Unsolved

On June 30, 1999, the body of 41-year-old Ricky McCormick was discovered near a cornfield in West Alton, Missouri. He’d last been seen alive five days earlier; now he was 15 miles from home though he owned no car. In his pockets were two handwritten notes (click to enlarge):

In the ensuing 18 years both the FBI’s Cryptanalysis and Racketeering Records Unit and the American Cryptogram Association have failed to find any meaning in these messages. In 2011 the FBI appealed to the public for their insights: If you have any you can contact them via this page.

“We are really good at what we do,” said CRRU chief Dan Olson, “but we could use some help with this one.”

November 22, 2017November 22, 2017 | Crime · Puzzles

Podcast Episode 178: Lateral Thinking Puzzles

Here are six new lateral thinking puzzles to test your wits and stump your friends — play along with us as we try to untangle some perplexing situations using yes-or-no questions.

See full show notes …

November 20, 2017November 20, 2017 | Podcast · Puzzles

The Grapevine

A problem from the British Columbia Colleges Senior High School Mathematics Contest, 2000:

Not all of the nine members on the student council are on speaking terms. This table shows their relationships — 1 means two members are speaking to each other, and 0 means they’re not:

  A B C D E F G H I
A - 0 0 1 0 0 1 0 0
B 0 - 1 1 1 1 1 1 1
C 0 1 - 0 0 0 1 1 0
D 1 1 0 - 1 0 1 0 1
E 0 1 0 1 - 0 1 0 0
F 0 1 0 0 0 - 0 0 1
G 1 1 1 1 1 0 - 0 0
H 0 1 1 0 0 0 0 - 0
I 0 1 0 1 0 1 0 0 -

Recently councilor A started a rumor, and it was heard by each councilor once and only once. Each councilor heard it from, and passed it to, another councilor with whom she was on speaking terms. If we count councilor A as zero, then councilor E was the eighth and last councilor to hear the rumor. Who was the fourth?

	Select Click for Answer
We know that A is at the start of the sequence, and E is at the end. Of the others, F and H are each speaking to only two other councilors. So F must pass the rumor between B and I (in one direction or the other, BFI or IFB), and H must pass it between B and C (BHC or CHB). Putting these fragments together, we must have either CHBFI or IFBHC. Also, this larger fragment can’t connect directly to either of the endpoints A or E, because neither C nor I is speaking to either of them. That means that the remaining councilors, D and G, must go on either end of our fragment before we can attach the endpoints. So the extended fragment is either GCHBFID or DIFBHCG, and with the endpoints A and E added we get the full sequence AGCHBFIDE or ADIFBHCGE. In either case, if A is counted as zero then the fourth person is B. (Crux Mathematicorum 27:3 [April 2001], 194.):

Click for Answer

November 16, 2017 | Puzzles

Murder Most Classificatory

In Anthony Boucher’s short story “QL 696.C9,” a librarian is found dead at her desk. She has been shot, and apparently spent her last moments typing the Library of Congress card catalog number that gives the story its title. The killer evidently saw nothing incriminating in this and so left it alone. The investigators have narrowed the list of suspects to junior librarian Stella Swift, children’s librarian Cora Jarvis, library patron James Stickney, and high school teacher Norbert Utter. Who did it?

	Select Click for Answer
In the Library of Congress system, QL refers to zoology, the 600s to vertebrates, 696 to birds, and .C9 to Cypseli — or swifts.

November 14, 2017 | Puzzles

A Mating Machine

elkies chess puzzle

A “no-brainer” by Harvard mathematician (and chess master) Noam Elkies. White has choices as to how to play, but no matter how he proceeds, he’ll wind up mating Black on the seventh move.

(From Tim Krabbé’s chess diary.)

November 10, 2017 | Puzzles

Black and White

tarrant chess problem

By Thomas Tarrant. White to mate in two moves.

	Select Click for Answer
1. Nf5! and 2. Rb5 will mate. From the Westminster Papers, 1877.

November 2, 2017November 1, 2017 | Puzzles

Coverup

A problem from the British Columbia Colleges Senior High School Contest for 2000:

If I place a 6 cm × 6 cm square on a triangle, I can cover up to 60% of the triangle. If I place the triangle on the square, I can cover up to 2/3 of the square. What is the area, in cm², of the triangle?

