Here is a word and math conundrum that might amuse the more omnium-gatherumists among the light-hearted posters of this AWAD forum ...

Jet and the Chick and her Three Little Daughters

Jet is a bud of mine. We are nerds, and we share two interests - women and mathematics. In that order in my case, and in the case of Jet, quite the opposite.

ME: (knocking on Jet's door) Knock! Knock! Knock!

JET: Hey amemeba, com'on in.
Hey buddy , did you notice that good-looking chick buck-dancing at Burly Earls last night? She is quite the (wink-smile) gay divorcee, but it seems that she has three children - all girls.

ME: Well yes, Jet, that might prove to be a problem. How old are her little girls?

JET: Well if you multiplied their ages by each other the product would be thirty-six.

ME: Com'on Jet, there are several combinations of three numbers that if multiplied would total thirty-six. I need more information to determine their ages.

JET: (reluctantly) OK, innumberignorant, the sum of their three ages is the same number as my apartment number - upon which you just knocked.

ME: Maybeso, Jet, I know well your apartment number, but I still need more information to determine the ages of the gay divorcee's three little girls.

JET: (sighs with resign) OK, amemeba, the oldest little girl has blue eyes.


QUESTION: HOW OLD ARE THE GAY CHICK'S LITTLE GIRLS?

buck-dancing?

buck-dancing?


CHORUS:

Charleston! Charleston!
Made in Carolina!
Some dance, some prance,
I'll say, there's nothing betta' than the
Charleston! Charleston!
Boy, how you can shuffle.
Ev'ry step you do, leads to something new,
Man I'm telling you, it's a lapazoo!
Buck dance, wing dance,
Will be a back number, but the
Charleston, the new Charleston,
That dance is surely a comer.
Sometime, you'll dance it
One time, that dance called
Charleston, made in South Caro-
Line!

('Bout eighty years old then, I'd say)

Nein, ja? Ja!

A lot of induction in this one, as well as deduction. The solution involves the reasoning that if this much information is enough to solve the problem, and if at the same time this much information is required to solve the problem, then only one of the potential solution sets will do!

well, if he'd said something about living on the first or second floor...

and thanks for the lyric.

NFI

stales

I dunno. I come up with five possiblities with only one of them eliminated by the eye-color proviso.

The import of the eye-color statement is

a) knowing the sum is not enough info to solve the problem, so there must be some sets of factors that have the same sum, and

b) the fact that there is an oldest.

The mislead-your-audience factor is the eye color, which has nothing to do with anything!

'nother name for clogging.

ah, clogging I know! thanks!

So after a false start (and the need to think a bit more cleverly), I think I got it. Am I right in assuming that there is, however, only one circumstance in which the product and sum can be the same using two different sets of three numbers, hence the blue-eyed tiebreaker?

still lost.

As far as I can tell, shanks, that's it exactly.

I'll post an explanation of the problem (as I see it) a little later on, under Answers to Puzzles. Assuming I can find it again!

I got a tip via PM(thanks!!), buthow do we know the apt number? what am I missing?

It doesn't really matter, any more than the blueness of the eyes matters. It's just that there is still not enough information to solve the puzzle even after knowing the sum...

Full answer is posted in I and A, "Answers to puzzles", located here: http://wordsmith.org/board/showflat.pl?Cat=&Board=announcements&Number=131223
at the bottom (today, anyway).

But it's not whited-out or anything, so only look if you really want to know.

huh

From a strictly logical point of view, even if there is a set of twins who are older than the third daughter, one of the twins will be older than the other. Not to mention that saying that one of them has blue eyes does not rule out the possibility that two, or even all three, have blue eyes. This means that even if Jet doesn't know which of the twins is older, if they both have blue eyes, then the older of them will.

However, if we assume that we have enough information* then there is a unique answer.

*You have a sphere. You drill a hole right through the center and measure the cylinder that the hole forms. How much volume is left in the sphere?


BTW: If you'd gone threaded, you could point directly to the post and not the whole thread, thus avoiding searching for the post as it becomes far up from the bottom.

http://wordsmith.org/board/showthreaded.pl?Cat=&Board=announcements&Number=131223

one of the twins will be older

nine minutes in the case of our twins...

left in the sphere

is this one of them x-y=z thangs?

>one of the twins will be older

does anyone know of twins whose birth-dates are celebrated separately??
-ron (reject-a-nit) obvious

From a strictly logical point of view, even if there is a set of twins who are older than the third daughter, one of the twins will be older than the other.

True of course, which is the reason I stipulated that the ages were reckoned in years. Indeed you don't even need to have twins to have two kids the same age; they could have been born eleven months apart...

The "hole-in-the-sphere" problem is nifty (whatever that means, but generally positive). If the diameter of the hole isn't given, and the problem is presented as having a unique solution, then the diameter must be irrelevant and you can choose any convenient figure you like. Including zero. But you do need the diameter of the original sphere!

The proof that the size of the hole is indeed irrelevant is left as an exercise for the reader...all you mathematicians out there listening?

diameter must be irrelevant

thanks doc, for giving me one more puzzle that makes me feel inadequate to the task... and is going to bug me all day long!

