What is the product of this series?

(*x* – *a*) (*x* – *b*) (*x* – *c*) … (*x* – *z*)

This yields to an insight, so I’ll withhold the answer.

What is the product of this series?

(*x* – *a*) (*x* – *b*) (*x* – *c*) … (*x* – *z*)

This yields to an insight, so I’ll withhold the answer.

From J. Newton Friend, *Numbers: Fun & Facts*, 1954:

A boy jumped onto one end of a piece of tree trunk lying on the top of a hill. Now the log happened to be exactly 13 feet long, an unlucky omen for the youth, and the impact caused it to begin rolling down the hill. As it rolled, he managed to keep himself upright on top and slowly walked across the log to the other end which he reached just as the log came to rest at the bottom of the hill, 84 feet from where it began to roll.

The log was 2 feet in diameter. How far did the boy actually travel and how far would he have travelled had the log been 3 feet in diameter?

From a 1987 Hungarian math contest for 11-year-olds:

How can a 3 × 3 × 3 cube be divided into 20 cubes (not necessarily the same size)?

“I used to flatter myself that I would immediately be able to see through any problem that might be asked of an 11-year-old,” writes University of Waterloo mathematician Ross Honsberger. “I don’t take anything for granted anymore!”

Point A is the center of one square and the vertex of another. The side of each square is 4 inches. What is the area of the shaded region?

By Sam Loyd. White to mate in two moves.

Here’s a long corridor with a moving walkway. Let’s race to the far end and back. We’ll both run at the same speed, but you run on the floor and I’ll run on the walkway, going “downstream” to the far end and “upstream” back to this point. Who will win?

This puzzle, by Les Marvin and Sherry Nolan, appeared in the *Journal of Recreational Mathematics* in 1977. “White to play in the adjoining diagram. If both players play optimally, will White win, lose, or draw?”

I don’t believe *JRM* ever published the solution. My stab: Either king is vulnerable to a check from the bishop file, and White will win a straight race. So I think Black must play defense. But if White attacks c7 with both knights and Black defends it doubly, then White can simply trade off all four knights (1. Nc7+ Nxc7 2. Nxc7+ Nxc7 bxc7) and the pawn will queen. So I think White wins.

This isn’t a very “mathematical” solution, but I can’t find a reliable alternative involving the parity of the knights’ moves, which seems to be what’s expected. Any ideas?

06/06/2014 UPDATE: A reader ran this position through a couple of strong chess engines and finds that it’s likely a draw — here’s one example:

[Event “?”]

[Site “?”]

[Date “????.??.??”]

[Round “?”]

[White “?”]

[Black “?”]

[Result “*”]

[FEN “k6n/Pp4n1/1P6/8/8/6p1/1N4Pp/N6K w – – 0 1”]

1.Nd1 Nf7 2.Nc2 Ne5 3.Nce3 Nd7 4.Nd5 Nxb6 5.Nxb6+ Kxa7 6.Nc8+ Ka6 7.Ne3 b5 8.Nd6 b4 9.Ne4 Nh5 10.Nc2 Kb5 11.Nxb4 Kxb4 12.Nxg3 Nxg3+ 13.Kxh2 Nf1+ 14.Kh3 Ne3 15.g4 Kb3 16.g5 Nd5 17.g6 Nf4+ 18.Kg3 Nxg6

There doesn’t seem to be a sure way for either side to reach a win. I suspect that Marvin and Nolan thought otherwise, but they were writing in 1977, without the benefit of computer analysis. Without a published solution, we can’t be sure.

(Thanks, Emilio.)

Here is a curious problem. We may safely assume that you had two parents; each of your parents had two parents, so that you had four grandparents. Arguing along similar lines you must have had eight great grandparents and so on. Assuming an average of three generations per century the number of your ancestors since the Christian Era began must have been nearly 1 trillion–

1,000,000,000,000,000,000 or 10

^{18}This is vastly more people than have ever lived on the Earth. What can we do about it?

— J. Newton Friend, *Numbers: Fun & Facts*, 1954

This one is slippery, so watch it closely.

A poor old lady, with little money and plenty of time, sat quietly one day trying to devise a plan for making a little change. She finally came up with a very clever idea. Taking an old necklace, which she knew was worth only $4, she went to a pawnshop and pawned it for $3. Then, on a street corner, she started a friendly acquaintance with a young man, finally persuading him to buy the pawnticket for only $2. Now, she had $5 altogether and thus had made $1 profit. The pawnbroker wasn’t out any money since he paid only $3 for a $4 item, and the young man paid only $2 to get the $4 necklace. Who lost?

— Raymond F. Lausmann, *Fun With Figures*, 1965

By Sam Loyd. White to mate in two moves.