Paying With a Gold Rod
The puzzle
You have hired someone for seven days and will pay them with a gold rod seven units long - one unit per day, paid at the end of each day.
The worker must hold exactly the right amount at the end of every day: one unit after day one, two after day two, and so on. They are happy to give change.
You may cut the rod twice. How?
Stuck?
One at a time, smallest first. Open only as many as you need — the whole point of the puzzle is the part you work out yourself.
Hint 1
Two cuts give three pieces. What three lengths add up to seven and can make every total from one to seven?
Hint 2
Giving change means a piece can come back to you.
Show the answer Hide the answer
Cut it into pieces of 1, 2 and 4. Then: day 1 give the 1; day 2 give the 2 and take the 1 back; day 3 give the 1; day 4 give the 4 and take back the 1 and the 2; day 5 give the 1; day 6 give the 2 and take the 1; day 7 give the 1.
How to get there
Two cuts produce three pieces, and the lengths 1, 2 and 4 are the ones that can be combined to make every total from 1 to 7 - which is the same reason three binary digits count to seven.
Day 1: hand over the 1-unit piece. The worker holds 1.
Day 2: hand over the 2 and take the 1 back. The worker holds 2.
Day 3: hand over the 1 again. They hold 1 + 2 = 3.
Day 4: hand over the 4 and take back both the 1 and the 2. They hold 4.
Days 5, 6 and 7 repeat the first three on top of the 4, giving 5, 6 and 7.
The technique worth keeping
Being able to take change back is what makes three pieces enough, and powers of two are the lengths that exploit it - every total is a sum of distinct powers of two, which is just its binary form. The same split answers "what weights let me measure any load" and a dozen questions like it.