Bar of Gold

[Tiberius]

Here is a probably easier puzzle than the last, this time involving a bar of gold.

You've got someone working for you for seven days and a gold bar to pay them. The gold bar is segmented into seven connected pieces. You must give them a piece of gold at the end of every day. If you are only allowed to make two breaks in the gold bar, how do you pay your worker?

My guess is computer literate people might get this before others.

Solution below:

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The question is solved by constructing a binary counter using the gold and the action of giving / receiving gold from the worker. You divide the bar into a segment of 1, 2, and 4. You then distribute the gold like so:

1st day: Give worker the 1 gold segment.
2nd day: Give worker the 2 gold segment, take the 1 gold segment back.
3rd day: Give worker the 1 gold segment.
4th day: Give worker the 4 gold segment, take 1 and 2 gold segments back.
5th day: Give worker the 1 gold segment.
6th day: Give worker the 2 gold segment, take the 1 gold segment back.
7th day: Give worker the 1 gold segment.

Its a simple binary clock, with bars of gold representing the powers of 2 that binary uses. To put it in a more technical way:

Let 000 be the clock at day 0. The right-most position represents 2^0 (1), the postion to the left of that represents 2^1 (2), and the position to the far left represents 2^2 (4).

With 000, none of the representations are filled.
With 001, the (1) representation is filled, and the total value is 1.
With 010, the (2) representation is filled, and the (1) isn't anymore, so the total value is 2.
With 011, both the (2) and the (1) representations are filled, so the total is 2+1 = 3.

etc...etc...

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[Eilonnwy]

Throw it on the ground really hard.

[Tiberius]

...and make more than two breaks in the bar? Not allowed I'm afraid...but amusing solution nonetheless

[Dotard]

Would this involve having the worker break the bar? Like if you broke it once with 2 pieces still attached, hand it to the worker and he breaks that in two and hands one back?

Some trickery like that?

[Zhalentine]

Here is a probably easier puzzle than the last, this time involving a bar of gold.

You've got someone working for you for seven days and a gold bar to pay them. The gold bar is segmented into seven connected pieces. You must give them a piece of gold at the end of every day. If you are only allowed to make two breaks in the gold bar, how do you pay your worker?

My guess is computer literate people might get this before others.

I'm not sure what counts as a break, so this may not be legal.

A-B-C-D-E-F-G

first you rip out D and you are left with A-B-C and E-F-G and then you put them on top of each other

A-B-C
E-F-G

then you remove B and F

[Tiberius]

Would this involve having the worker break the bar? Like if you broke it once with 2 pieces still attached, hand it to the worker and he breaks that in two and hands one back?

Some trickery like that?

No. To put it another way, the bar can only be broken into three pieces...ever. You have to work out a payment plan over 7 days so that you give him the same rate of gold a day.

I'm not sure what counts as a break, so this may not be legal.

A-B-C-D-E-F-G

first you rip out D and you are left with A-B-C and E-F-G and then you put them on top of each other

A-B-C
E-F-G

then you remove B and F

No. That would involve two breaks for D's removal (breaking CD and DE connections), and then another 2 for B, and another 2 for F.

You are only allowed 2 breaks (not "removals").

[leo-rcc]

1 part of 1 = A
1 part of 2 = B
1 part of 4 = C

day 1, give part A
Day 2, give part B, get part A back
Day 3, give part A
Day 4, give part C, get part A and B back
Day 5, Give part A
Day 6, Give part B, get part A back
Day 7, give part A.

[Tiberius]

I posted the solution and an explanation in the original post (hidden). leo has it spot on though

[Dotard]

Foul.

I asked about trickery and example may have said have him break it, which would be incorrect. However giving him some gold one day and having him give some back qualifys as trickery.

It's a trick question. Like saying you have three $20 bills and you have to give him same pay everyday from those $20s without exchanging them for smaller denominations.

Then to answer give him a $20 bill and have him give you change.

Basicly it is what I gathered from the "solution". Just get your 'change' back every other day.

It's a disingenuous problem.

[Tiberius]

Him breaking it would have been against the rule about "only 2 breaks". The only other rule was that every day you had to give him some gold.

It isn't a trick, it's called lateral thinking. It becomes obvious when you realise you have only 3 lots of gold and 7 lots to fit them into. You can't split 3 across 7 without breaking it down, so you have to think a bit more. Given that the only two people in this scenario are you and him, passing gold back and forth as "change" becomes logical, and you only need to figure out which breaks to make.

It's a bit like the fox, chicken, and wheat problem. A farmer needs to get them across the river, but the boat can only take him and one of the aforementioned items. Left alone, the fox wil eat the chicken, and the chicken will eat the wheat. The answer to the riddle involves taking items across and taking them back, so that you can transfer all items across without anyone getting eaten. Nowhere in the riddle does it say you can take things back across once they are on the other side, but it doesn't rule it out either.

No need to throw a tantrum when you can't figure out the answer.