Dead Center

In a shooting match, Andryusha, Volodya, and Borya each fired 6 shots, and each totaled 71 points.

Andryusha’s first 2 shots earned him 22 points, and Volodya’s first shot earned 3 points.

Who hit the bullseye?

	Select Click for Answer
The only way to partition the 18 shots into 3 equal sets of 6 is: 25, 20, 20, 3, 2, 1 25, 20, 10, 10, 5, 1 50, 10, 5, 3, 2, 1 The first row must be Andryusha’s, since it’s the only one that contains two numbers totaling 22. The score 3 appears only in the first and third rows, so the third must be Volodya’s. Thus he hit the bullseye. From Boris Kordemsky’s The Moscow Puzzles, 1956.

January 23, 2026January 23, 2026 | Puzzles

Vivid

The manuscript for Mozart’s Horn Concerto No. 4 in E-Flat Major, K. 495, is written in inks of four colors, red, green, blue, and black.

It’s not clear why. Possibly the composer was teasing his friend Joseph Leutgeb, the intended performer. But possibly it’s a code meant to inform the performance.

January 23, 2026 | Art

Sum of Squares

The sum of the first n square numbers is n(n+1)(2n+1)/6.

These sums comprise the square pyramidal numbers — each corresponds to the number of oranges that can be stacked in a square pyramid whose base has side n.

This visual proof, for n=3, shows that six square pyramids with n steps fit in a cuboid of size n(n + 1)(2n + 1).

(By CMG Lee.)

January 22, 2026January 22, 2026 | Science & Math

Credential Envy

In June 1936, when the University of Wisconsin bestowed an honorary doctorate of letters on actress Katharine Cornell, she received a telegram from Noël Coward:

DARLING DARLING DOCTOR KITTY,
THOUGH QUITE REASONABLY PRETTY
THOUGH UNDOUBTEDLY A STAR, DEAR
PLEASE REMEMBER WHO YOU ARE, DEAR.
WHY, IN LIEU OF ALL YOUR BETTERS,
SHOULD YOU HAVE DISTINGUISHED LETTERS?
THIS COMES FROM THE JEALOUS SOEL
OF YOUR SOMEDAY DOCTOR NOËL.

January 22, 2026 | Entertainment

Uh-Oh

https://commons.wikimedia.org/wiki/File:Circle_staircase_paradox.svg — Image: Wikimedia Commons

January 21, 2026 | Science & Math

Body Heat

London doctor Thomas Ellerby had strong feelings about the disposal of his remains — in his will of February 1827 he threatens to haunt his doctors if they don’t follow his instructions:

I bequeath my heart to Mr. W., anatomist; my lungs to Mr. R.; and my brains to Mr. F., in order that they may preserve them from decomposition; and I declare that if these gentlemen shall fail faithfully to execute these my last wishes in this respect I will come — if it should be by any means possible — and torment them until they shall comply.

The Lancet records that “[t]he gentlemen named — eminent in their day — rightly renounced the legacies left them, and it never appeared that they were, like St. Dunstan and other medieval saints, tromented by visitations from the other world.”

January 21, 2026 | Death

A Roman New Year

George Sicherman, of Sicherman dice fame, has designed a puzzle to welcome the new year: Print out the PDF, cut out the letters MMXXVI (the Roman numerals for 2026), and fit them into the tray provided so that no pieces with the same shape touch one another.

Sicherman’s website has a whole range of puzzles, most of them geometric.

(Thanks, William.)

January 20, 2026 | Puzzles

Succinct

In November 1943, as they planned a summit meeting, FDR wired Churchill to suggest that Cairo might be too dangerous a location. Churchill responded:

SEE ST. JOHN, CHAPTER 14, VERSES 1 TO 4.

These verses read:

Let not your heart be troubled: ye believe in God, believe also in me.
In my Father’s house are many mansions; if it were not so, I would have told you, I go to prepare a place for you.
And if I go and prepare a place for you, I will come again, and receive you unto myself; that where I am, there ye may be also.
And whither I go ye know, and the way ye know.

They met in Cairo.

January 20, 2026January 19, 2026 | History

Poser

Does the sequence of squares contain an infinite arithmetic subsequence?

	Select Click for Answer
Consider three squares in arithmetic progression, $a_{1}^{2}$ , $a_{2}^{2}$ , and $a_{3}^{2}$ . Then $a_{3}^{2} - a_{2}^{2} = a_{2}^{2} - a_{1}^{2}$ , or (a₃ – a₂)(a₃ + a₂) = (a₂ – a₁)(a₂ + a₁). We know that (a₂ + a₁) < (a₃ + a₂), so (a₂ – a₁) must be greater than (a₃ – a₂). Now suppose that $a_{1}^{2}, a_{2}^{2}, a_{3}^{2}, ...$ is an infinite arithmetic progression. Then (a₂ – a₁) > (a₃ – a₂) > (a₄ – a₃) > … . That’s impossible, since there’s no infinite decreasing sequence of positive integers. So the answer is no. From Arthur Engel’s excellent Problem-Solving Strategies, 2008.

Click for Answer

January 18, 2026 | Puzzles

Time Out of Joint

In 1582, as the Catholic world prepared to adopt the new Gregorian calendar, German pamphleteers lampooned the strife that attended the change:

The old calendar must be the right one for the animals still use it. The stork flies away according to it, the bear comes out of his hole on the Candlemas day of the old calendar and not of the Pope’s, and the cattle stand up in their stalls to honor the birth of the Lord on the Christmas night of the old and not of the new calendar. They also recognize in this work diabolical wickedness. The Pope was afraid the last day would come too quickly. He has made his new calendar so that Christ will get confused and not know when to come for the last judgment, and the Pope will be able to continue his knavery still longer. May Gott him punish.

“Inanimate objects were not so stubborn.” An Italian walnut tree that had reliably put forth leaves, nuts, and blossoms on the night before Saint John’s day under the old regime dutifully adopted the new calendar and performed its feat on the correct day in 1583. A traveler wrote, “I have today sent a branch, broken off on Saint John’s day, to Herr von Dietrichstein, who no doubt will show it to the Kaiser.”

(Roscoe Lamont, “The Reform of the Julian Calendar,” Popular Astronomy 28:1 [January 1920], 18-32.)

January 17, 2026 | History · Society

An idler's miscellany of compendious amusements

Author: Greg Ross