Togetherness

distance sums problem

What location on this line segment has the smallest sum of distances to the labeled points?

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Consider first the endpoints, A and K: The sum of the distances to these two points will be the same for any point between them. Similarly, any point between B and J will have the same total distance to points A, K, B, and J. Proceeding in this way and taking the points in pairs, we find that every point between E and G produces the same total distance to the outermost 10 labeled points. That leaves only F. So the answer is that point F has the minimum total distance to the labeled points. Strangely, this means that we can shift the points around without affecting this result. If points A through E crowded F closely on the left, and point K were 90 light-years away on the right, point F would still yield the smallest total sum, so long as the points appeared in the order shown above.

June 30, 2012February 5, 2013 | Puzzles

Black and White

By Herbert I. Ladd. In A Selection of Popular Two-Move Chess Problems (1887), Orestes Augustus Brownson calls this “a gem of purest ray serene.” White to mate in two moves.

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1. Ng6 and Black must lose by 2. Qd3#, 2. Qe5#, or 2. Qd6#.

June 29, 2012February 5, 2013 | Puzzles

Club Med

My doctor wants to establish a dosage for a new drug, so he gives me a bottle of 48 pills and tells me to take them throughout the month of June. I can take as many or as few as I like on any given day, so long as I take at least 1 pill each day. Show that there’s a sequence of consecutive days during which I take exactly 11 pills.

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We can solve this problem by an application of the pigeonhole principle. Get a calendar for the month of June and on each day record the running total number of pills I’ve taken so far in the month. So June 1 might read “1,” June 2 “5,” and so on. The entry for June 30 will read “48.” Write these numbers in red. Now revisit each date, add 11 to the red number, and enter the result in blue. In the example above June 1 now reads “1 12,” June 2 “5 16,” and so on. June 30 reads “48 59.” Now we’ve filled the calendar with 60 numbers, all of which fall in the range 1-59. This means that at least one number falls on two different days. In the interval between these two days I took 11 pills. By similar arguments it can be shown that intervals must exist during which I took precisely 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10 pills. Adapted from Kenneth R. Rebman, “The Pigeonhole Principle (What It Is, How It Works, and How It Applies to Map Coloring),” The Two-Year College Mathematics Journal, January 1979.

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June 28, 2012February 5, 2013 | Puzzles

Readership

Assume that these statements are true:

Not all Futility Closet readers are U.S. residents.
All walrus veterinarians who are not U.S. residents are not Futility Closet readers.

Do 1 and 2 imply that

Not all Futility Closet readers are walrus veterinarians?

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Yes. Statement 1 says that there are Futility Closet readers who are not U.S. residents. Let A be one of them. We’re claiming that A is not a walrus veterinarian, which would imply that statement 3 is true. Suppose the contrary, that A is a walrus veterinarian. Since he’s not a U.S. resident, statement 2 implies that he’s not a Futility Closet reader, which isn’t true. Adapted from Svetoslav Savchev, Mathematical Miniatures, 2003.

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June 27, 2012February 5, 2013 | Puzzles

Ones in a Million

Of the integers from 1 to 1,000,000, which are more numerous: the numbers that contain a 1 or those that don’t?

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To list the numbers that don’t contain a 1, imagine six spaces and fill each with the digits 0, 2, 3, 4, 5, 6, 7, 8, or 9. The number of ways of doing this is 9⁶. There’s one exception: 000000 doesn’t fall between 1 and 1,000,000. So the number of integers without 1 is 9⁶ – 1 = 531,440, and the number with 1 is 468,560.

June 25, 2012February 5, 2013 | Puzzles

The Figure 8 Puzzle

figure 8 puzzle

Can this loop of string be freed from its wire? Stewart Coffin, who devised the puzzle in 1974, writes, “I soon became convinced that this was impossible, but being a novice in the field of topology, I was at a loss for any sort of formal proof.” He published the challenge in a newsletter and has been receiving requests for a solution ever since. Adding to the confusion, in 1976 a British puzzle editor mistakenly claimed with that Coffin’s creation was equivalent to another puzzle with a known solution, and Pieter van Delft and Jack Botermans published an amusingly bewildering “solution” of their own in their 1978 book Creative Puzzles of the World.

In the meantime, fans around the world have continued to experiment, and mathematicians Inta Bertuccioni and Paul Melvin have both offered proofs that the puzzle is unsolvable. “Whoever would have guessed that this little bent piece of scrap wire and loop of string would launch itself on an odyssey that would carry it around the world?” Coffin writes. “Will it mischievously rise again, perhaps disguised in another form, as topological puzzles so often do?”

June 23, 2012June 22, 2012 | Puzzles · Science & Math

Black and White

kubbel chess problem

By L.I. Kubbel, 1941. White to mate in two moves.

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The quiet 1. Rg1 forces 1. … Kh5, permitting 2. g4#.

June 22, 2012February 5, 2013 | Puzzles

Odd and Even

A puzzle by Noboyuki Yoshigahara:

“An odd number plus an odd number makes an even number. An even number plus an odd number makes an odd number. An even number plus an even number is an even number. Right?”

“Yes.”

“An odd number times an odd number is an odd number. An even number times an odd number is an even number. Right?”

“Sure.”

“An even number times an even number is an odd number. Right?”

“Huh?”

“You don’t think so? An even number times an even number is an odd number.”

“Why?”

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AN ODD NUMBER PLUS AN ODD NUMBER has 26 characters, etc. Yoshigahara notes, “This puzzle works in Japanese as well.”

June 21, 2012February 5, 2013 | Puzzles

Circles and Squares

circles and squares 1

Here are three circles and two squares, inscribed successively as shown.

If the diameter of the largest circle is 10, what is the diameter of the smallest circle?

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There are a number of ways to solve this. The simplest is to rotate the smaller square 45 degrees: Now the radius of the largest circle (red), which we know to be 5, equals the side of the smaller square (green), because they’re the two diagonals of a square. So the diameter of the smallest circle is 5.

June 19, 2012February 5, 2013 | Puzzles

Heads of State

Ten senators are about to enter Congress when a barrage of snowballs knocks off their tophats. Each retrieves a hat at random. What is the probability that exactly nine of them receive their own hats?

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Zero. If nine receive their own hats, then the tenth must as well.

June 15, 2012February 5, 2013 | Puzzles