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1.
Consider the following puzzle:
You have a rectangular chocolate bar, made up of n identical squares of chocolate. You can take such a bar and break it along any row or column. How many times will you have to break the bar to reduce it to n single chocolate squares?
At first, this question might seem impossible. Perhaps we meant to ask for the smallest number of breaks needed. Let’s investigate.
(a)
Suppose you started with a 1Γ—3 bar. How many breaks would you need to reduce it to single squares?
(b)
If you had a 1Γ—4 bar, how many breaks are required?
If you had 4 squares arranged in a 2Γ—2 square, your first break would require you to break the chocolate into two 1Γ—2 bars. Then each of these would require more break(s), for a total of breaks to go from the 2Γ—2 to single squares.
(c)
A 6-square bar could either be a 1Γ—6 bar, requiring breaks, or a 2Γ—3 bar.
There are two ways to proceed now.
  1. Break the bar into two 1Γ—3 bars, each requiring more breaks, for a total of breaks.
  2. Break the bar into a 1Γ—2 bar and a 2Γ—2 bar. The 1Γ—2 bar takes more break(s) and the 2Γ—2 bar takes more break(s), for a total of breaks.
(d)
Based on the above data, what should our conjecture be for the number of breaks to reduce an n-square bar to single squares, in terms of n?
It will take breaks to reduce an n-square bar to single squares.
(e)
Do we believe this? Suppose you used one break to reduce the bar into two smaller bars, with a and b squares respectively. If the conjecture is correct, how many more breaks will it take to reduce the size a bar?
How many more breaks will it take to reduce the size b bar?
How many breaks is this all together, in terms of a and b, including the initial break?
But what is a+b? We got a and b by breaking the n squares in two pieces, so a+b=. This gives us a total number of breaks as .