Math Problem

[oznerol]

Actually, mugs is a bit right. I would imagine the "house next door" could be on either side - so that actually leaves multiple possibilities and by that basis no answer can be concluded.

The house number could be 12 or 15. The 3rd clue does not help in this case.

 

[mugs]

Originally posted by: TuxDave

Originally posted by: mugs

Originally posted by: TuxDave
The underlying assumption is that all pieces of information are useful. So telling #2 and having the salesman unable to figure out what the age is implies that knowing the sum of the ages AND the product of the ages must still have multiple solutions.

Want me to solve it for you or is that enough?

Or perhaps he doesn't know the sum, but rather two possible sums (two neighbors).

2 + 3 + 6 = 11
1 + 6 + 6 = 13

Of course evens and odds are usually on opposite sides of the street.

And I don't think the "eldest son" clue really rules out the possibility that the two oldest children are the same age, because they could be fraternal twins - a boy and a girl. (I'll grant that the author can assume twins to be the same age, even if one is born earlier)

Edit:
From your solution, 10 and 14 could be neighbors on either side of 12.

Seems like a poorly written problem overall.

I never said that the mother's house was #12? The bigger issue at hand is the fact that people tend to add information to the problem that's not explicitly written. This is an even BIGGER problem when the author of the question assumes that the reader should add unwritten information. You're adding that the house next door is in sequence to the the house he's at. That's not necessarily true.

I do have to agree that when you're trying to deduce and extract every piece of information, the writer has to be very careful on how they word things because it changes the whole meaning of the question. If the question was phrased as "the salesman looks around the corner and sees the house number and still has no clue" then it would be much clearer. We have technically added information that the salesman KNOWS what the next door house number which could be fixed with a small change.

As for the whole twins thing... yeah, that's always up to debate. It's a pretty clever way to try to give some information that sounds useless but still give useful information. If she said I have twins or I have no tie for oldest, that's kind of boring. Would be interesting to hear a new way of saying something random that's still useful.

I was just giving one possible example where the kids ages could add up to the house numbers of neighbors on either side of the house. If you remove the assumption that houses are numbered sequentially, you have to consider that her neighbors could be 10 and 21, or 16 and 38. That is, you can't rule them out without additional information. So you can't solve the problem unless you make assumptions. If the problem had specified that it was the neighbor to the left, or that there was only one neighbor, then we would know that there is only one number they could add up to.

The problem is, the author knows which clues we should read further into, but the reader has no way of knowing.


This reminds me of that stupid riddle posted by SVT Cobra/F22 Raptor (I think?) about the guys in the bar. People gave all sorts of possible solutions, all of which fit the parameters of the riddle, but the only acceptable solution was that the guys in the bar were actors in a play. We all just thought of possibilities that the author of the riddle hadn't thought of.

 

[z1ggy]

Originally posted by: TuxDave

Originally posted by: RedArmy

Originally posted by: Vegitto
The second and third clue don't add anything. They could be 36, 1 and 1 for all we know. You know, some people make a career out of playing the piano.

That's what I thought, the third question is definitely needed though...I just don't understand how the second one works...I don't understand the reasoning

Well, ok I guess I can spell it out:

So given that the product of the ages is 36, the salesman won't know how old they are because there are still multiple solutions. (I wrote the sum along side of it to continue my point)

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

The mom then said that the sum was something and the salesman is still unable to figure out the age because the house next door must've been #13. If it was any other sum then he'd be able to deduce what the ages were.

The 3rd clue implies there IS an oldest child and no "tie" for oldest leading to 2,2,9.

We have a winner.

 

[DrPizza]

From your solution, 10 and 14 could be neighbors on either side of 12.
Seems like a poorly written problem overall.

I would imagine the "house next door" could be on either side - so that actually leaves multiple possibilities and by that basis no answer can be concluded.

"The number of the house next door is the sum of their ages"

Some of you apparently need to take a course in discrete English before attempting a course in discrete mathematics. THE house next door, not "a" house next door. i.e. the sum is known.

TuxDave's solution is perfect, there are no ambiguities, only idiotic arguments. Christ, I can't believe I haven't seen "well, uhhh, what about 4, 5, and 1 1/2?" It's a simple fucking math problem, not an extensive legal document. You're just jealous that you couldn't figure it out, so are looking for reasons that it's not a perfect solution.