Okay guys, and while you're at it, prove this, will you?:

xn + yn = zn

Well, this sphere-with-a-hole puzzle is "under-determined". Try it for a six-inch sphere with a hole in it, and applying the principles above you should be able to solve it. (The volume of a sphere is 4/3 pi r-cubed, if you need it) (and note I didn't give the formula for the volume of a cylinder...; based on those blankety-blank principles above, that has to mean it too is irrelevant)

Okay guys, and while you're at it, prove this, will you?:
xn + yn = zn

I know what you mean, but it's already been done...not cleanly, but done. Cf. Andrew Weil.

you should be able to solve it

but we're going to end up with a formula, right, not an actual number? if that's the case, then I can give up right now... of course, I'll probably give up now anyway...

Well, it'd be something symbolic, like "36 times pi" (because pi always figures into working with circles and spheres), but if you want you could use 3.14 for approximately-pi and multiply it out and get a real number. The trick is to convince yourself that you are doing the right thing by selecting the calculation!

how can you figure out how much is left over if you don't know how much you're taking out?

I see I should restate the problem; there was an omission as presented.

"You have a sphere. You drill a hole right through the center and measure the cylinder that the hole forms. The side of that cylinder is six inches long [or three, or two, or whatever]. How much volume is left in the sphere?"

ASSUMING THAT THE PROBLEM HAS A SOLUTION -- not always a valid assumption but we have to make it -- since they've been so adamant about leaving out the other sizes, they mustn't really matter. The volume left behind (volume-of-sphere minus volume-of-hole) must be the same, no matter what the diameter of the sphere is: a larger-diameter hole will require a larger-diameter sphere to make the straight side measure six inches! And if that's so then you can specify whatever numbers makes your life easiest. Conveniently, the smaller the hole, the closer the remaining volume is to the volume of the sphere itself, and with a microscopically-small hole the volume removed is microscopic and negligible, and with a zero-size hole what's left is volume-of-sphere minus volume-of-hole, i.e. volume-of-sphere minus zero, or just plain ol' volume-of-sphere = 4/3 times pi times radius cubed. The diameter of the sphere will be the height of a zero-diameter hole, and the radius half of that, in this case 3 inches.

You have to take it as a leap of faith that the problem is honest and so we can make those assumptions. Or you can invoke a mathematician to prove it. (I'm a cardiologist, Jim, not a mathematician!) But the problem has been around long enough that I suspect it's true as presented.

(It does make it easier to have the problem stated correctly...)

So, eta, you were right the first time. As the problem was originally given, the answer is a formula: 4/3 times pi time the cube of (1/2 the height of the side of the hole).

And if we're reduced to having me as Resident Mathematician, it's a good thing this is a Words board!

You're right, wofa. The side (or height) of the cylinder is 6 inches. I left that out by mistake. The problem was given in the Mathematical Games section of Scientific American when Martin Gardner ran it, time back way back. It was stated that you didn't need higher math to solve the problem. Your reasoning about the arbitrariness of the diameter of the sphere was in the canonical answer.

Here's another one from the series of problems (which came about once every eight months or so):

Farmer Brown has ten pigs; seven of them are brown, two are white and one is pink. How many of them can say that they are not the same color as any other of Farmer Brown's pigs?

So ... I guess the answer isn't one - the pink one?

whew.

Farmer Brown has ten pigs; seven of them are brown, two are white and one is pink. How many of them can say that they are not the same color as any other of Farmer Brown's pigs?

Wrong answer One: Only one, the pink one.

Wrong answer Two: All of them can say it, but only one can say it and have it be true.

Right answer (I'll hazard a guess): None. Pigs can't talk!

Fap!

Oh, amemeba - see what you started?

if it's a problem not based strictly on math, it's prolly along the lines of: all pigs noses are pink, and the nose is the only thing you can see of yourself if you're a pig and so only the pink pig thinks he's the same color as everyone else... nein??

yeahbut®, which one has blue eyes? and what apartment does she live in? and how big's the hole?
and does she have any sisters?

Wait a minute wofahulicodoc, you left out two wrong answers...

_____________________________________________________
Question: Farmer Brown has ten pigs; seven of them are brown, two are white and one is pink. How many of them can say that they are not the same color as any other of Farmer Brown's pigs?
___________________________________________________

Wrong answer One: Only one, the pink one.

Wrong answer Two: All of them can say it, but only one can say it and have it be true.

Wrong answer three: Pigs can't talk.

Wrong answer four: Pigs can talk, it is seeing that they can't. Pigs see in "spectravision". Their eyes don't outline well and colors grade imperceptibly into the surrounding environment. They can't see rainbows.
Therefore none of Farmer Brown's mud rucking pigs can honestly claim uniqueness.

Wrong answer number five: Faldage unintentionally misstated the question.
When Faldage said..." How many of them can say that they are not the same color as any other of Farmer Brown's pigs?"
Think a minute. What are the odds that a farmer named Brown would have seven brown pigs?
Highy unlikely, huh? I think that Faldage meant to ask...
"How many of them can say that they are not the same color as any other of the farmer's brown pigs?"

In this case the answer is three - the two white pigs and the pink one.

Fess up Faldage!

I'll post the answer over to I&A in the go ahead and spoil your fun thread.