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Nerdy Math Problem - Page 4

Blogs > Byo
Post a Reply
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Monoxide
Profile Blog Joined January 2007
Canada1190 Posts
Last Edited: 2009-04-21 23:34:23
April 21 2009 23:18 GMT
#61
On April 22 2009 04:54 kingjames01 wrote:
Ah, I just got home and saw these. I got the same answer as above too.
I'll work on the integers one.

Here's one I just remembered when I was on my way home.

A shipwreck strands 10 people on a desert island whose surrounding waters are filled with the world's most deadliest man-eating sharks. These 10 people think using perfect logic. On the island they find a treasure chest filled with a 1000 gold pieces. They decide to divide the gold. Here is the process that they come up with. They each drew straws to figure out the order in which they will attempt to divide the gold. Whoever goes first can share the gold however that person chooses. The catch is that after the person has finished sharing the gold there is a vote. If the majority (greater than but not equal to 50%) of the people are not happy they can vote to have that person tossed off the island. It would then fall onto the person who was selected to be number 2 to attempt to share the gold, and so on.

Now the conditions are:
1) These people will choose life over death.
2) These people will choose more gold than equal or lesser amounts.
3) These people are perfect logicians.

Question: Can person one divide the gold in such a way that he survives the vote? If so, what is the maximum amount of gold that he can get? If not, show why.


Here is my answer
+ Show Spoiler +
This is how the first person should propose the distribution of the coins.
994 , 0, 1, 2, 0, 1, 0, 1, 1, 0

with the person getting 994 and then second getting 0 and so on.
kingjames01
Profile Blog Joined April 2009
Canada1603 Posts
April 22 2009 01:29 GMT
#62
On April 22 2009 08:18 Monoxide wrote:
Show nested quote +
On April 22 2009 04:54 kingjames01 wrote:
Ah, I just got home and saw these. I got the same answer as above too.
I'll work on the integers one.

Here's one I just remembered when I was on my way home.

A shipwreck strands 10 people on a desert island whose surrounding waters are filled with the world's most deadliest man-eating sharks. These 10 people think using perfect logic. On the island they find a treasure chest filled with a 1000 gold pieces. They decide to divide the gold. Here is the process that they come up with. They each drew straws to figure out the order in which they will attempt to divide the gold. Whoever goes first can share the gold however that person chooses. The catch is that after the person has finished sharing the gold there is a vote. If the majority (greater than but not equal to 50%) of the people are not happy they can vote to have that person tossed off the island. It would then fall onto the person who was selected to be number 2 to attempt to share the gold, and so on.

Now the conditions are:
1) These people will choose life over death.
2) These people will choose more gold than equal or lesser amounts.
3) These people are perfect logicians.

Question: Can person one divide the gold in such a way that he survives the vote? If so, what is the maximum amount of gold that he can get? If not, show why.


Here is my answer
+ Show Spoiler +
This is how the first person should propose the distribution of the coins.
994 , 0, 1, 2, 0, 1, 0, 1, 1, 0

with the person getting 994 and then second getting 0 and so on.


Nice! Good answer!
As a prize you receive one Schrute Buck! =)
Who would sup with the mighty, must walk the path of daggers.
Monoxide
Profile Blog Joined January 2007
Canada1190 Posts
Last Edited: 2009-04-22 01:57:49
April 22 2009 01:51 GMT
#63
On April 22 2009 04:02 meeple wrote:
Show nested quote +
On April 21 2009 23:15 stenole wrote:
On April 21 2009 22:37 Chromyne wrote:
On April 21 2009 22:27 stenole wrote:
On April 21 2009 21:06 meeple wrote:
At a party, a guest asks the age of the host’s three children. The host tells the guest the following information:

Assume each child is a whole number age;

The product of their ages is 72;

The sum of their ages is the same as my house number.

The guest goes outside and looks at the house number. Upon returning, the guest indicates that he still needs additional information. The host then says "Oops, I forgot to tell you that the oldest is the only one that likes vanilla ice cream." You now have all the information to determine the ages of the children. How old is each child?
Spent some time figuring out how the house number thing could help me out...+ Show Spoiler +
72 = 2*2*2*3*3

Possible combinations:
2,2,18 = 22
2,3,12 = 17
2,6,6 = 14
2,4,9 = 15
3,4,6 = 13
3,3,8 = 14

We can assume the guest would have been able to figure it out after seeing the house number - unless more of the age combinations add up to the same house number.
Because there is an oldest child, the two others have to be younger.

Ergo, the answer is 3, 3 and 8.


stenole beat me to it. I Got the same answer.


What is my prize? There has to be a win the internet prize. The difference between this riddle/puzzle and the one in the OP is that this one is carefully worded, it tells you that you have enough information to solve it, and it practically strings you along to a single answer by stating to use whole numbers, and giving you three different clues, all of which you know need to give you some sort of info.

Here is another unsolvable problem:

Assume you have 2 herds of cows that have accidentally joined into one herd, randomly dispersed over a circular area. There are two herder teams, A and B. A tries to herd as many of his cows west of the oval.

What is the fastest way for herding team B to get all their cows to the east:
A) B sits and watches A herd their cows westward.
B) B wants to herd cows towards the east.
C) B likes A's cows.


zizi yo... here's another in the same vein,

Find positive integers a1, a2, ... , an whose sum is 100 and whose product a1a2 ... an is maximal over every possible choice of n and every possible choice of a1, a2, ... , an.

Now this one is a optimization problem.


Here is my answer
+ Show Spoiler +
product = 2^50 then the integers are all 2s


not sure if that is correct...
edit : nvm thats incorrect..
ninjafetus
Profile Joined December 2008
United States231 Posts
Last Edited: 2009-04-22 03:24:56
April 22 2009 03:09 GMT
#64
On April 21 2009 15:28 kingjames01 wrote:If you're familiar with the term POLAR coordinates then you realize that means you're talking about a 2-D circle. I've been describing a 3-D sphere, thus I need spherical coordinates.


I know we're past this, and I understand the point you're making, but I'd like to make a small correction. Yes, 'polar' tends to refer to 2d more often and 'spherical' is specifically for 3d, but for we mathematicians out there (who don't need to have pesky physical situations to justify our work) there is a generalization to n-dimensional polar coordinates. Spherical is just the 3 dimensional case.
kingjames01
Profile Blog Joined April 2009
Canada1603 Posts
April 22 2009 03:44 GMT
#65
On April 22 2009 12:09 ninjafetus wrote:
Show nested quote +
On April 21 2009 15:28 kingjames01 wrote:If you're familiar with the term POLAR coordinates then you realize that means you're talking about a 2-D circle. I've been describing a 3-D sphere, thus I need spherical coordinates.


I know we're past this, and I understand the point you're making, but I'd like to make a small correction. Yes, 'polar' tends to refer to 2d more often and 'spherical' is specifically for 3d, but for we mathematicians out there (who don't need to have pesky physical situations to justify our work) there is a generalization to n-dimensional polar coordinates. Spherical is just the 3 dimensional case.


haha, you got me there! I rescind my statement. As you said, I was just pointing out that for physical round objects the usual convention is to use polar to refer to 2-D and spherical for 3-D. But yes, you are absolutely correct for the general n-dimensional case.
Who would sup with the mighty, must walk the path of daggers.
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