Puzzle of the day

Skogen · Post by **Skogen** » September 11, 2003, 2:58 pm

See if you can get this....and dont go looking for the answer on the internet!
Yes, there is enough info here to solve it.
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two MIT math grads bump into each other. they haven't seen each other in over 20 years.

the first grad says to the second: "how have you been?"
second: "great! i got married and i have three daughters now"
first: "really? how old are they?"
second: "well, the product of their ages is 72, and the sum of their ages is the same as the number on that building over there.."
first: "right, ok.. oh wait.. hmm, i still don't know"
second: "oh sorry, the oldest one just went to science camp with her entire class"
first: "wonderful! my oldest is the same age!"

problem: how old are the daughters?

Zamtuk · Post by **Zamtuk** » September 11, 2003, 3:03 pm

I got 9, 4, and 2

Post by **Sylvus** » September 11, 2003, 3:03 pm

2, 4 and 9.

Skogen · Post by **Skogen** » September 11, 2003, 3:10 pm

Nope.

Pilsburry · Post by **Pilsburry** » September 11, 2003, 3:11 pm

Whats the number on the building?

Skogen · Post by **Skogen** » September 11, 2003, 3:12 pm

Pilsburry wrote:Whats the number on the building?

You dont know it.....ALL the info you need is in what I gave

Kilmoll the Sexy · Post by **Kilmoll the Sexy** » September 11, 2003, 3:13 pm

How about 18, 4, and 1?

Kylere · Post by **Kylere** » September 11, 2003, 3:13 pm

The building number is 14

Zamtuk · Post by **Zamtuk** » September 11, 2003, 3:14 pm

yeah, that was why I edited, but saw sylvus had the same answer so re-edited...hrmm, ill solve this thing

Kylere · Post by **Kylere** » September 11, 2003, 3:14 pm

D'Oh easy 8, 3 and 3

Skogen · Post by **Skogen** » September 11, 2003, 3:14 pm

Kilmoll the Sexy wrote:How about 18, 4, and 1?

Nope.

Lohrno · Post by **Lohrno** » September 11, 2003, 3:14 pm

What is the number on "That building over there?"

-=Lohrno

Zamtuk · Post by **Zamtuk** » September 11, 2003, 3:16 pm

6, 6, 2

Post by **Sylvus** » September 11, 2003, 3:17 pm

Kylere's right. 8, 3, 3. 2, 6, 6 would be the other possibility, but saying the oldest one went with her whole class would have meant that both of the 6 year olds went.

Kylere · Post by **Kylere** » September 11, 2003, 3:18 pm

Nope Zamfuck has to be 8 3 and 3 he said his oldest daughter, not his tied for oldest daughter heh

Kylere · Post by **Kylere** » September 11, 2003, 3:18 pm

Now if only figuring out lame math puzzles was a salable job skill

Zamtuk · Post by **Zamtuk** » September 11, 2003, 3:19 pm

certainly a possibility.

Skogen · Post by **Skogen** » September 11, 2003, 3:19 pm

Kylere wrote:D'Oh easy 8, 3 and 3

Heard this one before, right?

Kilmoll the Sexy · Post by **Kilmoll the Sexy** » September 11, 2003, 3:20 pm

There is no logical correct answer that can be gleaned from this. There are way too many variables that are left open. These could be 60 year old men and the answer could be 36, 2, and 1. It could be anything. Without the known variable of the building number, this has no real answer.

Zamtuk · Post by **Zamtuk** » September 11, 2003, 3:21 pm

well, it did say MIT grads that haven't seen each other in over twenty years. That places them around 45-50. Though I am guessing if by over twenty it was between 20-25 or else they would have said over twenty-five years.

Kylere · Post by **Kylere** » September 11, 2003, 3:24 pm

Skogen wrote:
Kylere wrote:D'Oh easy 8, 3 and 3
Heard this one before, right?

Nope to be honest they first 5 times I read through it, I was all like WTF? Then boom it was there like a porn video on a 500 foot high screen.

