Cache and Carry

USC mathematician Solomon W. Golomb offered this problem in the Pi Mu Epsilon Journal, Fall 1971 (page 241):

Ted: I have two numbers x and y, where x + y = z. The sum of the digits of x is 43 and the sum of the digits of y is 68. Can you tell me the sum of the digits of z?

Fred: I need more information. When you added x and y how many times did you have to carry?

Ted: Let’s see. … It was five times.

Fred: Then the sum of the digits of z is 66.

Ted: That’s right! How did you know?

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Solution by Robert C. Gebhardt (from the Fall 1972 issue, page 349): If no carrying were involved, we could get the answer by just adding the two digit sums together, 43 + 68 = 111. But each carry increases the final digit sum by only 1 rather than the 10 that would otherwise result. So each carry reduces our expected final digit sum by 9. In this case there are five carries, so the digit sum of the total is 43 + 68 – (5 × 9) = 66.

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October 15, 2018October 14, 2018 | Puzzles

Black and White

By Rudolf L’Hermet. White to mate in two moves.

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1. Qa6! and Black can choose among four different mates: If he moves his knight, then 2. Qe6#. If he takes the rook, then 2. Qc4#. If he advances his pawn one square, then 2. Qd3#. If he advances his pawn two squares, then 2. Nf6#. From Benjamin Glover Laws, The Two-Move Chess Problem, 1890.

October 11, 2018 | Puzzles

The Hexagonal Tortoise Problem

In the 17th century, Korean aristocrat Choi Seok-jeong proposed a puzzle inspired by the pattern on a tortoise shell: Can you assign the numbers 1 to 30 to the vertices in this diagram so that each hexagon bears the same sum?

https://commons.wikimedia.org/wiki/File:Hexagonal-tortoise-problem.png — Image: Wikimedia Commons

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Image: Wikimedia Commons

October 10, 2018 | Puzzles

Common Sense

A quickie from Raymond Smullyan: On the Island of Knights and Knaves, knights always tell the truth and knaves always lie. Every inhabitant is either a knight or a knave. One day a visiting anthropologist comes across a native and recalls that his name is either Paul or Saul, but he can’t remember which. He asks him his name, and the native replies “Saul.”

From this we can’t know whether the native is a knight or a knave, but we can tell with high probability. How?

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If the native’s name isn’t Saul, it’s unlikely that he’d happen to choose that name for a lie. So he’s probably telling the truth, and is likely a knight (named Saul). Obligatory John Finnemore sketch:

October 7, 2018 | Puzzles

The Red Ball

An urn contains k black balls and one red ball. Peter and Paula are going to take turns drawing balls from the urn (without replacement), and whoever draws the red ball wins. Peter offers Paula the option to draw first. Should she take it? There seem to be arguments either way. If she draws first she might get the red ball straightaway, and it seems a shame to give up that opportunity. On the other hand, if she doesn’t succeed immediately then she’s only increased Peter’s chances of drawing the red ball himself. What should she do?

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Imagine that Peter and Paula simply take turns drawing balls, without bothering to inspect their color, until the urn is exhausted. Then afterward they look to see who has the red ball. The winner in this procedure will always be the same as in the original, so each player’s chance of winning will be the same as in the original game. If k is odd, then the total number of balls is even, and when the urn is empty each player will have the same number of balls, so each of them has a 1/2 chance of winning. But when k is even, then the total number of balls is odd, and the player who draws first will have an extra ball when the drawing is done. So, generally, Paula should accept Peter’s offer to draw first — it may help her, and at worst it leaves her chances unchanged. (From Wolfgang Schwarz, 40 Puzzles and Problems in Probability and Mathematical Statistics, 2008.)

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October 6, 2018October 9, 2018 | Puzzles

Line Limit

You own a goat and a meadow. The meadow is in the shape of an equilateral triangle each side of which is 100 meters long. The goat is tied to a post at one corner of the meadow. How long should you make the tether in order to give the goat access to exactly half the meadow?

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Draw a hexagon whose sides are 100 meters long. Now draw a circle, with the same center, whose radius r is the desired length of the tether: The difference in area between the hexagon and the circle is 6 times half the area of the meadow. The area of the circle is πr². The area of an equilateral triangle with side length a is $\displaystyle \frac{a^{2}\sqrt{3}}{4}$ . The area of a hexagon with side length a is $\displaystyle \frac{6a^{2}\sqrt{3}}{4}$ . For a = 100 the area of the hexagon is $\displaystyle 15000\sqrt{3}$ . If we want a circle half that size, then $\displaystyle \pi r^{2} = 7500\sqrt{3}$ and $\displaystyle r = \sqrt{\frac{7500\sqrt{3}}{\pi }} \approx 64.3037.$ (From Ian Stewart’s Cabinet of Mathematical Curiosities, 2009.)

