Math Problems Created 15 years ago2010-01-06 12:06:57 UTC by ZeG

The hotel owner asks you the following: It lives 3 guests in the hotel. If you multiply thier ages you will get 2450. The sum (using addition) of thier ages are equal to your age x 2.

The hotel owner also tells you that he is older than all the guests in the hotel. My question, How old is the Hotel Owner?

(I cant breach this one myself, dont have an answer. Help! )

XY = 2450
X + Y = 2Z where Z is your age

H > X
H > Y

It's a simple algebra problem. Isolate X or Y in one of your 2 equations, then substitute it in the other.

hmm.. so far I've come up with ALOT of disfunctional families, and some REALLY old numbers for 'your age'

all of which leaving the Hotel Owner roughly in the early/late 40's

(x * y * z) = 2450
(x + y + z) = (A * 2)

O > x
O > y
O > z

are you certain all of it is there? I've been discarding remainder numbers.. however not existing with any limitations.. really the Store Owner could be 135, and the couples child (guests) -24 @_@

lol, I'll try again later if someone else doesn't walk along and instantly solve it (i feel dumb now)

Im not that sharp, i know. But shoudlnt we have atleast 5 variables? Since its 3 guests, XYZ, The hotel owner, H, and yourself, U? So

XYZ = 2450

X+Y+Z = 2U (U is yourself)

H > X
H > Y
H > Z

And i dont really know how to proceed after this, All this above is statements, not equations. How would the equations look like?

/edit

GRRRR.. Soup it has 3 guests, if you multiply them together the number will be too high, and your guests will end up as 15, 13, and 9-1/2

hmm.. so far I've come up with ALOT of disfunctional families, and some REALLY old numbers for 'your age'

all of which leaving the Hotel Owner roughly in the early/late 40's

Quote:

(x * y * z) = 2450
(x + y + z) = (A * 2)

O > x
O > y
O > z

are you certain all of it is there? I've been discarding remainder numbers.. however not existing with any limitations.. really the Store Owner could be 135, and the couples child (guests) -24 @_@

lol, I'll try again later if someone else doesn't walk along and instantly solve it (i feel dumb now)

Im into this track that i may not never gain a exact age of this owner, just a hint like 80-85 (which is random numbers).

And yes, sorry. It IS 3 guests.

Lawl, 3 guests. I don't know why I thought it was only 2. Whoops.

do those work? it's been awhile since i mathed

wait! shit that top one isn't correct.. it ends up with Your Age being like 80

lol.. on hasteful review.. it seems only the teal equation holds any truth at all in that

I'm pissed off.. >_>

http://img442.imageshack.us/img442/5673/fuckq.jpg

I do all that and get something obvious, heh

Do "You" and the hotel owner count as guests?

...haha so complicated. Answers 137 says http://www.scribd.com/doc/24807401/Aalayance-Placement-Paper

I'm pissed off.. >_>

http://img442.imageshack.us/img442/5673/fuc
kq.jpg

I do all that and get something obvious, heh

Do "You" and the hotel owner count as guests?

No, we are the staff

GrimReafer I am sorry but wtf did you do there? I lol'd.
You multiplied that with 2 and you got practically the same thing.
B+C=2A ?

A+B+C=2age => A= 2age/(B+C)

I think this problem can't be resolved without a third know ecuation.

The sum (using addition) of thier ages are equal to your age x 2.

I don't know why but this seems very wrong to me. It means that all the ages are in function of your age. For me it's 32, but for satchmo for example it's over 80(he's 47 right? I don't remember) =)).
So for me, if all the guests would have the same age, all of them would have almost 11 years while for satchmo all of them would have about 26.

Either you're missing key information and would benefit from posting the exact problem statement or what you have here is essentially an unsolvable system of equations.

From the information you have provided you wind up having x, y and z related to each other as well as your age (which can be treated as a given, [I hope you know your own age]). The first complication comes from the fact that you only have two equations. This however is a well known complication, 2 equations with 3 unknowns. And this is usually solved using parametric equations.

There is however a third complication, and it comes from the actual question from the problem statement. The original question asks for the age of the hotel owner, and only states that he is older than all of the guests. This only accounts for a lower bound which means you will only be able to determine his minimum age, which is further complicated by the fact that you will not know the specific age of any of the hotel guests because of the parametric equations.

My advice would be to post the problem itself and the section from the book you are using. (As in, what is the title of the chapter and section of the book it is from). If it isn't from a book, what is the problem supposed to correspond to in your notes (Essentially what are you learning at the moment.)

And at what Unbreakable posted, that entire thing is gibberish. That 137 comes from problem 3 which from what I can tell pertains to absolutely nothing.

