Words and Numbers

If you write out the numbers from 1 to 5000 in American English (e.g., THREE THOUSAND EIGHT HUNDRED SEVENTY-THREE), it turns out that only one of them has a unique number of characters. Which is it? Spaces and hyphens count as characters.

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Of the nine nonzero digits, three require three characters (1, 2, 6), three require four characters (4, 5, 9), and three require five characters (3, 7, 8). This means that the last digit of the number we’re looking for must be zero, because any other digit would have counterparts of equal length. EIGHTY-ONE, for instance, is the same length as EIGHTY-TWO and EIGHTY-SIX. (It’s true that there are some unusually named numbers between 10 and 20, but it turns out that only one of these has a unique number of characters — 17, with nine — and 42 also uses nine characters.) The tens digit must also be zero, for similar reasons: 20, 30, 80, and 90 all use six characters; and 40, 50, and 60 use five. 10 uses three characters, but this matches 1, 2, and 6. 70 uses 7 characters, but so does 15. The discussion above regarding the units digit applies also to the hundreds digit, so that too is zero, and the candidates we’re left with are 1000, 2000, 3000, 4000, and 5000. Of these, only 3000 has a unique length, with 14 characters. (Proposed in Pi Mu Epsilon Journal Spring 1985 by Harry Herein; solved in PMEJ Spring 1986 by Bob LaBarre.)

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January 4, 2016January 10, 2016 | Puzzles

“The Peace Problem”

bondarenko peace problem

In 1984, at the height of the Cold War, Ukrainian chess journalist F.S. Bondarenko dedicated this puzzle to English chess editor A.J. Roycroft in the spirit of peace and goodwill.

White’s army is arranged as above. Add Black’s army (the standard complement of 8 pieces and 8 pawns) so that no piece of either color is under attack.

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December 31, 2015January 10, 2016 | Puzzles

Quickie

From Crux Mathematicorum, May 1998:

If x is x% of y, and y is y% of z, where x, y, and z are positive real numbers, what is z?

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Since y is y% of z, z must be 100. The situation is possible if and only if y is 100 as well.

December 26, 2015January 17, 2016 | Puzzles

Podcast Episode 86: Lateral Thinking Puzzles

Here are six new lateral thinking puzzles to test your wits and stump your friends — play along with us as we try to untangle some strange situations using only yes-or-no questions.

See full show notes …

December 21, 2015October 19, 2017 | Podcast · Puzzles

All Hands on Deck?

A reader named Hamp Stevens sent this conundrum to Martin Gardner, who published it in his Mathematical Magic Show (1965). Can these 25 cards be arranged to form five poker hands, each of them a straight or better (that is, straight, flush, full house, four of a kind, straight flush, or royal flush)? If it’s possible, find the five hands; if it’s not, prove that it’s impossible.

“This ingenious puzzle is quickly solved if you go about it correctly,” Gardner wrote. “A single card is the key.”

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The key is the 4 of hearts. There is no 3 or 5 present, so the 4 of hearts can’t be part of a straight. There are only three other hearts, so it can’t be part of a flush. It’s the only 4 in the set, so it can’t be part of a full house. And the set contains no four cards of the same value, so the four of hearts can’t be the fifth card accompanying four of a kind.

December 18, 2015January 3, 2016 | Puzzles

Black and White

carpenter chess problem

By George Edward Carpenter. White to mate in two moves.

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1. Qh3! Ke4 2. Rc4# From the Dubuque Chess Journal, 1873.

December 17, 2015December 27, 2015 | Puzzles

Hard Target

In a special football game, a team scores 7 points for a touchdown and 3 points for a field goal. What’s the largest mathematically unreachable number of points that a team can score (in an infinitely long game)?

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Eleven. For any integer n greater than that, the possible remainders when n is divided by 3 are 0, 1, and 2. If the remainder is 0, then we can get to n by scoring field goals. If the remainder is 1, then n – 13 has remainder 0 and we can reach n by scoring one touchdown and some number of field goals. And if the remainder is 2, then n – 14 has remainder 0 and we can get to n by scoring two touchdowns and some number of field goals. Put simply, all integers n greater than 11 can be written in the form n = 7a + 3b for some nonnegative integers a and b.

