## Monday, March 28, 2016

### Solution to Spring 2016 POTW #2 (Weird? You Bet!)

See here for the original post of the puzzle.

Trial and error may work here, but I don't recommend it. Instead:

• Suppose Adam starts with \$A and Brendan with \$B
• Adam loses first game, pays out to double Brendan's money.
• New results: Adam has \$(A-B)\$ and Brendan has \$(2B).
• Brendan loses second game, pays out to double Adam's money.
• New results: Adam has \$(2(A-B))=\$(2A-2B) and Brendan has \$(2B-(A-B)) = \$(3B-A).
• By assumption: 120 = 2A-2B = 3B-A.
• Solve this system of equations to get A=150 and B=90.
Playing it out, just to check:
• Start: Adam has \$150, Brendan has \$90.
• Brendan loses. New results: Adam has \$120, Brendan has \$120.
(Here's a bonus challenge for you readers: Is there anything special about \$120? Could this puzzle be solved for any ending amount? And what if there were 3 players and each one lost a round? Try to generalize!)

## Sunday, March 27, 2016

### Spring 2016 POTW #3: Stop, Clock, and Roll

Submissions due by midnight on Sunday, April 3, 2016.

On this New Year's Day (January 1, 2016), at precisely 12:00 noon, I set up three clocks in my office. Let's call them Clock A, Clock B, and Clock C.

The next day, at precisely noon, I checked on them and discovered that Clock A had gained exactly 1 minute, Clock B had lost exactly 1 minute, and Clock C was right on time.

Assuming that the same rate of gaining/losing time continues on these clocks, When is the next date on which all three clocks will show precisely noon at the same moment? (That is, your answer should be a specific date, like "May 6, 2025" or something.)

(Use the submission box below to submit your answer. No need to explain: a correct answer will suffice for 10 points. However, feel free to explain your answer if you want a chance at some partial credit, in the event that your answer is incorrect.)

## Saturday, March 12, 2016

### Spring 2016 POTW #2: Weird? You Bet!

Submissions due by midnight on Sunday, March 27, 2016.

Two players, Adam and Brendan, play high-stakes cribbage. They both entered with a certain amount of money, and they decided that the loser of a game should double his opponent's money. (For example, if Adam (with \$10) lost to Brendan (with \$6), then Adam would pay out \$6 to double Brendan's money to \$12, leaving Adam with just \$4.)

Adam lost the first game, and then Brendan lost the second game, with each loser paying out according to the rule described above. After these two games and payouts, they miraculously discovered that both of them had exactly \$120.

How much money must each player have started with, before the two games, to end up in this situation?

(Use the submission box below to submit your answer. No need to explain: a correct answer will suffice for 10 points.)

## Monday, March 7, 2016

### Solution to Spring 2016 POTW #1 (Vinyl Buys)

See here for the original post of the puzzle.

Say x and y are the costs of the two records. I bought them for x+y=100 dollars and then sold them for 1.2x+0.8y=110 dollars. (Note the 20% markup on x and 20% discount on y.) Solve this simple system of two equations with two unknowns and you'll find x=75 and y=25.