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September 23, 2018September 24, 2018 | Puzzles

Black and White

I just ran across this in Benjamin Glover Laws’ The Two-Move Chess Problem, from 1890. It’s by G. Chocholous. White is to mate in two moves.

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If White withdraws his light-squared bishop along the b1-h7 diagonal, then Black is facing two mate threats: White can move his king off the a-file, discovering mate by the a8 rook, or he can move his dark-squared bishop off the first rank, discovering mate by the h1 queen. A few sample lines: 1. Bf5 Rxa8+ 2. Ba7# 1. Bf5 Rxh1 2. Kb3# 1. Bf5 Rh4 [ready to block a rook check or take the queen] 2. Bh2# 1. Bf5 Rd8 [ready to block a queen check or take the rook] Bd4# (If 1. Bf5 c2 then 1. Bd4#!) At first it looks as though Black is doomed, but he does have one resource. If after e.g. 1. Bf5 he plays 1. … Rg8 then White has no way to mate on the next move: If the white king discovers check, Black can simply take the rook, and if the g1 bishop discovers check by the queen, the rook can interpose. The answer is 1. Bg6!, the only initial bishop move that blocks the rook’s route to g1 and keeps both mate threats alive.

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September 20, 2018 | Puzzles

Cube Route

Created by Franz Armbruster in 1967, “Instant Insanity” was the Rubik’s Cube of its day, a simple configuration task with a dismaying number of combinations. You’re given four cubes whose faces are colored red, blue, green, and yellow:

https://commons.wikimedia.org/wiki/File:Instant_insanity_Cube.png — Image: Wikimedia Commons

The task is to arrange them into a stack so that each of the four colors appears on each side of the stack. This is difficult to achieve by trial and error, as the cubes can be arranged in 41,472 ways, and only 8 of these give a valid solution.

One approach is to use graph theory — draw points of the four face colors and connect them to show which pairs of colors fall on opposite faces of each cube:

https://commons.wikimedia.org/wiki/File:Instant_sanity_graph.png — Image: Wikimedia Commons

Then, using certain criteria (explained here), we can derive two directed subgraphs that describe the solution:

https://commons.wikimedia.org/wiki/File:Instant_insanity_final.png — Image: Wikimedia Commons

The first graph shows which colors appear on the front and back of each cube, the second which colors appear on the left and right. Each arrow represents one of the four cubes and the position of each of the two colors it indicates. So, for example, the black arrow at the top of the first graph indicates that the first cube will have yellow on the front face and blue on the rear.

This solution isn’t unique, of course — once you’ve compiled a winning stack you can rotate it or rearrange the order of the cubes without affecting its validity. B.L. Schwartz gives an alternative method, through inspection of a table, as well as tips for solving by trial and error using physical cubes, in “An Improved Solution to ‘Instant Insanity,'” Mathematics Magazine 43:1 (January 1970), 20-23.

September 18, 2018September 17, 2018 | Puzzles · Science & Math

Self-Study

For a puzzlers’ party in 1993, University of Wisconsin mathematician Jim Propp devised a “self-referential aptitude test,” a multiple-choice test in which each question except the last refers to the test itself:

1. The first question whose answer is B is question

(A) 1
(B) 2
(C) 3
(D) 4
(E) 5

2. The only two consecutive questions with identical answers are questions

(A) 6 and 7
(B) 7 and 8
(C) 8 and 9
(D) 9 and 10
(E) 10 and 11

3. The number of questions with the answer E is

(A) 0
(B) 1
(C) 2
(D) 3
(E) 4

The full 20-question test is here, the solution is here, and an interesting collection of solving routes is here.

(Jim Propp, “Self-Referential Aptitude Test,” Math Horizons 12:3 [February 2005], 35.)

September 12, 2018September 11, 2018 | Puzzles

Forefathers

A problem from the 1996 mathematical olympiad of the Republic of Moldova:

Twenty children attend a rural elementary school. Every two children have a grandfather in common. Prove that some grandfather has not less than 14 grandchildren in this school.

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Consider the grandfather with the largest number of grandchildren, and suppose, for a contradiction, that that number is less than 14. There’s a child at the school who isn’t his grandchild, and each of his grandchildren shares a grandfather with that child. They can’t all share the same grandfather, because then that man would surpass our “maximal” grandfather. So there’s some third grandfather to divide that duty with him. Now, every student we haven’t spoken about shares a grandfather with one that we have. And that leaves room for no fourth grandfather; every student in the school must count some two of these men as his grandfathers. We have 40 grandfatherings to assign to these 20 students, and 40 = 3 × 13 + 1, so try as we might, we can’t assign 13 or fewer to each of the three men. One of them must have at least 14 grandchildren. (This is all laid out more rigorously at the link below, page 371.) (Olympiad Corner, Crux Mathematicorum 27:6 [October 2001], 357-373.)

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September 11, 2018September 10, 2018 | Puzzles