Kilmoll the Sexy · Post by **Kilmoll the Sexy** » September 11, 2003, 3:26 pm

But everything is by assumption. There are only 2 facts given. X*Y*Z=72 and X+Y+Z=V where V is unknown. The fact of 20 years means nothing, since the man could have had a daughter during or even before he last saw the other guy. You can mathematically figure out x, y, and z if you are given the unknown variable, but you can shoot holes in anything given as an answer without knowing V.

Skogen · Post by **Skogen** » September 11, 2003, 3:28 pm

Kilmoll the Sexy wrote:There is no logical correct answer that can be gleaned from this. There are way too many variables that are left open. These could be 60 year old men and the answer could be 36, 2, and 1. It could be anything. Without the known variable of the building number, this has no real answer.

Nope...the key to it all lies in the fact that the 2nd guy could not get the age after looking at the number on the building. That means there is more than one combination of numbers that sum to the building number, and also have a product of 72. After factoring 72, we find that the ONLY two combos of three numbers that have equal sums are 2,6 &6, and 3,3 & 8
Since we know there is an oldest daughter, that dumps the 2, 6, & 6 possiblities. so 3, 3, & 8 it is!

Fattyfat · Post by **Fattyfat** » September 11, 2003, 3:34 pm

2 6 6 is still a possibility just not a likely one.

what if 1 daughter was born in feb and the next was born 9 months later?

Skogen · Post by **Skogen** » September 11, 2003, 3:36 pm

Fattyfat wrote:2 6 6 is still a possibility just not a likely one.

what if 1 daughter was born in feb and the next was born 9 months later?

Yep! Thats a possiblity However, I covered that by saying that ALL the information is availble to solve it. If 2,6, & 6 were an answer, Not enough info is given.

Extra points to Fattyfat.

Kilmoll the Sexy · Post by **Kilmoll the Sexy** » September 11, 2003, 3:36 pm

That still makes no sense. That was the first thing I explored with the building number, but it still does not hold true. You only can have ONE oldest child, whether they are twins or single birth children. Unless the mother pulls off some incredible stunt and hatches one out her ass at the same time.....but discounting that fact, he could still have the oldest child be 6 and have another also be 6. Not to mention the fact that it is also possible to have 2 children be the same age as the gestation period in humans is roughly 9 months.

Sometimes people seem to go out of their way to write puzzles that have an answer that can be disproven. It was obviously not an MIT grad that wrote this one.

edit: damn you fatty

Skogen · Post by **Skogen** » September 11, 2003, 3:39 pm

Kilmoll the Sexy wrote:That still makes no sense. That was the first thing I explored with the building number, but it still does not hold true. You only can have ONE oldest child, whether they are twins or single birth children. Unless the mother pulls off some incredible stunt and hatches one out her ass at the same time.....but discounting that fact, he could still have the oldest child be 6 and have another also be 6. Not to mention the fact that it is also possible to have 2 children be the same age as the gestation period in humans is roughly 9 months.

Sometimes people seem to go out of their way to write puzzles that have an answer that can be disproven. It was obviously not an MIT grad that wrote this one.

edit: damn you fatty

don't be bitter that you couldn't get it.

If one is born in january, and the next is conceived right away and delivered in october. next october both siblings will be one year old.

Zamtuk · Post by **Zamtuk** » September 11, 2003, 3:42 pm

extra points to me for coming up with that answer first!

Fattyfat · Post by **Fattyfat** » September 11, 2003, 3:51 pm

now that you started posting a puzzle I went looking for some here's one:

The warden meets with 23 new prisoners when they arrive. He tells them, "You may meet today and plan a strategy. But after today, you will be in isolated cells and will have no communication with one another.

"In the prison is a switch room, which contains two light switches labeled A and B, each of which can be in either the 'on' or the 'off' position. I am not telling you their present positions. The switches are not connected to anything.

"After today, from time to time whenever I feel so inclined, I will select one prisoner at random and escort him to the switch room. This prisoner will select one of the two switches and reverse its position. He must move one, but only one of the switches. He can't move both but he can't move none either. Then he'll be led back to his cell.

"No one else will enter the switch room until I lead the next prisoner there, and he'll be instructed to do the same thing. I'm going to choose prisoners at random. I may choose the same guy three times in a row, or I may jump around and come back.

