House of CardsDate: 12/02/2001 at 07:49:12 From: Jo Smith Subject: House of cards problem Can you find a rule for working out the number of cards you need to build a house of cards of any size? For example, you would need 15 cards to build a 3-tier house, but I need a quick way to find out how many you would need for a 100-tier house. Date: 12/03/2001 at 01:28:49 From: Doctor Jeremiah Subject: Re: House of cards problem Hi Jo. It depends on how you make your house. Is this how you would make a three-tier house? /\ / \ / \ / \ / \ ------------------ /\ /\ / \ / \ / \ / \ / \ / \ / \ / \ ------------======------------ /\ /\ /\ / \ / \ / \ / \ / \ / \ / \ / \ / \ / \ / \ / \ If that is so, then you need (for the vertical cards): 1 pair of cards for the first (top) tier, 2 for the next, and 3 for the next, etc. The total of these cards is 1+2...+n-1+n, which is equal to n(n+1)/2 pairs of cards. That means there are n(n+1) cards in the vertical part. And you need (for the horizontal cards): 0 pair of cards for the first (top) tier, 1 for the next, 2 for the next, etc. The total of these cards is 0+1...+n-2+n-1. Notice that every term is 1 smaller than before. That means the sum of these cards is n(n+1)/2 - n or: n(n+1)/2 - n (n^2+n)/2 - 2n/2 (n^2+n-2n)/2 (n^2-n)/2 n(n-1)/2 So the vertical and horizontal added together is: n(n+1) + n(n-1)/2 2n(n+1)/2 + n(n-1)/2 (2n^2+2n)/2 + (n^2-n)/2 (2n^2+2n+n^2-n)/2 (3n^2+n)/2 n(3n+1)/2 And when n = 3: n(3n+1)/2 = 3(9+1)/2 = 30/2 = 15 So what would it be for n = 100? - Doctor Jeremiah, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/