[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.

[Guest]

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

[Guest]

ljb has the right answer

[Guest]

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

[pdqkemp]

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!

[Guest]

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!

[Guest]

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.

[pdqkemp]

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!