Results 11 to 20 of 23
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10-24-2005, 07:48 PM #11Member
I came up with an awesome math/logic problem stoned.
Excuse me...by my calculations...every individual can have kids only with one other individual during their lives not 2 as I said above
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10-24-2005, 07:48 PM #12Senior Member
I came up with an awesome math/logic problem stoned.
oh my head hurts
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10-24-2005, 07:51 PM #13Member
I came up with an awesome math/logic problem stoned.
And well I'm really ashamed but if I'm making a mistake here could you gimme the slightest of hints about what it may be?
Pwease
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10-24-2005, 07:52 PM #14Senior Member
I came up with an awesome math/logic problem stoned.
dude you left your weed in the math department didnt you. thats why you have these trippy problems
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10-24-2005, 09:06 PM #15Senior Member
I came up with an awesome math/logic problem stoned.
Originally Posted by Caruso329
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10-25-2005, 12:18 AM #16OPSenior Member
I came up with an awesome math/logic problem stoned.
Haha okay guys, stop trying to read so deep into this. This is just a fucking HYPOTHETICAL situation. I just came up with those rules (18 and 36 breeding for example) to add some structure to the problem and to make it more difficult.
And every male can (and should, if you want to figure out the answer) have sex with every female in the other family that he isn't related to in some way. If they share the same aunt or are cousins for example, they cannot breed. Same goes for step-relations.
Originally Posted by OlympianBud
-Because I said they could!If you really want a reason, then they're too busy raising the children they have now to have any more until their current ones have matured.
-You're on to something there Olympian, I can't really answer this without giving it away.
Good luck, and I'll probably post the solution later tonight or tomorrow.
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10-25-2005, 10:56 AM #17Senior Member
I came up with an awesome math/logic problem stoned.
[1] How many generations can these 2 families live for until there are no more possible unrelated combinations of males and females able to produce?
Three generations: The original four, their kids, and grandkids.
[2] How many people will have been alive after everyone dies out?
The four originals, their 16 children, and the 128 grandkids. Total - 148.
[3] What will have been the most people alive at one time?
148 - the originals being 72, their first children being 54, their second children being 36. The first set of grandkids being 18, and the second set having just been born.
[4] How many years will these people survive for?
In total, from the origainl four at 18, to the grandkids deaths; 127 years.
[EDIT] Not sure about the answers to three and four now.
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10-25-2005, 11:12 AM #18Senior Member
I came up with an awesome math/logic problem stoned.
[EDIT] Update to last post.
[3] 4 originals, 16 kids, 64 grandkids.
[4] 84 people alive.
My brain aches, please tell me I got it right?!
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10-25-2005, 11:33 AM #19OPSenior Member
I came up with an awesome math/logic problem stoned.
ohh psychopixi you are soooo close. Solution to come after I'm done with my classes today.. so in maybe 7-8 hours? Haha, soo close psychopixi. You're definitely on the right track, but the 16 (2nd generation) will not produce 64 children. I think you're just trying to turn it into a simple equation, which it's not that simple. Go back and check some of your pairings with the 2nd generation. Remember, if Jack mates with Jill, then you have to make sure Jill's dad, mom, aunts, uncles, grandparents, great grandparents, great great etc. are not tied to Jack's side of the family at all.
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10-25-2005, 12:14 PM #20Senior Member
I came up with an awesome math/logic problem stoned.
Okay... so I have 7 or so hours to crack this. *determined* how about this...
How about the children of the original four mate with each other - to give 64, and with the one of the original four who is of the opposie sex and not a parent, giving an extra 64.
Say the original four are Ma, Fa, Mb and Fb.
Ma and Fa have M1, M2, F1, and F2.
Ma and Fb have M3, M4, F3 and F4.
Mb and Fa have M5, M6, F5 and F6.
Mb and Fb have M7, M8, F7 and F8.
That's 16 kids.
Ma and Fa's kids have to mate with Mb and Fb's kids to ensure no incest occurs.
Mb and Fa's kids have to mate with Ma and Fb's, for the same reason.
M1 mates with F7 and F8, four kids with each gives eight kids.
M2 mates with F7 and F8, four kids with each gives eight kids.
Total 16, so far.
F1 and F2 mate each mate with M7 and M8.
16 kids from those unions giving us 32 grandkids now.
M3 and M4 each mate with F5 and F6, giving us 16 more kids, and a running total of 48 grandchildren.
F3 and F4 each mate with M5 and M6, giving us yet another batch of 16 grandkids, and a running total of 64.
How'm I doing so far?
THEN; M1 and M2 mate with Fb (Fa was their mother). This gives us eight more kids.
F1 and F2 mate with Mb (Ma was their father). Eight more.
Running total is now at 64 plus 16, is 80.
M3 and M4 mate with Fa, and F3 and F4 mate with Mb.
Running total is 80 plus 16, is 96.
M5 and M6 mate with Fb, and F5 and F6 mate with Ma.
M7 and M8 mate with Fa, and F7 and F8 mate with Ma.
Running total is now 96 plus 32, is 128.
That's 128 grandchildren, 16 children and the 4 originals. 148 people.
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