Pill Poser

I have a bottle of 250 pills. I take 1.5 pills a day. I take out two, take one, break the other in half, take one of the halves and put the other half back in the bottle. How many days will it be until the bottle contains the same number of whole pills and half pills? How many whole pills will be there on that day? If there will never be same number of halves and wholes, how many days till halves first exceed wholes and how many wholes will there be on that day?

Edit: I will not take out three halves on any day until I have no wholes, i.e., until the wholes are gone.

Originally posted by JS357
I have a bottle of 250 pills. I take 1.5 pills a day. I take out two, take one, break the other in half, take one of the halves and put the other half back in the bottle. How many days will it be until the bottle contains the same number of whole pills and half pills? How many whole pills will be there on that day? If there will never be same number of halves ...[text shortened]... ll not take out three halves on any day until I have no wholes, i.e., until the wholes are gone.

if I have understood the problem, I reckon on the 84th day there will be 82 whole pills and 84 half pills (the day before, the 83rd, there would be 84 whole and 83 halves)

IF I have understood the problem!

Originally posted by st dominics preview
if I have understood the problem, I reckon on the 84th day there will be 82 whole pills and 84 half pills (the day before, the 83rd, there would be 84 whole and 83 halves)

IF I have understood the problem!

You got it. Count halves H up by one per day D and wholes W down by two per day D, and see when the difference is minimized. This is brute force.

Using algebra:

one any day, after dosing, H=D, W=250-2D

set H=W

then D=250-2D

subtract D

0=250-3D

3D=250

D=250/3

D=83.33...

The .33... means there is no day on which W=H.

cheers

my method exactly

one could, of course, set as a puzzle 'when both are equal' if you make the number of pills a multiple of 3 (say 300)