Jump to content
BrainDen.com - Brain Teasers
  • 0


superprismatic
 Share

Question

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.

Link to comment
Share on other sites

7 answers to this question

Recommended Posts

  • 0

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

Link to comment
Share on other sites

  • 0

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 by jerbil
Link to comment
Share on other sites

  • 0

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?!? :wacko: 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. :huh:

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! :blush:

Link to comment
Share on other sites

  • 0

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?!? :wacko: 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. :huh:

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! :blush:

Link to comment
Share on other sites

  • 0

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.

Link to comment
Share on other sites

  • 0

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! :duh: I wasn't thinking about A & B picking TWO items as making a difference, but now I understand.

Thanks again! :D

Link to comment
Share on other sites

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

Guest
Answer this question...

×   Pasted as rich text.   Paste as plain text instead

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

Loading...
 Share

  • Recently Browsing   0 members

    • No registered users viewing this page.
×
×
  • Create New...