What would you rather see? "The sum of their ages is equal to the number on the house located next door, more specifically, the only dwelling located AS SHOWN ON MAP OF RECORD OF SURVEY, IN THE CITY OF IDIOTS, COUNTY OF MORONS, STATE OF DENIAL, FILED IN BOOK 113 PAGES 90 AND 91 OF RECORDS OF SURVEY IN THE OFFICE OF THE COUNTY RECORDER OF SAID COUNTY, MORE PARTICULARLY DESCRIBED AS FOLLOWS: THE SOUTHWEST QUARTER OF THE NORTHEAST QUARTER OF FRACTIONAL SECTION 4, TOWNSHIP 1 NORTH, RANGE 14 WEST, OVER-ANALYZED MERIDIAN, IN THE CITY OF IDIOTS, COUNTY OF MORONS, STATE OF DENIAL, ACCORDING TO THE OFFICIAL PLAT OF SAID LAND.
EXCEPTING THE EASTERLY 50 FEET OF SAID LAND. ALSO EXCEPTING THAT PORTION OF SAID SOUTHWEST QUARTER OF THE NORTHEAST QUARTER OF SECTION 4, LYING NORTHERLY OF A LINE PARALLEL WITH AND DISTANT NORTHERLY 750.00 FEET MEASURED AT RIGHT ANGLES FROM THE CENTERLINE OF THE EAST-WEST HIGHWAY ROUTE 123, SAID CENTERLINE BEING DESCRIBED AS FOLLOWS: BEGINNING AT THE INTERSECTION OF SILLINESS WAY (100.00 FEET WIDE) WITH THE CENTERLINE OF FOOLISH AVENUE, BEING THE SOUTHEAST CORNER OF SAID SOUTHWEST QUARTER OF THE NORTHEAST
QUARTER OF SECTION 4; THENCE ALONG SAID CENTERLINE OF SILLINESS WAY, SOUTH 1°00?12? WEST 621.13 FEET TO ITS INTERSECTION WITH THE EASTERLY PROLONGATION OF THE CENTERLINE OF SAID HIGHWAY; THENCE ALONG SAID PROLONGATION AND SAID
CENTERLINE, NORTH 89°03?06? WEST TO THE WESTERLY LINE OF SAID HIGHWAY."

 

[Lamont Burns]

So much angst.

 

[waggy]

Originally posted by: Lamont Burns
So much angst.

i think one of his goats fainted on him while trying to have fun with it.

 

[GasX]

Did DrPizza forget his blood pressure medication today?

 

[DrPizza]

Originally posted by: Mwilding
Did DrPizza forget his blood pressure medication today?

lmfao, as a matter of fact, I haven't taken it in 2 days. My wife says she can always tell when I haven't taken it because I get incredibly irritable.

 

[Juked07]

Originally posted by: RedArmy

Originally posted by: z1ggy
6, 3, 2...? lol Is this a college course?

I'd be very surprised if there was a high school that taught Discrete Mathematics (not saying it isn't possible, but it would be rare)

Therefore, to answer your question about it being a college course, yes.

I took a discrete math course in HS. I also took multivariable calc...

 

[Lamont Burns]

Originally posted by: DrPizza

Originally posted by: Mwilding
Did DrPizza forget his blood pressure medication today?

lmfao, as a matter of fact, I haven't taken it in 2 days. My wife says she can always tell when I haven't taken it because I get incredibly irritable.

No way!

 

[biggestmuff]

9, 2 and 2 was my first guess.

 

[Unmoosical]

Moral of the problem, the mother fails.

"She informs him that she has 3 children but does not wish to disclose information about them to a complete stranger" yet still gives enough information (and some, being the piano) to the stranger to figure out the ages.

 

[RedArmy]

Originally posted by: Juked07

Originally posted by: RedArmy

Originally posted by: z1ggy
6, 3, 2...? lol Is this a college course?

I'd be very surprised if there was a high school that taught Discrete Mathematics (not saying it isn't possible, but it would be rare)

Therefore, to answer your question about it being a college course, yes.

I took a discrete math course in HS. I also took multivariable calc...

?

Nice response DrPizza, I was thinking the same thing, but wasn't going to put it in such a...pronounced way.

 

[Paperdoc]

TuxDave got the most probable answer, by far. To quibble, though, he made an assumption: the statement, "3) Her eldest child plays the piano" is interpreted as a definitive statement that there is only one "eldest child", and that eliminates the possibility that the twins are older than the third sibling. Hence the 2,2,9 combo wins over the 1,6,6 combo. That's a reasonable inference from Statement #3.

But just suppose the mother is canny enough to realize what the insurance salesman is doing with his questions, and the piano player is the 6-year-old twin who was born minutes before the other. If we recognize that possibility, then the original problem as presented contains insufficient information to distinguish between two equally possible solutions.

