Three for Three

A problem from the February 2006 issue of Crux Mathematicorum:

In a set of five positive integers, show that it’s always possible to choose three such that their sum is a multiple of 3.

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This solution is by Geneviève Lalonde. Any integer must fall into one of three groups: It’s a multiple of 3. It’s one less than a multiple of 3. It’s one more than a multiple of 3. In our set of five numbers, either it’s the case that all three of these groups are represented or it’s not. If all three are represented, then the set includes numbers of the form 3a – 1, 3b, and 3c + 1. The sum of these is 3a – 1 + 3b + 3c + 1 = 3(a + b + c), which is a multiple of 3. If not all three groups are represented, then one of the groups must contain at least three numbers from our set of five. If there are three numbers in the first group, they’ll have the form 3a, 3b, and 3c. If there are three numbers in the second group, they’ll have the form 3a – 1, 3b – 1, and 3c – 1. If there are three numbers in the third group, they’ll have the form 3a + 1, 3b + 1, and 3c + 1. In each case, the sum of the three numbers is a multiple of 3.

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June 29, 2022June 28, 2022 | Puzzles

Black and White

carpenter chess problem

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

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1. Qa8 and the queen will mate. From Chess Problems, 1886.

June 26, 2022June 25, 2022 | Puzzles

Clairvoyance

When Raymond Smullyan was teaching probability at Princeton, he told one class about the birthday paradox — the fact that if there are 23 people in a room, the chances are greater than 50 percent that at least two of them share a birthday. There were only 19 students in the classroom, so he said that the chance that two of them shared a birthday was quite small.

One boy said, “I’ll bet you a quarter that two of us here have the same birthday.”

Smullyan thought about that for a moment and said, “Oh, of course! You know the birthday of someone else here as well as your own!”

The boy said, “No, I give you my word that I don’t know the birthday of anyone here other than my own. Nevertheless I’ll bet you that there are two of us here who have the same birthday.”

Smullyan took the bet and lost. Why?

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Two of the students were identical twins. “Boy, did the class have a good laugh!” (From Smullyan’s final book, A Mixed Bag, 2016.)

June 25, 2022June 24, 2022 | Puzzles

Two Trains

Two trains set out at 7 a.m., one headed from A to B and the other from B to A. The first reaches its destination in 8 hours, the second in 12. At what hour will the two trains pass one another?

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The ratio of the trains’ speeds is 3:2, so in a given length of time they’ll cover distances in that proportion. Thus they’ll meet 3/5 of the way from A to B. The faster train will reach that point in 3/5 of 8 hours, or at 11:48 a.m. (Another approach: In each hour of travel the trains cover 1/8 and 1/12 of the distance AB, respectively, diminishing the distance between them by 1/8 + 1/12 = 5/24 of the total distance between A and B. Thus they’ll meet after 24/5 hours, or, again, 4 hours 48 minutes.)

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June 14, 2022June 13, 2022 | Puzzles

Square Meal

chomp example

In the game Chomp, two players begin with a rectangular grid. The first player chooses any square and removes it from the grid, together with all the squares above and to the right of it. The second player chooses one of the remaining squares and removes that, together with all the squares above it and to its right. The two take turns in this way until one of them is forced to remove the last square. That player loses.

If the starting grid is square, can either player force a win?

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The first player can force a win by choosing the square immediately northeast of the lower left corner square: This reduces the grid to a big L. Now the first player simply aims for symmetry: If the second player removes three squares from one leg of the L, the first removes three squares from the other. Eventually the second player will be forced to take the last square and lose the game. From Paul Halmos, Problems for Mathematicians, Young and Old, 1991. Similar: Breaking Bad.

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June 13, 2022June 12, 2022 | Puzzles

Black and White

kling chess problem

By Joseph Kling. White to mate in two moves.

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1. Rf6 forces 1. … Kd4, and then 2. Rf3 is mate. From The Chess Euclid, 1849.

June 6, 2022June 5, 2022 | Puzzles

Collared

A puzzle from R.M. Abraham’s Diversions & Pastimes, 1933:

A prisoner escapes from Dartmoor Prison and has half-an-hour’s start of two warders and a bloodhound who race after him. The warders’ speed is 4 miles per hour; the dog’s 12 miles per hour, but the prisoner can only do 3 miles per hour. The dog runs up to the prisoner and then back to the warders, and so on back and forth until the warders catch the prisoner. How far does the dog travel altogether?

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The dog travels 18 miles. The prisoner has a head start of 1.5 miles, and traveling at a relative speed of 1 mile per hour, the warders will catch him in 1.5 hours. If the dog runs at 12 mph, it will cover 18 miles in that time.

June 3, 2022June 2, 2022 | Puzzles

Mixed Emotions

A brainteaser by S. Ageyev, from the November-December 1991 issue of Quantum:

Suppose that we change the signs of 50 of these numbers such that exactly half the numbers in each row and each column get a minus sign. Prove that the sum of all the numbers in the resulting table is zero.

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Call the original table A and represent it as the sum of two tables B and C: Now if we assign minus signs correspondingly to all three tables as described in the problem, then the sum of each row in Table B and each column in Table C will be zero. It follows that the sum of all the numbers in Tables B and C combined, and hence in Table A, is zero.

May 26, 2022May 25, 2022 | Puzzles

Riddle

When Louis Philippe was deposed, why did he lose less than any of his subjects?

Because, while he lost only a crown, they lost a sovereign.

— Edith Bertha Ordway, The Handbook of Conundrums, 1915

May 23, 2022May 23, 2022 | Puzzles

Birds of a Feather

A problem from the February 2006 issue of Crux Mathematicorum:

Prove that if 10a + b is a multiple of 7 then a – 2b must be a multiple of 7 as well.

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This solution is by Eric Zhang. If 10a + b is a multiple of 7, then 10a + b = 7n, where n is an integer. Rearrange that as b = 7n – 10a and substitute it into a – 2b and you get a – 2b = a – 2(7n – 10a) = 21a – 14n, which is a multiple of 7.

May 20, 2022May 20, 2022 | Puzzles