superprismatic Posted September 26, 2009 Report Share Posted September 26, 2009 There are two piles, containing between 100 and 1000 items. A is allowed to choose any two items from the first pile, and B is allowed to choose any two from the second pile. If A has only two-thirds as much choice as B, how many items are in each pile? SUPERPRISMATIC'S LAST WALTER PENNEY PUZZLE: The Den has done a spectacular job of solving all but 1 (#40: Un-halving a Message) of Walter's puzzles. I'm sure that #45 will be solved. And I'm willing to bet that you people are so tenacious that #40 will also be solved within a few weeks. I hope you enjoyed his puzzles. I'll try my best to make up a few challenging ones each month or so. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted September 26, 2009 Report Share Posted September 26, 2009 Thanks for the Penny puzzles, Super. I've really enjoyed them. I joined about halfway through, so now I'm going to check out the older ones. I'm still trying to figure #40 out - it's easy to get stuck going down a dead end. 441 C 2 = 97020 = 2/3 * 145530. 145530 = 540 C 2 ==> First pile 441, Second 540 Quote Link to comment Share on other sites More sharing options...
0 Guest Posted September 27, 2009 Report Share Posted September 27, 2009 ljb has the right answer Quote Link to comment Share on other sites More sharing options...
0 Guest Posted September 27, 2009 Report Share Posted September 27, 2009 (edited) It is a simple matter to find that, if the numbers to be found are m and n respectively, then 3(2m-1)2 = 2(2n-1)2 for which the number pair (n; m) = (540; 441) is the only possibility for 99 < n < 1001. For your interest, the number pairs (1; 1), (6; 5), (55; 45) are the relevant pairs for n < 100, while (5,341; 4,361), (52,866; 43,165), (5,180,280; 4,229,681) are the next three solutions for n >100. Edited September 27, 2009 by jerbil Quote Link to comment Share on other sites More sharing options...
0 pdqkemp Posted September 27, 2009 Report Share Posted September 27, 2009 Ok I have a request. Would someone that understands those "correct" answers please explain a little more thoroughly for those of us with wee brains, please?!? I'm serious, by the way. I love these puzzles, but I know I don't quite have what it takes to solve them 90% of the time. USUALLY, I can at least understand the answers though....but not this time. For me, if A had 200 to choose from and B had 300 to choose from (for example), then I would think that A has 2/3 as much to choose from. I know this is wrong just from seeing the answers folks have shown so far, but I'm just not sure what I'm misunderstanding. The answer given for A is NOT 2/3 of the answer given for B. So.....any "explanation for dummies" would be greatly appreciated! Quote Link to comment Share on other sites More sharing options...
0 Guest Posted September 27, 2009 Report Share Posted September 27, 2009 Ok I have a request. Would someone that understands those "correct" answers please explain a little more thoroughly for those of us with wee brains, please?!? I'm serious, by the way. I love these puzzles, but I know I don't quite have what it takes to solve them 90% of the time. USUALLY, I can at least understand the answers though....but not this time. For me, if A had 200 to choose from and B had 300 to choose from (for example), then I would think that A has 2/3 as much to choose from. I know this is wrong just from seeing the answers folks have shown so far, but I'm just not sure what I'm misunderstanding. The answer given for A is NOT 2/3 of the answer given for B. So.....any "explanation for dummies" would be greatly appreciated! Quote Link to comment Share on other sites More sharing options...
0 Guest Posted September 27, 2009 Report Share Posted September 27, 2009 Hey pd...For the 200/300 example you stated that's just number of items, we're talking about choices... A or B is only going to walk away with 2 items regardless. So what size are the piles if A doesn't have as many choices as B So his answer is: 441 objects, choosing 2 (which is 441*440/2) = 97020 50% more is 145530. so X choose 2 = 145530, (X * (X-1))/2, X = 540 http://am40sw07.blogspot.com/2007/02/probability-using-choose-formula.html I'm guessing he got it through trial and error? You could do this: 1.5(A*(A-1))=B*(B-1) 100<=A+B<=1000 but A and B can only be whole numbers...not sure how to do that on my calculator. Quote Link to comment Share on other sites More sharing options...
0 pdqkemp Posted September 28, 2009 Report Share Posted September 28, 2009 Hey pd...For the 200/300 example you stated that's just number of items, we're talking about choices... A or B is only going to walk away with 2 items regardless. So what size are the piles if A doesn't have as many choices as B So his answer is: 441 objects, choosing 2 (which is 441*440/2) = 97020 50% more is 145530. so X choose 2 = 145530, (X * (X-1))/2, X = 540 http://am40sw07.blogspot.com/2007/02/probability-using-choose-formula.html I'm guessing he got it through trial and error? You could do this: 1.5(A*(A-1))=B*(B-1) 100<=A+B<=1000 but A and B can only be whole numbers...not sure how to do that on my calculator. Thank you, ajrettke! That makes sense to me now! I wasn't thinking about A & B picking TWO items as making a difference, but now I understand. Thanks again! Quote Link to comment Share on other sites More sharing options...
Question
superprismatic
There are two piles, containing
between 100 and 1000 items. A is
allowed to choose any two items
from the first pile, and B is
allowed to choose any two from
the second pile. If A has only
two-thirds as much choice as B,
how many items are in each pile?
SUPERPRISMATIC'S LAST WALTER
PENNEY PUZZLE: The Den has done
a spectacular job of solving all
but 1 (#40: Un-halving a Message)
of Walter's puzzles. I'm sure
that #45 will be solved. And
I'm willing to bet that you
people are so tenacious that
#40 will also be solved within
a few weeks. I hope you enjoyed
his puzzles. I'll try my best
to make up a few challenging
ones each month or so.
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