There are $128$ coins of two different weights, $64$ each. How can one always find two coins of different weights by performing no more than $7$ weightings on a regular balance?
There are $8$ coins of two different weights, $4$ each. How can one always find two coins of different weights by performing two weightings on a regular balance?
The second part is easy.
Take any 4.
Just divide into groups of 2 that is two on each balance.
If both arms weigh same
Then take any arm and weigh its constituents on a balance separately.
If again same then all the 4 from the original were of one type so take any of them and one from left over.
Else they wont weigh same and we are done.Trivially.
Else
Take one from each arm.
If they weigh different then done.
Else the weigh same.Then obviously the other coins from balances wont weigh the same.Hence done.
You have to apply the same idea for the 1st part but i am getting a bound of 6 so i think there is some mistake somewhere