## A couple of good puzzles

1) Let

\[A = 1! \times 2! \times 3! \times \dots \times 119! \times 120!\]Can you cross out a factorial (\(n!\)) out of the product in \(A\) so that the remaining product (\(A/n!\)) is a square?

*Solution (hover over)*:

\begin{align} A &= 2 \times 4 \times \dots \times 120 \times (1!)^2 \times (3!)^2 \times (5!)^2 \times \dots \times (119!)^2 \\

&= 60! \times 2^{60} \times (1!)^2 \times (3!)^2 \times (5!)^2 \times \dots \times (119!)^2 \end{align} So \(A / 60!\) is a square.

2) Every day a man arrives at a station and is picked up by his wife. His wife drives from home, picks him up and drives back home with him. One day he arrived one hour earlier than usual and started walking home. He didn’t communicate to the wife that he had arrived earlier, so it didn’t affect her routine. She saw him on her way to the station, picked him up, turned around (in no time) and drove him home. They arrived 20 minutes earlier than usual. How long did the man walk for?

*Solution (hover over)*:

Let’s say the man usually arrives at 17. Since the wife drove for 20 minutes less than usual in total, she drove for 10 minutes less each way. So she picked him up at 16:50. Since the man arrived at 16, he walked for 50 minutes.

Thanks to the guys who shared these puzzles with me!