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The Armchair
08-02-2004, 03:56 PM
You're sitting at the 2/6 table at the Excal. You, being a cordial 2+2er, know to answer logic puzzles in white-on-white text, as to not give it away for your fellow boardmates. A seat opens on your left, and a guy you've never seen before sits down. (We'll call him "Left.") But the guy on your right ("Right") does know him -- they were former college roommates!

Putting collusion issues aside (the guy on the right is a total fish), you stay at the table.

And then the strangest conversation happens:

Left: Hey, Right, how's it goin! Long time no see!
Right: Hey! How's by you?
Left: Good, good... on vacation with the fam. Wife's upstairs with the three kids.
Right: Three? Wow, congrats, how old are they?
Left: I'm not going to tell you.
Right: ???

At this point, you're struggling mightily to figure out how Right successfully said "???," given the lack of vowels. But no matter -- the conversation just got interesting.

Left: I'm sure you can figure it out.
Right: Uh. . . how.
Left: I'll give you clues, of course.
Right: Of course.

(Apparently, they used to do this all the time. It's amazingly that either of them ever got far enough into a relationship to have children.)

Left: Multiple their ages together and you'll get 36, then --
Right: Okay, so, they're . . . wait a sec, I can't figure it out yet!
Left: Maybe you shouldn't interrupt, then? In any event, you're right, you can't figure it out yet. You also need to know that if you add their ages together, it equal our old dorm room mumber.
Right: Hey, neato. So they're . . . wait, I also need to know --
Left: The oldest one can read.
Right: Cool. Thanks.

At this point, you're confused. Left, always looking to screw with someone, turns to you and says:

Left: I'll pay your next three blinds if you can tell me how old my kids are. Five if you can tell me how you figured it out.

You answer. . . ?

(Edit in italics)

aloiz
08-02-2004, 04:14 PM
Ok I'll post in whit this time.
<font color="white">
The last question asked I deduced to be whether the twins were older or younger than the third child. Given that the answer was older the only combination is 6,6,1.
</font>

aloiz

The Armchair
08-02-2004, 04:26 PM
Given what I wrote, you're right, but I changed it to make it so you're wrong. The reason I changed it is because your reasoning is incomplete due to an omission on my part.

AndysDaddy
08-02-2004, 04:27 PM
My response in White:
<font color="white"> The answer is 6, 6, and 1. The only way that the unanswered (to us) dorm room question could ellicit right's response is if two such sums added up equally. The combos of 6x6x1 and 2x2x9 both add up to 13. The unasked question was "Are the twins younger or older than their sibling". Since they are older, the answer must be 6,6, and 1.</font>

The Armchair
08-02-2004, 04:33 PM
Five blinds to you. Perfect answer (at least before my edit).

AndysDaddy
08-02-2004, 04:37 PM
Your edit does not change my answer, and changes the reasoning only superficially. In fact, your change seems to make the answer ambiguous. More in white:
<font color="white">Six year olds can certainly be expected to be able to read, if only a bit, as can nine year olds. While your statment "The oldest can read" (my emphasis) implies that the oldest is not a twin, in fact one of the twins is older (as any older twin will probably tell you).</font>
So now I am confused as to what the correct reasoning is.

aloiz
08-02-2004, 04:46 PM
Now I'm slightly confused. Before you've changed it I don't see how my reasoning is incomplete given what you wrote. Now given what you've added I'm slightly less sure of the answer. Here's what I got:
<font color="white">
All three number combinations that multiply up to 36:
6 6 1
9 2 2
2 3 6
4 3 3
36 1 1
9 4 1
12 3 1
18 2 1
Only two of those combinations add up to the same number (661 and 922), and since we know that the dorm room number was insufficient information these are the only two possiblities. The oldest one can read I assume that means 9 2 2, but plenty of kids read at age 6, and one twin is always older than the other so I'm not exactly sure. Maybe I'm missing something.
</font>


aloiz

The Armchair
08-02-2004, 04:55 PM
<font color="white"> (a) If it's enough information for the roommate, then it cannot create ambiguity. Even though one twin is technically older than the other, that's clearly irrelevant. </font>

kyro
08-02-2004, 05:05 PM
i didn't look at the answers, so therefore no chance of me stealing anyone's glory...

my guess is 2,2,9

Right couldn't tell the ages of the kids even after Left told him the sum was their dorm number. So of the 8 possibly ways to factor, only two make sense, as they are the only two combinations that have the same sum.

1,6,6
2,2,9

My final guess is really a leap of faith. Left says "oldest child" which suggests one child can be the oldest. Since the first combination has two "oldest children" the second combination seems to be the only possible answer.

How'd I do? <font color="white"> </font>

jwvdcw
08-02-2004, 05:16 PM
<font color="white">I guess I"m totally retarded, but where does it say that theres twins? It seems as if everyone is getting this from the 'add their ages equals our dorm room number, but I just don't get it. 6+6+1= 13..what does that tell you?</font>

aloiz
08-02-2004, 05:37 PM
Response in white:
<font color="white">
There's only two three number combinations that multiply to 36 and add up to the same number, 922 and 661. Since the dorm room number was not enough information we know that these are the only two possibilities. Armchair changed his original post, which originally had the response to the last question as "yes they're older" in which case you assume the unasked question was refering to the twins being older or younger than the third child. Now that he changed it to "yes the oldest can read" I think there's more ambiguity in the question than before. However using the last clue you're suppose to eliminate the 661 as a possiblity. </font>

aloiz

jdl22
08-02-2004, 05:43 PM
<font color="white"> Just thinking out loud through this. So the product of the three is 36. Numbers that multiplied together equal 36 are 36x1,18x2,12x3,9x4,6x6

That leaves these possible combinations (with sum after):
1,1,36 - 38
2,1,18 - 21
2,2,9 - 13
2,3,6 - 11
3,1,12 - 16
3,3,4 - 10
4,1,9 - 14
4,3,3 - 10
6,1,6 - 13
6,2,3 - 11

There are some duplicates here so the shortened list is:
1,1,36 - 38
1,2,18 - 21
1,3,12 - 16
1,4,9 - 14
1,6,6 - 13
2,2,9 - 13
2,3,6 - 11
3,3,4 - 10

Since the guy had to qualify his statement that must mean that there are two possibilities for solutions to the problem, ie two possible 3 number sets that add up to their dorm room number whose product is 36. That could only mean that their dorm room number is 13. Hence the children are either two, two, and nine or one, six, and six. At this point we have a problem. Nine year olds and six year olds can read. If he considers people with the same number of years as the same age then clearly they are 2, 2, and 9. If not then it could be either.</font>

jwvdcw
08-02-2004, 05:49 PM
Ok I get how you get down to 2 possibilites now....but the 'the older one can read' still makes no sense to me since 1.theres twins and 2.Can't 6 year olds and 9 year olds read?

The Armchair
08-02-2004, 06:14 PM
It's a nitpick, but an entirely fair one. Of course, it's also a requiste part of the question that we're treating the kids' ages as integers; therefore, there's no difference between 6 and 6. The answer must be 9, 2, 2.

If you want to claim ambiguity, that's fine, but you should either treat the kids as integers or not, and not mix-match.

In any event, the ambiguity is my fault. The original italicized line said "They're older," which is fine, but it guts the question. That's a not-so-clear reference to the existance of twins, and the only set of twins older than the third child is 6, 6, 1. I should have said "they're younger," which would have made the options 36,1,1 9,2,2 and 4,3,3. That'd require the full analysis.

GuyOnTilt
08-02-2004, 06:50 PM
Booooo! Something hard please!!!

GoT