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

Readouts

sallows readouts puzzle

A puzzle by Lee Sallows. In this readout from a computer-driven electronic display, the digits in the fifth row have been obscured. What are they?

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They’re the same as the numbers in the first row. Added colors make it easier to see that each number in row n counts how many segments of each type (color) occur in row n – 1. Remarkably, the loop closes after four cycles. (Thanks, Lee.)

May 16, 2022May 15, 2022 | Puzzles

Staffing

A problem from Crux Mathematicorum, April 2006:

A group of people must be formed into committees. Show that the number of possible committees that can be formed with an odd number of members is the same as the number that can be formed with an even number of members. (Assume that a committee with no members and one that includes everyone are both allowed.)

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This solution is by Geneviève Lalonde. Call a committee an even committee if it has an even number of members and an odd committee if it has an odd number of members. If the total number of people in the group is odd, then for each even committee the total number of remaining people is odd (and thus constitutes an odd committee). And vice versa: For each odd committee, the number of people who aren’t in that committee is even and constitutes an even committee. So there’s a one-to-one correspondence between the even and odd committees, and their numbers must be equal. If the total number of people in the group is even, then consider one person (call him John). For each even committee C that doesn’t contain John, there’s a corresponding odd committee made up of everyone who’s not in C except for John. And for each even committee C that contains John, there’s a corresponding odd committee made up of John plus everyone who’s not in C. So again there’s a one-to-one correspondence between the even and odd committees, and we conclude that their numbers are equal.

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May 5, 2022May 4, 2022 | Puzzles

Black and White

taylor chess problem

By J. Paul Taylor. White to mate in two moves.

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1. Be4! and mate must follow. From Elementary Chess Problems, 1880.

April 29, 2022April 28, 2022 | Puzzles

Can Do

A puzzle by Sam Loyd:

John the milkman has two 10-gallon cans full of milk. Two customers have a 5- and a 4-quart measure and want 2 quarts put into each measure. How can he accomplish this?

“It is a juggling trick pure and simple, devoid of trick or device, but it calls for much cleverness to get two exact quarts of milk into those measures employing no receptacles of any kind except the two measures and the two full cans.”

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Call one of the 10-gallon cans A and the other B. Fill the 5-quart pail from can A. Pour the 5-quart pail into the 4-quart pail. Empty the 4-quart pail into can A. Pour the 5-quart pail into the 4-quart pail. Fill the 5-quart pail from can A. Fill the 4-quart pail from the 5-quart pail. Empty the 4-quart pail into can A. Fill the 4-quart pail from can B. Pour the 4-quart pail into can A. This fills can A and leaves 2 quarts in the 4-quart pail. “Thus the milkman has supplied each of his customers with exactly two quarts of milk and solved his perplexing problem.” From Loyd’s Cyclopedia of Puzzles, 1914. See also Well Done.

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April 28, 2022April 27, 2022 | Puzzles

Perspectives

eureka elevation puzzle

A puzzle from Eureka, March 1957:

These drawings depict an object as seen from above and from the front. What does it look like when viewed from the side? The object has no curved surfaces, and all hidden edges in such drawings are represented by dotted lines.

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(From the Eureka Digital Archive, published under a Creative Commons license.)

April 22, 2022April 21, 2022 | Puzzles

An Observant Anthropologist

A puzzle from the 1998 Moscow Mathematical Olympiad, via Peter Winkler’s excellent Mathematical Puzzles, 2021:

An anthropologist is surrounded by a circle of natives. Each native either always lies or always tells the truth. The anthropologist asks each native whether the native to his right is a liar or a truth teller. From their answers, she’s able to deduce the fraction of the circle who are liars. What is the fraction?

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The key is to notice that if all the liars were changed to truth-tellers and vice versa, none of the answers would change. So if the fraction x can be determined, it must have the property that x = 1 – x, and x must be 1/2.

April 20, 2022April 19, 2022 | Puzzles

Two Weighings

A problem from the Leningrad Mathematical Olympiad: You have a set of 101 coins, and you know that it contains one counterfeit coin X. The 100 genuine coins all have the same weight, which is different from that of X. Using only two weighings in an equal-arm balance, how can you determine whether X is heavier or lighter than the genuine coins?

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Divide the coins arbitrarily into piles of 30, 30, and 41 coins, calling these respectively A, B, and C. Now weigh A against B. If these balance, then X must be in C; combine A and B and weigh 41 of these coins against the 41 coins in C and you’ll find out whether X is heavy or light. If A and B don’t balance, then one of those groups contains the bad coin. For definiteness, suppose that A goes down in the weighing. That means either that A contains the bad coin and it’s heavy or that B contains the bad coin and it’s light. So determining where X is will also tell us its relative weight. Divide A in half and weigh these halves against each other. If they balance then the coins are all good and X must be in B. If they don’t balance then X is in A. (Obviously the same plan will work with certain pile sizes other than 30, 30, and 41, but that arrangement will do the job.)

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April 9, 2022April 8, 2022 | Puzzles