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December 16, 2015December 27, 2015 | Puzzles

Pleased, I’m Sure

A problem by Atlantic College mathematician Paul Belcher:

Anna and Bert invite n other couples to a dinner party. Before the meal begins, some people shake hands. No one shakes hands with their own partner, no one shakes hands with themselves, and no two people shake hands with each other more than once. Afterward, Anna asks all the other 2n + 1 people how many times they shook hands, and she gets a different answer from each of them. How many times did Anna shake hands?

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Altogether there are 2n + 2 people at the party. None shakes hands with themselves or their partner, so the maximum number of handshakes that one person could have participated in is 2n, and the minimum is 0. Anna receives a different answer from each of the 2n + 1 people she asks, so the answers she gets must be 0, 1, 2, …, 2n. Let the person who shook hands 2n times be called Max and the person who shook hands 0 times be called Minie (making no assumptions about gender). Max must have shaken hands with everyone but his or her own partner. Since Max has shaken hands with Anna, Max cannot be Bert and Max’s partner must be Minie. Now remove Max and Minie and all their handshakes. This decreases by 1 the number of handshakes that each of the others participated in, and leaves us facing essentially the same problem but with n – 1 other couples. If we repeat this process, eventually we’re left with Anna and Bert as the only two people remaining. In that situation, neither of them shakes hands. And since one handshake disappeared as each of the other n couples was removed from the problem, both Anna and Bert have shaken hands n times. (Paul Belcher, “But What Colour Are Anna’s Eyes?”, Mathematical Gazette 87:509 [July 2003], 322-323.)

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December 15, 2015December 27, 2015 | Puzzles

A Christmas Challenge

Here’s a unique challenge for the holidays — one of the United Kingdom’s intelligence agencies, GCHQ, is distributing the puzzle above on its Christmas card this year. (See GCHQ’s website for details and a high-resolution grid.)

The puzzle is a nonogram: Each row and column bears a string of numbers that indicates the lengths of consecutive runs of black squares that will appear there when the grid has been completed. For example, “3 3” in the eighth row means that in the finished puzzle two shaded sections of 3 squares each will appear somewhere along its length. Some squares in the grid have already been shaded to get you started.

“By solving this first puzzle players will create an image that leads to a series of increasingly complex challenges,” notes the agency. “Once all stages have been unlocked and completed successfully, players are invited to submit their answer via a given GCHQ email address by 31 January 2016. The winner will then be drawn from all the successful entries and notified soon after.” The agency invites players to make a donation to the U.K.’s National Society for the Prevention of Cruelty to Children if they’ve enjoyed the puzzle.

(Thanks, Alex.)

02/08/2016 UPDATE: The answers have now been revealed — see the link at the bottom of this post.

December 12, 2015February 8, 2016 | Puzzles

Product Recall

A problem from the 2004 Harvard-MIT Math Tournament:

Zach chooses five numbers from the set {1, 2, 3, 4, 5, 6, 7} and tells their product to Claudia. She finds that this is not enough information to tell whether the sum of Zach’s numbers is even or odd. What is the product that Zach tells Claudia?

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420. Giving the product of the five chosen numbers is equivalent to giving the product of the two unchosen numbers. The only possible products that are produced by more than more than one pair of numbers are 12 (produced by {3, 4} and {2, 6}) and 6 (produced by {1, 6} and {2, 3}). In the second case, the sum of both pairs is odd, so the sum of the five chosen numbers would have to be odd too. So the first must be the case, and product of Zach’s five chosen numbers is $\displaystyle \frac{1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7}{12}=420.$ (From Titu Andreescu et al., 104 Number Theory Problems, 2007.)

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December 8, 2015December 20, 2015 | Puzzles · Science & Math