It is impossible methinks.. in the end you might be able to get the ages of all the guests, but you cant tell the age of the owner when you only have a lower limit of what his age might be.

Or we could all be mis-interpreting it like I did the first time >_>

Edit: Got ninja'd

Edit: Also Striker.. I know it was a circle, but I was just writing

Sometimes it helps you see stuff.

There is however a third complication, and it comes from the actual question from the problem statement. The original question asks for the age of the hotel owner, and only states that he is older than all of the guests. This only accounts for a lower bound which means you will only be able to determine his minimum age, which is further complicated by the fact that you will not know the specific age of any of the hotel guests because of the parametric equations.

Indeed. I wont get his exact age.

Here is the EXACT question i have:

Eric, that drives a hotel in the locals just hired a new chef, Sara. When Sara demands to know the age of the guests to make up a good menu. Eric says the following:

It lives 3 guests here. If i multiply all thier ages togheter, i get 2450. The sum of thier ages is equal to your age times two.

Sara complains this info aint enough, then Eric says that he is Older than all the guests.

A couple of minutes later have Sara, that knows Erics age, solved the problem.

How old is Eric?

Can't go wrong with Cheezeburgers.

And no matter what Eric's age is, I've solved to the conclusion, that he is a total F*ck.

This is a logic problem, not a math problem. It's impossible to solve it with equations like others have tried. Also, the wording of the problem was very important.

2450 has 5 prime factors (2,5,5,7,7). This means that the ages must be broken up into products of these 5 numbers. Because there are 3 guests, the logical breakdown is that one of these guests is a child aged 2, 5, or 7, and the other two guests are the products of the remainders.

These are the possible breakdowns in this situation:

2 - [35,35] or [25,49] or [5,245] or [7,175]
5 - [10,49] or [14,35] or [2,245] or [5,98] or [7,70]
7 - [10,35] or [14,25] or [2,175] or [5,70] or [7,50]

Now for the second bit of info: the sum is 2 times Sara's age. Without knowing that age, the only thing this gives us is that the sum must be even. BUT the kicker here is that Sara says that this isn't enough information, so there must be 2 or more combinations that add to twice her age. Let's find the sums (in order, getting rid of unrealistic ages):

72,76
64,54,108,82
52,46,82,64

Duplicates are: 64, 82

So Sara is looking at one of these two combinations:
[5,10,49] or [7,7,50]
[5,7,70] or [7,5,70]

The second one are identical, so the duplicate is:
[5,10,49] or [7,7,50]
(Also, this means that Sara is 32)

So the last bit of info must let Sara know which is the valid combination. Eric says that he is older than all guests, and Sara knows the answer after that. This means that one of these combinations has a value older (or equal to) than Eric. This would be [7,7,50] as 50 is the oldest age.

This means that the guest combination is
[5,10,49]
So Eric is older than 49. BUT he must not be older than 50, or the previous cancellation wouldn't be valid.

THEREFORE:
Eric is 50.

QED

You should go to NASA.

That was quite discouraging.. although I get it now >_>

See, it was meteors.

I come home from school to find...

God dammit. I hate you all.

You know a perfectly fine pair of tits just walked past this thread, right?

That was quite discouraging..

Seconded.

Penguin: How did you think to use prime number factors?

2 - [35,35] or [25,49] or [5,245] or [7,175]
5 - [10,49] or [14,35] or [2,245] or [5,98] or [7,70]
7 - [10,35] or [14,25] or [2,175] or [5,70] or [7,50]

How?

Otherwise, this really does make sense. I knew it would be impossible to solve, to much unknown info. Your explination does make sense tho. Goodie!

Those are the combinations that you can multiply 7,7,5,5,and 2 without one of the numbers (the ones to the left of the dash).

Those are the combinations that you can multiply 7,7,5,5,and 2 without one of the numbers (the ones to the left of the dash).

Aye ofc, i should've seen that. I blame all the christmas food!

Penguinboy; Not much of a logical geek, so where'd you get your numbers!?

2 - [35,35] or [25,49] or [5,245] or [7,175]
5 - [10,49] or [14,35] or [2,245] or [5,98] or [7,70]
7 - [10,35] or [14,25] or [2,175] or [5,70] or [7,50]

By multiplying two of the given primes ( 2, 5, 5, 7, 7)
10 = 2 x 5
14 = 2 x 7
35 = 7 x 5
25 = 5 x 5
49 = 7 x 7
etc

And the ages needs to consist of primes since they together needs to multiply into 2450.
2 x 5 x 5 x 7 x 7 = 2450.

