((2 * 6) + 10) * 8) / 4)) + 20)
6 -> ((((20 + (8 * 12)) / 4) - ((2 * (6 + (10 * 8))) / 4)) + 20)
8 -> ((((20 + ((8 * 12) / 4)) - ((2 * 6) + (10 * 8))) / 4) + 20)
9 -> (((20 + ((((8 * 12) / 4) - (2 * (6 + 10))) * 8)) / 4) + 20)

[Zag]

Nicely done, rm!

I'm a little gratified to see that my approach was correct, even if I failed at the execution of it. It's a very clever idea to build a puzzle around, and, I'm bet, darn hard to find a find a set that works.

It's a really well constructed puzzle, lostdummy.

[Scurra]

Ditto. At least I noticed the odd thing, even if I didn't understand what it meant (obvious in retrospect, which is always the hallmark of a good puzzle.)
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New definitions: COFFEE - someone who is coughed upon}

[lostdummy]

Good work, that is correct solution ;p

And Zag, you are correct that it was not easy to find set that works. I had idea about using brackets after reading that 1020yr puzzle, even if that one did not require anything more than basic math.

Then I decided to make 'real' puzzle based on bracket idea, and initially I thought to find single solution within certain year span (like decade) and set puzzle to find that exact solution.

I started doing it manually (btw, I found WolframAlpha is good place to evaluate expressions of type "(2*(3+4)-5)", just need to type in search bar). But to be 100% sure that solution will be unique, it required not only finding solution for every solvable year in first century or so, but provind that 'unsolvable' years are really unsolvable - which is practically impossible to do by hand.

So I made quick program to list all solutions (ie for all possible combinations of brackets). And found out that my initial idea wont work, since every decade in first centuries had more years as solution than years without solution.

Then I realized I can reverse problem and ask for number that is not solution, if I can find decade with only one missing. And conveniently enough, first decade was about only one with that property.

Only part that I was worried about was if it will be too hard to find solution manually. If I had to solve such puzzle, I would be among first to decide to write program solution if too hard by hand, but good puzzle should be doable by hand. And seeing that RM solution has similar 'programmatic' signature as my program solutions (brackets around each pair, including around whole expression and around obvious ones like "(((2 * 6) + (10 * 8)) / 4) " ) there is chance he did what I would do and solved it by program. Which is still great (my favorite) approach, but is not exactly best for 'proofing manual solvability' of puzzle since it leaves above question open.

One way to 'test' if its reasonably solvable manually would be if some of you find some different solutions for already listed numbers (there is different solution for "1" and "4" at least), or just try to find solution manually without looking at listed solutions.

If it proves too hard, then there is option of reducing number of elements in formula from 10 to something lower (although it will require again finding suitable decade with just one number missing).

Anyway, I would be interested to hear your opinions if it is too hard to solve manually or not ;p

[Zag]

I only spent about 15 minutes on it, after coming to the epiphany of what was needed. I almost didn't say it because I thought that if it wasn't the answer, I should make a puzzle that did use that trick.

Anyway, I don't think it is too hard, at least for this crowd, but it probably is for a general audience puzzle. There are probably a couple thousand combinations? But many of those can be pruned out as unreasonable when doing it manually. Of course, I knew someone would attack it programmatically, and I would have once I had time if rm hadn't.

[ralphmerridew]

I think the number of answers is (ignoring redundancies) the 10th Catalan number ( http://en.wikipedia.org/wiki/Catalan_number ), 16796.

[lostdummy]

[lostdummy]

Just found another nice setup, much easier for manual solving :

My nephew is teenager NOT "two plus three times four minus three plus two times eight divided by four" years old. How old is he?