 

[QED]

Originally posted by: Paperdoc
TuxDave got the most probable answer, by far. To quibble, though, he made an assumption: the statement, "3) Her eldest child plays the piano" is interpreted as a definitive statement that there is only one "eldest child", and that eliminates the possibility that the twins are older than the third sibling. Hence the 2,2,9 combo wins over the 1,6,6 combo. That's a reasonable inference from Statement #3.

But just suppose the mother is canny enough to realize what the insurance salesman is doing with his questions, and the piano player is the 6-year-old twin who was born minutes before the other. If we recognize that possibility, then the original problem as presented contains insufficient information to distinguish between two equally possible solutions.

Yes, you're right-- you are quibbling. Now stop it.

What makes you think you have enough liberty to assume the mother is trying to "con' the salesman when it is not mentioned in the problem at all, yet completely discard the "eldest child" clue which was specifically a part of the problem?

 

[silverpig]

Originally posted by: DrPizza
Some of you apparently need to take a course in discrete English before attempting a course in discrete mathematics. THE house next door, not "a" house next door. i.e. the sum is known.

TuxDave's solution is perfect, there are no ambiguities, only idiotic arguments. Christ, I can't believe I haven't seen "well, uhhh, what about 4, 5, and 1 1/2?" It's a simple fucking math problem, not an extensive legal document. You're just jealous that you couldn't figure it out, so are looking for reasons that it's not a perfect solution.

Actually, it's a valid point. How is the sum known?

Imagine the guy goes up to house 12 and sees house 10 and 14 beside him. The information in the riddle still works.

Also, imagine he goes up to house 12 and sees 13 and 11 beside him (to count for the other possible house numbering).

 

[apepooooop]

twins are a possibility for 6,6,1. But the gestational period of a human is only 9 months, it is conceivable for the female body to produce eggs within 6 weeks. That could make siblings who would be the same age for a couple months out of the year. That would produce a definitively older 6 year old.

 

[TuxDave]

If you all have time to bicker about this problem, you all have time to solve a similiar problem that I used a similiar strategy to solve:

Let x and y be two numbers with 1 < x < y and x+y <= 100.
Suppose Sally is given the sum and Paul is given the product.

Paul: I do not know the two numbers.
Sally: I knew that you didn't know the two numbers.
Paul: Now I know the two numbers.
Sally: Now I know the two numbers.

What are the two numbers?

 

[QED]

Originally posted by: silverpig

Originally posted by: DrPizza
Some of you apparently need to take a course in discrete English before attempting a course in discrete mathematics. THE house next door, not "a" house next door. i.e. the sum is known.

TuxDave's solution is perfect, there are no ambiguities, only idiotic arguments. Christ, I can't believe I haven't seen "well, uhhh, what about 4, 5, and 1 1/2?" It's a simple fucking math problem, not an extensive legal document. You're just jealous that you couldn't figure it out, so are looking for reasons that it's not a perfect solution.

Actually, it's a valid point. How is the sum known?

Imagine the guy goes up to house 12 and sees house 10 and 14 beside him. The information in the riddle still works.

Also, imagine he goes up to house 12 and sees 13 and 11 beside him (to count for the other possible house numbering).

The word "THE" indicates a specific home with no possibility of ambiguity. Therefore, there is only one house next door.

You cannot rightly say "There is a fire at the Starbuck's at the intersection of Fifth Avenue and Tenth Street" if there is more than one Starbucks at that intersection-- you would have to say "There is a fire at A Starbucks at the intersection of..." or "There is a fire at ONE OF THE Starbucks at the intersection of..."

 

[Ichigo]

Originally posted by: TuxDave
If you all have time to bicker about this problem, you all have time to solve a similiar problem that I used a similiar strategy to solve:

Let x and y be two numbers with 1 < x < y and x+y <= 100.
Suppose Sally is given the sum and Paul is given the product.

Paul: I do not know the two numbers.
Sally: I knew that you didn't know the two numbers.
Paul: Now I know the two numbers.
Sally: Now I know the two numbers.

What are the two numbers?

x = 4,,
y = 13,,

 

[TuxDave]

Originally posted by: Ichigo

Originally posted by: TuxDave
If you all have time to bicker about this problem, you all have time to solve a similiar problem that I used a similiar strategy to solve:

Let x and y be two numbers with 1 < x < y and x+y <= 100.
Suppose Sally is given the sum and Paul is given the product.

Paul: I do not know the two numbers.
Sally: I knew that you didn't know the two numbers.
Paul: Now I know the two numbers.
Sally: Now I know the two numbers.

What are the two numbers?

x = 4,,
y = 13,,

Good googling skills. (or scripting which is what I would've done)