(a) 22 4/5
(b) 24
(c) 36
(d) 40
(e) 60

	Select Click for Answer
The answer is (d). The key is to see that these measures are one and the same: The triangle overlaps the square by the same amount that the square “underlaps” the triangle, so these areas are maximized at the same moment. If A is the area of the triangle, then 0.6A = 2/3 × 36, and A = 40 cm². From “The Skoliad Corner,” Crux Mathematicorum, December 2000.

October 18, 2017October 17, 2017 | Puzzles

Black and White

shinkman chess problem

“We now give what is acknowledged to be the finest two-move problem extant,” wrote J.H. Blackburne in the Strand in 1908. “It is by the American expert, W.A. Shinkman, and is also claimed by G.E. Carpenter, a fellow countryman of his. Here we have not only a difficult key-move, but also beauty of theme and artistic construction, the three essential qualities necessary to a perfect problem.”

White to mate in two moves.

	Select Click for Answer
“It is solved by B to R 4th [that is, 1. Ba4], moving the B from a strong position, where it defended a Kt, to, apparently, a useless square. In reply, Black has a choice of five moves”: 1. … f5 2. Qg8# 1. … f6 2. Ndc7# 1. … e4 2. Qxe4# 1. … Kxd5 2. Bb3# 1. … d6 2. Nbc7# “Truly a marvellous composition.” (“Bridge and Chess Problems,” Strand, March 1908.)

October 12, 2017October 11, 2017 | Puzzles

The Fifth Element

pmej matrix puzzle

Charles W. Trigg offered this puzzle in the Fall 1977 issue of Pi Mu Epsilon Journal (PDF):

This square array contains the first 25 positive integers. Choose five, no two from the same row or column, so that the largest of the five elements is as small as possible, and justify your choice.

	Select Click for Answer
Sidney Penner of Bronx Community College of CUNY, found the solution: 1, 3, 6, 8, 11. In the array, crossing out all the numbers higher than 11 eliminates all the numbers in the third row except 8. Choosing 8 means we have to forswear every other number in the fifth column, which leaves only one possibility in the fourth row, 3. And choosing 3 eliminates every other candidate in the first column, so the only available entry in the first row is 11. At the request of problem editor Leon Bankoff, Jeanette Bickley ran the following BASIC program on a DEC PDP 11/70 computer, and it confirms Penner’s solution:

Click for Answer

October 11, 2017October 10, 2017 | Puzzles

The Impossible Puzzle

Dutch mathematician Hans Freudenthal proposed this puzzle in 1969 — at first it appears impossible because so little information is given.

X and Y are two different whole numbers greater than 1. Y is greater than X, and their sum is no greater than 100. S and P are two logicians; S knows the sum X + Y, and P knows the product X × Y. S and P both reason perfectly, and both know everything I’ve just told you.

S says, “P does not know X and Y.”
P says, “Now I know X and Y.”
S says, “Now I also know X and Y.”

What are X and Y?

	Select Click for Answer
X = 4 and Y = 13. S knows the sum is 17, and P knows the product is 52. S makes a list of all the addends that could produce 17 and fulfill the other conditions above: 2 + 15 (product: 30) 3 + 14 (product: 42) 4 + 13 (product: 52) 5 + 12 (product: 60) 6 + 11 (product: 66) 7 + 10 (product: 70) 8 + 9 (product: 72) S knows that P is thinking of one of the products listed above; she can say that P doesn’t know X and Y because each of these products has multiple solutions (e.g. 30 = 2 × 15 = 3 × 10). P has been building his own list — in his case, he’s listing pairs of factors that produce 52. It’s a short list — there are only two possibilities that meet the conditions: 2 × 26 4 × 13 When S makes her first announcement, P understands that her candidate list is like the one above — it’s one in which all of the products are ambiguous. And this eliminates the first candidate on P’s list. If X = 2 and Y = 26, then their product would be 52, as P expects, but S would know that their sum was 28, and she couldn’t then claim that P didn’t know the numbers — her list of suspects would look like this: 2 + 26 (product: 52) 3 + 25 (product: 75) 4 + 24 (product: 96) 5 + 23 (product: 115) … That fourth item, X = 5 and Y = 23, gives a product of 115, which has no other set of factors that meet the requirements. So if 28 were the sum, S couldn’t have declared so confidently that P was still in the dark. Realizing this, P can eliminate 2 × 26 from his list, leaving only one possibility: X is 4 and Y is 13. He says, “Now I know X and Y.” And by reconstructing his thinking, S can determine these values herself. (Thanks, Jon.)

Click for Answer

September 24, 2017 | Puzzles · Science & Math