2 + 8 + 3 + 0! - 1 = 13

]]>If we say they all do not like throwing pirates overboard then C can still propose 100,0,0 as D will not get any gold either way and we now know he doesn't like throwing other pirates overboard.

So we get to A being able to propose 97,0,1,2,0 or 97,0,1,0,2 whether the rule is pirates like throwing other pirates overboard or do not like it.

However, if we are not sure whether pirates like throwing other pirates overboard and in particular C does not know whether D would want to throw C overboard if the gold coin distribution is the same then it is safer for C to offer D a gold coin and the proposed solution given of 97,0,1,0,2 is correct.

]]>http://www.mathsisfun.com/puzzles/5-pirates.html

but as you increase the number of pirates, the patterns get intricate and worth seeing.

First a couple of additional rules, because otherwise the solutions are not unique:

Only whole number of gold coin solutions can be proposed and they don't trust any other agreement.

If a pirate can bribe two or more different pirates for the same number of gold coins and doesn't need to bribe them all then he prefers to bribe the highest ranking pirate.

My answer

Let me know if you do not agree.]]>

That is more wishful thinking than rational logic. If A proposal does not get enough votes, B will make a proposal B 99, C 0, D 1, E 0 and that will be voted through because D won't be offered any by C. D prefers 1 to 0 so votes with B giving the two votes needed for that distribution to be agreed. Therefore, C will not get to have a turn at making a proposal. So if C or E vote down A's proposal of 98,0,1,0,1 then they will get nothing. Therefore it makes sense for them to vote to accept 1 gold coin.

]]>Solution:

- The mirror inverts the clock 180 deg around the 12-6 hrs vertical Axis.

- Let Tm and Td denote the mirror and departure times, respectively,

Then the effect of the inversion is such that:

Tm+Td = 12 for all times around the vertical axis

Or Td = 12 -Tm ........................(1)

- Let Ta denote the arrival time

- Given:

Td+20/60 =Ta ......................(2)

Tm=Ta-2.5 ..........................(3)

Now, substituting (3) into (1) and substituting for Td in (2), we get

Tc =12-Tc+2.5+20/60

=> Tc = 7.416666666667 hours

= 7hrs + (0.41666667*60min)

= 7:25

1. Nd3+(discovered) Kc4 (forced)

Kc4 isn't forced, as Black has an alternative with Rc5 to block White's queen. That leads to mate in four.

]]>Welcome to the forum!

Thanks for your answer. I see now that your solution is the obvious one. Amazing that we all missed it.

Bob

]]>Thanks so much for your help. I do appreciate it.

Here's what I discovered:

And oh, I agree with you. This is no puzzle. It's magic; math magic.

Thanks again.

Yours ever grateful,

math9maniac.