House of Cards

Date: 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/