So for example; 5 - [14,35]
5 x 2 x 7 x 5 x 7 = 2450 = 5 x 14 x 35

That's some beautiful maths and logic you used there penguin.

I still don't understand what made you think to start finding prime-number factors in the first place.

2450 is an even number ending in zero...

meaning its OBVIOUSLY divisable by prime numbers (2 being one of them)

im certain if i wasnt spending every day idolizing women and anime artists i'd have thought of that sooner

I've never heard that before.

Aren't all numbers ending in zero even?

Edit: God dammit.

i was mour or less refering to it ending in zero.. meaning it's divisable by 5 AND 2

This is a logic problem, not a math problem. It's impossible to solve it with equations like others have tried. Also, the wording of the problem was very important.

2450 has 5 prime factors (2,5,5,7,7). This means that the ages must be broken up into products of these 5 numbers. Because there are 3 guests, the logical breakdown is that one of these guests is a child aged 2, 5, or 7, and the other two guests are the products of the remainders.

These are the possible breakdowns in this situation:

2 - [35,35] or [25,49] or [5,245] or [7,175]
5 - [10,49] or [14,35] or [2,245] or [5,98] or [7,70]
7 - [10,35] or [14,25] or [2,175] or [5,70] or [7,50]

Now for the second bit of info: the sum is 2 times Sara's age. Without knowing that age, the only thing this gives us is that the sum must be even. BUT the kicker here is that Sara says that this isn't enough information, so there must be 2 or more combinations that add to twice her age. Let's find the sums (in order, getting rid of unrealistic ages):

72,76
64,54,108,82
52,46,82,64

Duplicates are: 64, 82

So Sara is looking at one of these two combinations:
[5,10,49] or [7,7,50]
[5,7,70] or [7,5,70]

The second one are identical, so the duplicate is:
[5,10,49] or [7,7,50]
(Also, this means that Sara is 32)

So the last bit of info must let Sara know which is the valid combination. Eric says that he is older than all guests, and Sara knows the answer after that. This means that one of these combinations has a value older (or equal to) than Eric. This would be [7,7,50] as 50 is the oldest age.

This means that the guest combination is
[5,10,49]
So Eric is older than 49. BUT he must not be older than 50, or the previous cancellation wouldn't be valid.

THEREFORE:
Eric is 50.

QED

And this is why I will never learn any programming language ever.

Probably could have solved it with a simple matrix too.

Probably could have solved it with a simple matrix too.

Impossible with the given information.

Also..

It's a simple algebra problem.

Stop saying simple

I still don't understand what made you think to start finding prime-number factors in the first place.

If i multiply all thier ages togheter, i get 2450. The sum of thier ages is equal to your age times two.

That sentence gives us the information necessary to understand that we can use primes to solve it

Impossible with the given information.

Bleh, it was a a stab in the dark anyway.

Stop saying simple

Wat.

I actually now believe that if I knew a lot of mathematics and physics, I could calculate the angular velocity of a shit when somebody does his job in a toilet, only by hearing the sounds.
Mathematics... magic !

Man, I did not know there were so many savants here!

we should do more, but not math problems, some logic ones, Since Penguiboy solved that one, he canpick one

example;

Mary's mum has four children.
The first child is called April.
The second May.
The third June.
What is the name of the fourth child?

And this is why I will never learn any programming language ever.

I don't know half that shit they're talking about, but I like to think I'm a decent programmer.

Mary's mum has four children.
The first child is called April.
The second May.
The third June.
What is the name of the fourth child?

Mary

Dunno.. thats the kind of problem for the person without the proper attention span.

I enjoyed the first problem because it was a solid logic problem, and not one of those gay 'logic' problems that make you have to half-assedly memorize the names of tons of people and dumb little details. I dont like those.

Mary's mum has four children.
The first child is called April.
The second May.
The third June.
What is the name of the fourth child?

Aw now I feel stupid

I love mathematical and logical problems as the one in this thread. It's the only reason I started studying math in the first place

Juliette ?

[EDIT] On second thought. Facepalm. It's Mary.

Juliette is a pretty name.. I might name my 2nd girl juliette (the first one is going to be named 'fionna destiny')



lets see..

here's one i heard from a television program

Annette and the Queen are attending the same party..

Annette is wearing a formal blue dress, and the queen is wearing a combination synthetic silk gown that can take the shape of all other dresses in know existance

the party begins at 5 sharp, and it will take Annette 4 hours to prepare for the party..

with this information.. At what time will the queen arrive at the party, and how how many dress designs will the queen's gown revert through before resembling the dress Annette is wearing..

as a bonus question, how many dress designs will the queen's gown be able to revert through before overheating?

Hahah, Look Around You = win, psilous.