question and answer thread.

But this saves 99(.33) people:

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Originally posted by solarflare
But I think my latest idea would work, I just didn't describe it well enough. The first person is able to say whether the difference between the number of red hats and blue hats is a multiple of three or one more or one less.

Assuming it's a multiple of three, then the second person seeing the same difference among the 98 people (excluding himself and the first person) has a white hat. If it's one more then he has a blue hat, and if it's one less he has a red hat. The same argument works for when the original number is one more or one less than a multiple of three, but with a shift.

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Once the answer is ackl...could some please explain it clearly Maybe with pictures ..

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Originally posted by solarflare
But this saves 99(.33) people:

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ARg, you are of course right :) I just skimmed that post and didn't realize that you did post the answer :)

Congratulations.

good job ..solarflare ... amazing thinking

So when we're talking about canibals ;) I have a new one :p

Canibals capture two men and tell them that they are going to be eaten, but also given a chance to escape. Both men are going to be told each a number A and B, (B=A+1) (1<=A<B<=10). Each has to guess the other's number. Each has the right to say no more than 3 times "I don know your number", one time "Your number is..." and nothing else. The canibals will point one of them to begin. What must be the two men's strategy so they would guess each-other's numbers with the phrase "Your number is...", and thus escape?

oops! forgot to mention :rolleyes:
One cannot say two things in a row, they must take turns.

Its just iterating out the possibilities. If the first person has 1 or 10, then they know the other person's number is 2 or 9 respectively. Otherwise, if the first person has neither 1 nor 10, they say "I don't know your number". Then the second person knows the first has neither 1 nor 10, so if they have 2 or 9, they know the other has 3 or 8 respectively. Otherwise, they say they don't know either. Widdling down the choices the possibilities are:

First: 1, 10
Second: 2, 9
First: 3, 8
Second: 4, 7
First: 5, 6

and that's it. Somewhere along the line, one of the two will have one of the options for their turn and know the other's number. In a worst case, both people will have said that they didn't know the other's number twice (the last step possible will always have the First asked knowing the result). So the option for 3 dont-knows is one too many!

Without a calculator or computer (in other words just using algebra and/or calculus) prove that

pi^e > e^pi

Or "pi raised to the power of e is less than e raised to the power of pi"

claim: x^e < e^x for all x > e.

since ln(x) is strictly increasing on the interval (0,infinity) and so
on the interval (e,inifinity) it is enough to show that

eln(x) < x for all x > e.

consider f(x) = x - eln(x)

f(e) = 0;

f'(x) = 1 - e/x which is positive when x > e and so f(x) is
increasing on the interval [e,infinty) .

The result proven

If the pi is partially eaten (represented by pi prime), however, pi'<e, thus the solution does not hold.

Darn, I did not consider the notorious partially eaton pi problem:(

I hope we can assume that Pi (3.14...) is bigger than e (2.72...). The first part is proving that for x > e we have x > ln(x) e:
x > e
x - e > 0
1 - e/x > 0 (we can divide by x since it's >0)
Now we integrate the function 1 - e/x on the interval [e, infinity[ and we get:
x - ln(x) e
Since its derivative is positive (meaning increasing function) and its value at e is 0, we have for x > e:
x - ln(x) e > 0
x > ln(x) e

Now we insert Pi into the function:
Pi > ln(Pi) e
ln(e) Pi > ln(Pi) e (since ln(e) = 1)
Taking the exponential of both sides (both are > 0), we get
e^Pi > Pi^e

[edit: arg I took too long to write it up :( ]

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Originally posted by Yves M
[edit: arg I took too long to write it up :( ]

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Must have been because of this wasted step:

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ln(e) Pi > ln(Pi) e (since ln(e) = 1)

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;)

Yes. Those proofs are sufficient. This was an interesting problem I saw earlier today. The proof I took was very similar to Yves'.

- Kevin

The proofs are actually identical given the fundamental theorem
of calculus, however Yves proof contains a slight abuse of notation.
You find a particular antiderivative. Remember getting points off
if you forget the arbitrary constant. ;)