# Gold Bar Puzzle

You’ve got someone working for you for seven days and a gold bar to pay him. The gold bar is segmented into seven connected pieces.

You must give them a piece of gold at the end of every day.

What and where are the fewest number of cuts to the bar of gold that will allow you to pay him 1/7th each day?

The fewest number of cuts will be 2.

If we cut it into 3 part where the parts are 1, 2 and 4 segments long.

Day 1 – Give 1 segment bar

Day 2 – Give 2 segment bar ask for 1 segment bar back

Day 3 – Give 1 segment bar

Day 4 – Give 4 segment bar ask for 1 + 2 back

Day 5 – Give 1 segment bar

Day 6 – Give 2 segment bar ask for 1 segment bar back

Day 7 – Give 1 segment bar

It only takes two cuts to create three segments.

I am a digital hardware engineer — Here goes my solution

i’ll require 3 bit to give id to each bar 1st

000

001

010

011

100

101

110

111

Here if you know each bit represent weights of 4, 2 and 1 i.e, 111 means 4*1+2*1+1*1 = 7

Now you know min bar length of 4, 2 and 1 we can use to represent / distribute all 7 pieces.

421

001 –> give bar1 1st day (weight1 is 1 and rest are 0)

010 –> take bar1 back (see weight1 is 0) and give bar 2 (weight2 is 1 )

011 –> give bar1 back again

100 –> take both bar1 , bar2 and give bar 4.

101 –> give bar1 again.

110 –> take bar1 back and give bar2 again.

111 –> give bar1 as well. We are done 🙂

Just 2 cuts.

First one between 1 and 2 and second between 3 and 4.

lets number them from 1 to 7

So you have 1 2-3 4-5-6-7 in this formation and connection

Day 1- Give 1

Day 2- Give 2-3 take back 1

Day 3- Give 1 also

Day 4- Take all back and give 4-5-6-7

Day 5- Give 1 with the rest

Day 6- Take back 1 and give 2-3

Day 7- Give 1.

So we need only two cuts

### Your Answer

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