Not for the faint hearted

Logic Level 5

I have 3 hourglasses: the first measured 13 minutes, second measured 24 minutes, third measured 41 minutes.

If I want to measure 101 minutes, what is the minimum number of times I need to turn the hourglasses?

This is a continuation of this problem and this other problem .

The answer is 7.

1 solution

Francis Kong
Nov 11, 2017

Let $T_x$ be the hourglass measured $x$ minutes.

My algorithm to get 7 flips:

Flip $T_{24}$ and $T_{41}$ (3 total up to now)
When $T_{24}$ is done, flip it again (3 total up to now)
When $T_{41}$ is done, flip $T_{13}$ (4 total up to now)
Your 101 minute starts when $T_{24}$ is done.
When $T_{13}$ is done, flip it again (5 total up to now)
When $T_{13}$ is done, flip $T_{41}$ (6 total up to now)
When $T_{41}$ is done, flip it again (7 total up to now)
Your 101 minute ends when $T_{41}$ is done.

I am not sure how to prove 6 to be impossible.