# Ok, my prof is at it again. This time it's an impossible math riddle problem...

Discussion in 'After Hours Lounge (Off Topic)' started by Jonny K, Apr 10, 2003.

1. ### Jonny K Second Unit

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Another bonus assignment. If I can keep getting these right (with the help of my friends on the internet), I might ace this course with bonuses alone!

Anyway:
"
Jane goes to her friends house for a barbecure. Her friend greets her at the door and Jane responds "What a nice group of children playing in your yard". Her friend answers "They actually came from 4 families - mine is largest, my sisters smaller, my brothers smaller still, and my cousin's the smallest. The product of the numbers is my house number. But there were not enough people to play baseball.

Jane asks "could you tell me if you cousin's family has only one child?". Her friend answers, and Jane then responds by telling her the number of people in each family.

How many children in each family?
"

Note that we don't know the house number or the answer to the question. I confirmed this with the prof.

Here's the math translation (as given by our prof):

X1 > X2 > X3 > X4 >= 1

X1 + X2 + X3 + X4 < 18

X1*X2*X3*X4 = House #

X1 represents the largest family, while X4 is the smallest. 18 is the number of people you need to play baseball.

Now here's what I have so far...

X4 can either be 1 or 2. If it's greater than 2, then the minimum you'd get would be:

6 > 5 > 4 > 3

And 6 + 5 + 4 + 3 = 18 which is not less than 18, so you can't have this case.

Looking at the remaining two cases (X4 = 1 or 2), I get many possible combinations!

X4 = 1:

4 + 3 + 2 + 1 = 10
5 + 3 + 2 + 1 = 11
...
11 + 3 + 2 + 1 = 17

etc etc.

X4 = 2:

5 + 4 + 3 + 2 = 14
6 + 4 + 3 + 2 = 15
...
8 + 4 + 3 + 2 = 17

etc etc.

So without knowing the house number, how in the WORLD do I solve this darn problem! I know a guy who's on the DEAN'S list at the University (he's super smart), and he has no clue either...

Any idea?

Thanks.

Jonny K.

2. ### Sebastian Second Unit

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7 5 3 1
Product 105
Sum 16
She lives on an odd St.
I did some math on the other possible combination (not all) and they all came up even.
So the only # that gives you odd answer (product) is 7*5*3*1
Do not have the time to do the rest of the possible combinations.

3. ### ken thompson Second Unit

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I'm not sure why you think the answer needs to generate an odd product.

4. ### TimDoss Second Unit

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You can't figure it out without being given the house number... I've seen this before and the house number was
given as "120". The answer being 5, 4, 3, and 2.
If X4 equals 1 then there are a couple of
ways to come up with the answer.
Hope this helps.

5. ### Artur Meinild Screenwriter

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The point is that Jane knows the house number is 105. In fact, she didn't even need to ask if the cousin had only one child, since 1-3-5-7 is the only possibility that could generate an odd number where the sum is under 18 and each number is bigger than the one before it.

6. ### Leila Dougan Screenwriter

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I don't have time to work it out, but. . .

there are two different things going on here:

1) the number of PEOPLE in each family
2) the number of CHILDREN in each family

house number = product of children in each family
can't play baseball = less than 18 people, not just children

There is Jane's friend, the sister, the brother, and the cousin. . we don't know about spouses, though. That automatically adds 1 person to each family, so. . x4 >= 2

7. ### Paul Bond Stunt Coordinator

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I agree mostly with TimDoss.

We can safely assume that she knows what her friend's house number is, and since she obviously has an incredible gift for mathematics, she is able to determine what combinations of children would result in the house number

If the number of children in the cousin's family is 1, then there are two possible combinations which generate this number.

If the number of children is 2, then there is only one possible combination. Since the answer to her question gave her what she needed to know to provide the solution, then the cousin must have 2 children.

I think with what all has been said here, you should know or be able to figure it out from here. (I always hate to give away the whole thing.)

Paul

8. ### Jonny K Second Unit

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Cool. I'll work on it.

I was thinking that the answer must have something to do with the fact that she only needed to ask one question to figure out the answer...

Jonny K.

9. ### Jeremy Stockwell Supporting Actor

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I believe the work's already been done, my friend.

JKS

10. ### Artur Meinild Screenwriter

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11. ### TimDoss Second Unit

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There should be no assumptions in a math problem, even if
it is somewhat of a riddle, with any assumptions you're
going to have multiple correct answers depending on what
you've assumed... I see absolutely no reason to think that
it is being playfully worded to make you think to include or
exclude any parents. Besides, Jane and two parents in each
family come out to 9 people making it possible to only have
8 children... 4+3+2+1=10 unless we are to assume that some
of them are divorced, or maybe they mean that there are
enough people to play baseball, but some of the kids aren't
athletic enough to play very well.

12. ### BrianW Cinematographer

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Dang it, Paul beat me to it.

13. ### Jonny K Second Unit

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Impressive.

Thanks.

Jonny K.

14. ### TimDoss Second Unit

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Brian... so have you ever tried to match wits with a sicilian when death is on the line??

15. ### Cees Alons Moderator Moderator

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Of course she knows the housenumber: the whole conversation is held at the door! All she has to do is look up. I seriously cannot understand where the assumption comes from that the housenumber is odd. That's absolutely a false lead, therefore.

I agree with Paul Bond and BrianW. You don't need the housenumber yourself. All you need to know is that Jane, knowing that number, needed one more answer, which tells you (1) she needs it to solve the puzzle and (2) getting a specific answer gave her the solution as well.

Cees

16. ### BrianW Cinematographer

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17. ### Chris Lockwood Producer

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The bus driver's name was Bob.

18. ### Christopher P Supporting Actor

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I wish the internet was around 10 years ago so I could get help with my homework when I was in school.

Chris

19. ### TimDoss Second Unit

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Ok, I'm bored, here are all the possibilities I could come
up with... I trust that Jonny won't just copy everything down for an easy A

2345 -- 120 -- 14
2346 -- 144 -- 15
2347 -- 168 -- 16
2348 -- 192 -- 17
2356 -- 180 -- 16
2357 -- 210 -- 17
2456 -- 240 -- 17

1234 -- 24 -- 10
1235 -- 30 -- 11
1236 -- 36 -- 12
1237 -- 42 -- 13
1238 -- 48 -- 14
1239 -- 54 -- 15
12310 - 60 -- 16
12311 - 66 -- 17
1245 -- 40 -- 12
1246 -- 48 -- 13
1247 -- 56 -- 14
1248 -- 64 -- 15
1249 -- 72 -- 16
12410 - 80 -- 17
1256 -- 60 -- 14
1257 -- 70 -- 15
1258 -- 80 -- 16
1259 -- 90 -- 17
1267 -- 84 -- 16
1268 -- 96 -- 17
1345 -- 60 -- 13
1346 -- 72 -- 14
1347 -- 84 -- 15
1348 -- 96 -- 16
1349 -- 108 -- 17
1456 -- 120 -- 16
1457 -- 140 -- 17
1356 -- 90 -- 15
1357 -- 105 -- 16
1358 -- 120 -- 17

So 120 is the only product that is obtainable using a one or a two for the value of X4.
I still say for a definitive answer we need to also know what the house number is.

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