Black and White

hermanson chess problem

By H. Hermanson, Schachvärlden, 1930. White to mate in two moves.

	Select Click for Answer
1. Qb1! and Black is doomed. If he captures the rook or advances the pawn then the queen gives mate, and if he moves the rook anywhere along the eighth rank then White opposes it with his own rook, mating with the bishop.

May 24, 2012February 4, 2013 | Puzzles

Pride of Place

If we made a dictionary of all the integers, arranging their English names in alphabetical order, which would come first and which last?

	Select Click for Answer
EIGHT would come first and ZERO last.

May 22, 2012February 4, 2013 | Puzzles

Boxing Match

boxing match

A large rectangle is partitioned into smaller rectangles, each of which has either integer height or integer width (or both). Prove that the large rectangle itself must have integer height or integer width.

	Select Click for Answer
Place the rectangle’s lower left corner at the origin of a checkerboard grid each of whose squares has side 0.5. Now, because each of the smaller rectangles has at least one integer side, each will be colored half white and half black. Consequently the large rectangle itself must be half white and half black. Now suppose that the large rectangle does not have integer width. What heights will permit it to contain equal areas of black and white? Only those with precisely an even number of checkerboard rows — the integers. From Stan Wagon, “Fourteen Proofs of a Result about Tiling a Rectangle,” American Mathematical Monthly, Vol. 94 (1987), pp. 601-617, via Peter Winkler, Mathematical Puzzles: A Connoisseur’s Collection (2004).

Click for Answer

May 21, 2012February 4, 2013 | Puzzles

“Cherchez Le Tigre!”

find the tiger

An amusement card from France, circa 1880. Where is the tiger?

	Select Click for Answer

May 20, 2012February 4, 2013 | Puzzles

Black and White

harley-watney chess problem

By B. Harley and C.G. Watney, Good Companion, 1921. White to mate in two moves.

	Select Click for Answer
1. Qc8 Kh6 2. Qh8#. (If 1. … h3 then 2. Qxh3#.)

May 18, 2012February 4, 2013 | Puzzles

One of a Kind

You’re about to roll five regular dice. Which is more likely, rolling no sixes or rolling exactly one six?

	Select Click for Answer
The chance of rolling no sixes is 5/6 × 5/6 × 5/6 × 5/6 × 5/6 The chance of rolling exactly one six is 1/6 × 5/6 × 5/6 × 5/6 × 5/6 … times 5, because the six can appear on any one of the five dice. So the probabilities are equal.

May 17, 2012February 4, 2013 | Puzzles

Sweet and Low

At age 10, Nob Yoshigahara won the grand prize in a newspaper competition with this puzzle:

Given a sack of sugar, an unbalanced scale, and two 5-pound weights, measure exactly 10 pounds of sugar.

	Select Click for Answer
Put both weights in one pan and pour sugar into the other until it balances. Then remove the weights and pour sugar into the first pan until it balances again. The first pan will contain exactly 10 pounds of sugar.

May 16, 2012February 4, 2013 | Puzzles

Leftovers

A problem from Litton Problematical Recreations, which attributes it to Fermat circa 1635:

What is the remainder upon dividing 5^999,999 by 7?

	Select Click for Answer
When successive powers of 5 are divided by 7, the remainders form a repeating series: 5¹ / 7 = 0 remainder 5 5² / 7 = 3 remainder 4 5³ / 7 = 17 remainder 6 5⁴ / 7 = 89 remainder 2 5⁵ / 7 = 446 remainder 3 5⁶ / 7 = 2232 remainder 1 5⁷ / 7 = 11160 remainder 5 5⁸ / 7 = 55803 remainder 4 5⁹ / 7 = 279017 remainder 6 5¹⁰ / 7 = 1395089 remainder 2 … The 999,999th term of the series is 6.

Click for Answer

May 14, 2012February 4, 2013 | Puzzles

Red White and Blue

I place four balls in a hat: a blue one, a white one, and two red ones. Now I draw two balls, look at them, and announce that at least one of them is red. What is the chance that the other is red?

	Select Click for Answer
It’s 1 in 5. Six possible pairings can be drawn from the hat: red 1 / red 2 red 1 / white red 2 / white red 1 / blue red 2 / blue white / blue We know that the last has not been drawn. That leaves 5 possible combinations, so the possibility of red 1 / red 2 is 1 in 5. “Many people cannot accept that the solution is not 1 in 3,” writes Erwin Brecher, “and of course it would be, if the balls had been drawn out separately and the color of the first ball announced as red before the second had been drawn out. However, as both balls had been drawn together, and then the color of one of the balls announced, then the above solution, 1 in 5, must be the correct one.”

Click for Answer