## Search This Blog

### Gold Bar Fewest Cut Puzzle

Gold Bar Fewest Cut Puzzle - 11 August

A worker is to perform work for you for seven straight days. In return for his work, you will pay him 1/7th of a bar of gold per day. The worker requires a daily payment of 1/7th of the bar of gold. 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?

1. 2 cuts would suffice.

2 cuts should be made such that we have 3 pieces of gold bar. 1 bar, 2 piece bars and 4 piece bars.

Day 1: Give worker 1 bar
Day 2: Take back 1 bar and give 2 piece bar
Day 3: Give 1 bar [Making 2 + 1 = 3]
Day 4: Take back 1 & 2 piece bar and give 4 piece bar
Day 5: Give 1 piece bar [Making 4 + 1 = 5]
Day 6: Take back 1 piece and give 2 piece
[Making 5 - 1 + 2 = 6]
Day 7: Give 1 piece bar

The above is based on the assumption that worker would use the gold bar given to them only after the end of 7th day of work.

2. good! @Gaurav! it was superb

3. i would give him the equivalent market price of 1/7 of a bar of gold in cash as to not ruin my bars of gold = 0 cuts :)

4. melt the bar and pour him liquid

5. I disagree with the answer being 2. surely the fact he wants to be paid 1/7th a day suggests he is going to spend it, otherwise why wouldn't he just wait for the week to finish?

Based on this, the answer is 0 cuts, the Man just trades his 1 solid gold bar at the bank for 7 smaller gold bars.

1. What if the worker wants to leave your job after 3 days.

6. It's defn a minimum of 2 cuts, those of you that disagree are jokers and don't fully understand the essence of maths related problems. I know there is a possibility of the worker not bringing back the 1/7 piece of gold, and also the employer could use other methods of payment. In problem solving you have to try and make no additional assumptions, just deal with the problem and that's it.
Some of my students asked me,

What if he was ill on some days?

One student said, zero cuts because he could die in a car accident...

Etc... Assumptions are endless!

We need to look at the heart of the problem, what is it trying to introduce?

It introduces the numbers,

1, 2, 4, 8, 16, 32, 64, 128... etc

Using the above numbers you can make any number you wish.
Example: 39 = 1 + 2 + 4 + 32

YOU TRY SOME... This is a series of lessons for any KS3 group, links into how to write Binary Numbers, electronic and gates ( AND, OR, NOT gates, etc...)

This is a beautiful problem, embrace it :-)

1. but aren't riddles supposed to test more than just mathematical smarts? by thinking outside the box? if we only ever taught mathematical intelligence by itself, people would then struggle to apply it to real world problems.

Just saying, in the real world this would be the LEAST smart solution to the one bar of gold problem.

7. Exactly...great problem and an intelligent answer! like it :)

8. heyyyyyyoooo

9. need 6 cuts only

10. bwaaaaaaaah.

11. It's simple every day he will give the worker 1/7 per day so by the end of the week the worker would get 7/7 so it is like this:
1/7 x7/1 =7/7

12. |-------| <---gold bar
|-| |--| |--| |--| each - is a 1/7
day one give him |-|
day 2 take back |-| and give one of the |--|
day 3 give back |-| he now have |--||-|
day 4 take back |-| give second |--| for total of |--||--|
day 5 give |-| for total |--||--||-|
day 6 take back|-| and give last |--| now there is |--||--||--|
day 7 give him all of the bar left and your done

1. Dear 2 Cuts is enough.
|-------| into |-|,|--|,|----| like.

13. yes only two cuts

14. I think it would take at least three cuts.

Before any cutting, the bar of gold is in one piece of 7/7.

cut 1. After cutting off 3/7, the bar is in two pieces that are 4/7 and 3/7.

cut 2. Then cut both pieces at the same time making the 3/7 piece a 1/7 piece and 2/7 and the 4/7 piece into two 2/7 pieces. Now you have three 2/7 pieces and one 1/7.

cut 3. Finally, cut the reminding 2/7 pieces in two at the same time making six 1/7 pieces and adding the 1/7 from cut two, you have seven 1/7 pieces.

15. Presuming that the worker is going to spend his 1/7 of a gold bar each evening (thereby invalidating the 2 cut idea), 4 cuts would suffice:

Cut 1/7 off the end of the bar
Cut the remainder into 6 equal pieces equalling 1/7, which can be done in three cuts - 1 horizontal and 2 vertical (and vice versa)

I don't mean to have a go at those who like the 2 cut idea - it's very clever and very mathematically sound, just not very logical from a realistic viewpoint of events in that if the worker did not spend his gold each evening you could simply give him the whole bar at the end of the week, 0 cuts.