Simple Enough

I think this appeared originally in Ed Southall’s Twitter feed. What fraction of the square is shaded?

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The answer is 1/3, but there’s a beautiful variety of solutions in the discussion.

July 5, 2019July 4, 2019 | Puzzles

“A Chess Packing Problem”

In 2006 Martin Gardner asked: Can you arrange the 16 non-pawn pieces in a standard chess set on a 5 × 5 board so that no piece attacks a piece of the opposite color? As in a conventional game, the two bishops of each color must stand on squares of opposite colors.

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Cameron Byer, a second grader at Calvary Christian Academy in Harrisonburg, Va., sent this solution: The problem was originally proposed and solved by Mamikon Mnatsakanian in Soviet Science & Life in 1976, Issue 2. (Andy Liu, “Problem Section,” Math Horizons 14:2 [November 2006], 40-42.)

June 25, 2019 | Puzzles

The Dark Half

A puzzle from the 1997 Ukrainian Mathematical Olympiad:

Cells of some rectangular board are coloured as chessboard cells. In each cell an integer is written. It is known that the sum of the numbers in each row is even and the sum of numbers in each column is even. Prove that the sum of all numbers in the black cells is even.

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To begin with, notice that the sum of the numbers in the white cells in any given row or column must have the same parity (odd or even) as the sum of the numbers in the black cells. That’s necessary because we know that the sum of the whole row (or column) is even: If the white total is even, then the black total must be even, and if the white total is odd then the black total must be odd. Since that’s so, and since we’re interested only in parity, then in evaluating the black squares over the board as a whole we can make that substitution — in a given row or column, we can substitute the parity of the white sum for the parity of the black sum. Then let’s calculate the sum of the board’s black cells by considering each column in turn, and we’ll choose to sum the black cells in the odd-numbered columns and the white cells in the even-numbered columns. But that’s just the same as summing every second row on the board, and we already know that each of those totals is even. So the final sum must be even. (Via R.E. Woodrow, “Olympiad Corner,” Crux Mathematicorum 29:2 [March 2003], 87-103.)

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June 25, 2019June 24, 2019 | Puzzles

Black and White

By K. Makovsky. White to mate in two moves.

In The Two-Move Chess Problem (1890), Benjamin Glover Laws calls the first move here “ideal” and “splendid.” “[I]t is not always a composer’s good fortune to strike a vein which is susceptible of such an excellent opening move as is illustrated in this problem.” What is it?

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1. Qh4! and the bishop will mate. (Benjamin Glover Laws, The Two-Move Chess Problem, 1890.)

June 23, 2019 | Puzzles

Decomposition

This verse is known as “Lord Macaulay’s Last Riddle.” Lord Macaulay was Thomas Babington Macaulay (1800-1859), though his authorship of the riddle is uncertain. What’s the answer?

Let us look at it quite closely,
‘Tis a very ugly word,
And one that makes one shudder
Whenever it is heard.

It mayn’t be very wicked;
It must be always bad,
And speaks of sin and suffering
Enough to make one mad.

They say it is a compound word,
And that is very true;
And then they decompose it,
Which, of course, they’re free to do.

If, of the dozen letters
We take off the first three,
We have the nine remaining
As sad as they can be;

For, though it seems to make it less,
In fact it makes it more,
For it takes the brute creation in,
Which was left out before.

Let’s see if we can mend it —
It’s possible we may,
If only we divide it
In some new-fashioned way.

Instead of three and nine, then,
Let’s make it four and eight;
You’ll say it makes no difference,
At least not very great;

But only see the consequence!
That’s all that need be done
To change this mass of sadness
To unmitigated fun.

It clears off swords and pistols
Revolvers, bowie-knives,
And all the horrid weapons
By which men lose their lives;

It wakens holier voices —
And now joyfully is heard
The native sound of gladness
Compressed into one word!

Yes! Four and eight, my friends!
Let that be yours and mine,
Though all the hosts of demons
Rejoice in three and nine.

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Manslaughter. (Thanks to Anton Lewis.)

June 18, 2019June 18, 2019 | Puzzles

Tight Squeeze

Above: From Paris, 1927: a novelty car that can “sidle” into parking spaces.

Below: Someone was actually working on this in the 1950s (thanks, Martin):

A related puzzle from The Chicken From Minsk, Yuri B. Chernyak’s 1995 collection of math and physics problems: Why is it easier to parallel-park a (conventional) car by backing into the space rather than pulling in directly?

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Imagine pulling out of the space. With the front wheels turned sharply to one side, the center of rotation is close to the car’s rear end; the front of the car swings out of the space, avoiding the car in front, and then the rear follows it. Because the center of rotation is so close to the rear of the car, it would be hard to back out of a tight space. Now play the scenario in reverse: By backing into the space, the driver is putting the least maneuverable end of the car into position first and can then rotate the rest into place.

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June 13, 2019June 17, 2019 | History · Puzzles · Technology

A Narrative Alphametic

James Mayfield gives this clever expression in the August 2006 issue of Word Ways:

BEAR + RARE + ERE = RHYME

It’s an alphametic: If each letter is taken to represent a different digit, it encodes a valid equation. What’s the equation?

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If A=5, B=7, E=9, H=0, M=8, R=1, and Y=3, then: 7951 1519 + 919 ----- 10389 Mayfield gives 94 additional narrative alphametics, including the remarkable BRUTUS + STABS = CAESAR, in the article. He provides no answers, but if you get stuck you can derive them using this nifty calculator. (James Mayfield, “Some Narrative Alphametics,” Word Ways 39:3 [August 2006], 179-181.)

June 12, 2019 | Puzzles

Cutting Up

Is it possible to divide a circle into n parts of equal area using only a straightedge and compasses?

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Yes. Draw a diameter, divide it into equal parts, and then draw semicircles on those divisions: Image: mathslinks.net

June 7, 2019 | Puzzles

Counter Play 2

sallows counter puzzle 1

Lee Sallows sent this clever puzzle, a followup to one we presented in 2010:

In the square shown above, any 3 counters in a straight line sum to 15.

Puzzle: Reposition the counters (again, one to each cell) to produce a new square again showing eight collinear triplets summing to 15, but with 1 now placed in a corner square.

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(Thanks, Lee!)

June 6, 2019June 6, 2019 | Puzzles

Black and White

wainwright chess problem

A densely imaginative little problem by Joseph C.J. Wainwright. White to mate in two moves.

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1. Rg6! and Black can choose from four striking finishes: 1. … Bxf7 2. Nf6# 1. … Kxf7 2. gxh8=N# 1. … Kxh7 2. g8=Q# 1. … Rxh7 2. Rf8# From the Brisbane Courier, Aug. 16, 1919.

June 5, 2019June 4, 2019 | Puzzles