# Medium Puzzles

### Sample 1

We have a group of 20 coins, one of which is a counterfeit coin.  We know that the false coin is lighter than all of the other coins.  Given a simple scale, what is the fewest number of weighings needed to determine which coins is fake?

Solution:

The minimum number of weighings needed to determine the fake coin is 3.  Consider the tree below, in which we follow a “branch” if that side of the scale was lighter than the other.  In all cases, it will take a maximum of three weighings to determine the false coin.

[diagram included in full question set]

### Sample 2

The Disappearing Dollar:

Three men went to a motel to get a room. The clerk told them that the price of the room was \$30. Each man gave the clerk \$10 to pay for the room. Later the clerk realized he had overcharged the men and gave the bellhop \$5 to give back to the men. The bellhop decided that \$5 was too hard to divide equally among the three men so he gave each man \$1 and kept the other \$2. Each man originally paid \$10 and got \$1 back so they paid \$9 each. Three times \$9 is \$27 and the bellhop kept \$2.

Where did the other dollar go?

Solution:

There really isn't a missing dollar. The riddle fools you into thinking that what the men paid plus what the bellhop received should equal \$30.

The three men paid \$27, the motel received \$25, and the bellhop received \$2. Total paid is \$27 (\$9+\$9+\$9) and the total received is \$27 (\$25+\$2). The books balance and there is no missing dollar.

It is only a coincidence that what the men paid plus what the bellhop received comes close to the original \$30 figure. Consider if the men had initially paid \$60 (\$20 each) for the room and the manager gave the bellhop \$35 to return to the men. The bellhop could have given each man \$10 back and kept \$5. Each man has now paid \$10 (\$30 total) the bellhop has \$5. \$30+\$5=\$35 and we see that now we're so far off from the original \$60 that you don't think to look for the "missing" \$25.