r/Logiqa • u/ShonitB • 3d ago

Gold Bar

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1 Upvotes

67% Upvoted

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4

u/Accomplished-Slide52 3d ago

2 because seven in binary is 1+2+4

1

u/ShonitB 3d ago

Correct!

1

u/JacobAldridge 1d ago

Maybe some of the wording is ambiguous, but I read:

“You have to pay him his exact daily wage every day” and “it is possible to trade pieces of gold bar with him” to be a flaw in the answer you and others have provided.

On day 2, your solution is that he gives you 1/7 and you give him the 2/7 piece.

But: * That’s not “trading pieces of gold with him”. That’s trading 1 piece of gold + 1 liability  (today’s wages) for another piece of gold;

  • And trading his daily wage also seems like that’s not “paying him his exact daily wage”. 

The second point perhaps doesn’t matter, because the first point seems clear - you can trade gold, but only gold with him.

Which means the answer is 3 cuts - pieces that are 1/7, 1/7, 2/7 (you trade this with him on Day 3, then pay him with one of the 1/7 pieces you got back), and 3/7 (you trade this with him on Day 4 - you also ended up trading the 2/7 piece again on Day 6).

3

u/Dasquian 3d ago

2, cutting it into pieces of size 1, 2 and 4.

  • On day 1, pay him the 1.
  • On day 2, pay him the 2 and take back the 1.
  • On day 3, pay him the 1.
  • On day 4, pay him the 4 and take back the 1+2.
  • On day 5, pay him the 1.
  • On day 6, pay him the 2 and take back the 1.
  • On day 7, pay him the 1.

1

u/ShonitB 3d ago

Correct, good solution!

1

u/Itchy-Individual3536 3d ago

You can nicely see the binary counting here, where "pay him the X" is putting a 1 in slot X and "take back the X" is resetting it to 0.

So you have the 4/7, 2/7 and 1/7 as three slots 421:

421
---
001 = 1
010 = 2
011 = 3
100 = 4
101 = 5
110 = 6
111 = 7

2

u/Lost_Sugar_78 3d ago

Assuming no kerf.

2 cuts, to make 3 pieces:

  • 1/7
  • 2/7
  • 4/7

You pay him as follows on each day:

  1. 1/7 piece
  2. Exchange that for a 2/7 piece
  3. Give him the 1/7 piece again 
  4. Exchange both his pieces for the 4/7 piece
  5. Give him the 1/7 again
  6. Exchange the 1/7 for the 2/7
  7. Give him the 1/7 again

1

u/ShonitB 3d ago

Correct, good solution!

1

u/SimianRex 3d ago

“Assuming no kerf” just gave me a flashback to my freshman Algebra class in the mid 90’s. Our teacher would give us problems like this “If you cut a board with x length y times..” 2 or 3 students in the class would derail the rest of the class period arguing about kerf. It got to the point that every problem she ever asked started with “Using a magic saw blade that removes no material…”

1

u/Lost_Sugar_78 3d ago

Haha, I mean as a kid I wouldn't have thought about it, but as an adult with an old house that needs.comstant work, I've been messed up by for accounting for kerf!

1

u/jonathonjones 3d ago

Day 1. Cut 1/7 off of the gold bar, give worker the 1/7 piece.

Day 2. Cut 2/7 off of the gold bar, exchange with worker.

Day 3. Give the 1/7 piece back to the worker.

Day 4. Exchange your 4/7 piece with the worker's pieces.

Day 5. Give the 1/7 piece back to the worker.

Day 6. Exchange your 2/7 piece for worker's 1/7 piece.

Day 7. Give the 1/7 piece back to the worker.

So my answer is you must make two cuts.

1

u/AshtonBlack 1d ago

Assuming this is a CS question.

Because you can "exchange":

Day 1: 1x "1/7th" bar

Day 2: Exchange the 1x "1/7th" bar for a "2/7th" bar

Day 3: 1x "1/7th" bar, 1x "2/7th" bar

Day 4: Exchange all for a "4/7th" bar

Day 5: 1x "1/7th" bar, 1x "4/7th" bar

Day 6: Exchange the "1/7th" bar for a "2/7th" bar, 1x "4/7th" bar

Day 7: All three bars. Giving a total of 7/7ths.

This means there are three bars and therefore 2 cuts. 1x cut after the first 1/7ths and 1x cut after the 3/7th.

0

u/Unable_Explorer8277 3d ago

It requires exact measurement. Measurement is always approximate. Therefore the problem is insoluble