"But, given enough time, everyone will eventually visit the switch room as many times as everyone else. At any time anyone of you may declare to me, 'We have all visited the switch room.' and be 100% sure.

"If it is true, then you will all be set free. If it is false, and somebody has not yet visited the switch room, you will be fed to the alligators."

What is the strategy they come up with so that they can be free?

Lohrno · Post by **Lohrno** » September 11, 2003, 3:56 pm

Fattyfat wrote: What is the strategy they come up with so that they can be free?

When a prisoner gets out of his cell, he should kick the Warden's ass, steal his keys, and set everyone free!

Oh! I am smart!

-=Lohrno

Kilmoll the Sexy · Post by **Kilmoll the Sexy** » September 11, 2003, 3:58 pm

Skogen wrote:don't be bitter that you couldn't get it.

If one is born in january, and the next is conceived right away and delivered in october. next october both siblings will be one year old.

I am not bitter. They have flawed language in their puzzle. GIGO.

Fattyfat · Post by **Fattyfat** » September 11, 2003, 3:58 pm

oh btw. i don't know the answer iu'm working it also.! I think i'm close

Kylere · Post by **Kylere** » September 11, 2003, 4:01 pm

Ooo the switch thing is rough, you would have to set up a series of rules based on the position of the switches when the first prisoner entered the room. Hmm

I do not deem this worth thinking through any further without being paid for it.

Skogen · Post by **Skogen** » September 11, 2003, 4:01 pm

Kilmoll the Sexy wrote:How about 18, 4, and 1?

Who goes to science camp at 18! Wait, some total MIT geeks daughter I guees!

Kilmoll the Sexy · Post by **Kilmoll the Sexy** » September 11, 2003, 4:09 pm

My train of thought after discounting the 6/6/2 , 8/3/3 ages was that the oldest daughter could have been a teacher and taking her class to a science trip. That would have been 36/2/1, but I went with a college age girl instead....mainly because she could have been hawt.

Skogen · Post by **Skogen** » September 11, 2003, 4:09 pm

Kilmoll the Sexy wrote:But everything is by assumption. There are only 2 facts given. X*Y*Z=72 and X+Y+Z=V where V is unknown. The fact of 20 years means nothing, since the man could have had a daughter during or even before he last saw the other guy. You can mathematically figure out x, y, and z if you are given the unknown variable, but you can shoot holes in anything given as an answer without knowing V.

Solving by X unknowns and X equations is the wrong approach.

Faerin · Post by **Faerin** » September 11, 2003, 4:32 pm

There are 2 switches. Switch A is used as a counter, switch B as a dummy.
1 of the 23 inmates is declared leader, he will count the number of people to have flipped the switch, and will be the only one to decide whether or not all inmates have entered the room.

If you enter the room for the first or second time, and switch A is off, turn it on. Otherwise flip switch B. Each inmate other than the leader must do this twice, since the starting position of switch A isn't known. Every other time, you will flip the dummy.

If the leader enters and sees Switch A on, he turns it off and adds to the total inmates that have flipped the switch, if it's off, no one new has been in since his last visit, and he flips the dummy.

When the leader counts 44, he declares all inmates have visited the room.

Skogen · Post by **Skogen** » September 11, 2003, 4:33 pm

Faerin wrote:There are 2 switches. Switch A is used as a counter, switch B as a dummy.
1 of the 23 inmates is declared leader, he will count the number of people to have flipped the switch, and will be the only one to decide whether or not all inmates have entered the room.

If you enter the room for the first or second time, and switch A is off, turn it on. Otherwise turn on switch B. Each inmate other than the leader must do this twice, since the starting position of switch A isn't known. Every other time, you will flip the dummy.

If the leader enters and sees Switch A on, he turns it off and adds to the total inmates that have flipped the switch, if it's off, no one has been in since his last visit, and he flips the dummy.

When the leader counts 44, he declares all inmates have visited the room.

teh win!

Zamtuk · Post by **Zamtuk** » September 11, 2003, 4:36 pm

hrmm, i could figure it out, but that will take a great deal of time. I think Faerin is on to something, but do not think it is totally correct.

VeeshanVault

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