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There is a large bar of Swiss chocolate laid out in an array of 6×8 squares. Typically, the bars are shared by breaking them along the ridges. If you break the bar initially on a horizontal ridge, the break is of length 6, and if you break on an initial vertical ridge, the break of of length 8. For example, you could start with a vertical break on the second ridge and get a 2 × 8 and a 4 × 8 piece, than then break the 4 × 8 piece on the 4th horizontal ridge and get two 2 × 2, and two 4 × 4 pieces. You cannot stack the pieces, you can only break one piece at a time. If you want to end up with the bar completely broken up, what is the method of breaking which has the fewest breaks.

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  • $\begingroup$ Can you give an example of two methods of breaking it up completely that have different numbers of breaks? (This is a trick question, as is the original problem.) $\endgroup$ – saulspatz Aug 28 '18 at 8:11
  • $\begingroup$ I don't have much idea about it. $\endgroup$ – Devendra Singh Rana Aug 28 '18 at 8:14
  • $\begingroup$ After the break, we have two pieces. After the second break, we have three pieces, ... $\endgroup$ – saulspatz Aug 28 '18 at 8:15
  • $\begingroup$ I got your point t think. $\endgroup$ – Devendra Singh Rana Aug 28 '18 at 8:16
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Initially, there is one piece of chocolate. When you have finished, there will be 48. Observe that each break splits a piece into two pieces, and so increases the total number of pieces by 1. To reach 48 pieces, you therefore need 47 breaks, and all methods require this number.

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  • $\begingroup$ How can you make sure that 47 is the minimum no. Of breaks. $\endgroup$ – Devendra Singh Rana Aug 28 '18 at 8:17
  • $\begingroup$ Because each break increases the total number of pieces by one, and so after one break there are two pieces, after two breaks there are three, after three breaks there are four, and so on until after 47 breaks, there are 48 pieces as desired. $\endgroup$ – Benedict Randall Shaw Aug 28 '18 at 8:19
  • $\begingroup$ @DevendraSinghRana Because you must break the bar to increase the number of pieces, and you only increase by one piece per break. As Benedict says, there is no way to get to from $1$ to $48$ pieces without $47$ breaks. $\endgroup$ – saulspatz Aug 28 '18 at 8:20
  • $\begingroup$ Oh yes you definitely right. $\endgroup$ – Devendra Singh Rana Aug 28 '18 at 8:20

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