Javelin

A puzzle by Steven T., a systems engineer at the National Security Agency, from the NSA’s September 2016 Puzzle Periodical:

Three athletes (and only three athletes) participate in a series of track and field events. Points are awarded for 1st, 2nd, and 3rd place in each event (the same points for each event, i.e. 1st always gets “x” points, 2nd always gets “y” points, 3rd always gets “z” points), with x > y > z > 0, and all point values being integers.

The athletes are named Adam, Bob, and Charlie.

Adam finished first overall with 22 points.
Bob won the Javelin event and finished with 9 points overall.
Charlie also finished with 9 points overall.

Question: Who finished second in the 100-meter dash (and why)?

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First, realize that the total points awarded is 40. Given that we are dealing with integers that means the total number of Events (TE) times the total number of points awarded in an event (TP) must equal 40. We can quickly eliminate some possibilities.

Events (TE)	Points (TP)
1	40	Not possible since we know there are at least two events.
2	20	Not possible since Adam and Bob both won 1 event. There would be no way for Bob and Charlie to finish with the same points.
4	10	A possibility.
5	8	A possibility.
8	5	Not possible, since 1st > 2nd > 3rd > 0 means the minimum number of points awarded HAS to be 6 (3, 2, 1).
10	4	Same reason.
20	2	Same reason.
40	1	Same reason.

So we now know it was either 4 events with 10 points awarded in an event or 5 events with 8 points awarded. Let’s look at the 4/10 situation first. The possible points for each place with 10 overall points available is (all other combinations are impossible because of the 1st > 2nd > 3rd > 0 constraint):

5, 3, 2 Not possible as there is no way to get to 22 points in 4 events (greatest possible points would be 20).
5, 4, 1 Same reason.
6, 3, 1 Not possible, since no combination of 4 numbers can get to 22 points (3 × 6 + 3 = 21, 4 × 6 = 24).
7, 2, 1 A possibility — let’s look further.

Can we get Adam to 22? Yes. Adam can finish 1st 3 times and 3rd once (in the Javelin). 3 × 7 + 1 = 22.

Can we get Bob to 9? Nope. Bob won the Javelin (7 points). There’s no way to get to 9 with the remaining event/point combinations. Sooooo …

There must be 5 events with a total of eight points (we’re getting close). Let’s look at the possible points (again, the other combinations are not possible because of the 1st > 2nd > 3rd > 0 constraint):

4, 3, 1 Not possible as there is no way to get to 22 points (4 × 5 = 20 is maximum)
5, 2, 1 Looks like this is the winner. Let’s check.

Can we get Adam to 22? Yes. Adam can finish 1st 4 times and 2nd once (4 × 5 + 2). Since we know Bob finished 1st in the Javelin, Adam must have finished 2nd.

Can we get Bob to 9? Yes. Bob finished 1st in the Javelin (5 points) and finished 3rd 4 times (4 ×1) for a total of 9 points.

Can we get Charlie to 9? Yes. Charlie finished 3rd in the Javelin (1 point) and must have finished 2nd in the 4 other events (4 × 2) for a total of 9 points.

Since Charlie finished 2nd in EVERY EVENT OTHER THAN THE JAVELIN, he MUST have finished 2nd in the 100-meter dash. Here is a little table:

	Javelin	Event 2	Event 3	Event 4	100m Dash	Total
Adam	2nd (2 pts)	1st (5 pts)	1st (5 pts)	1st (5 pts)	1st (5 pts)	22 pts
Bob	1st (5 pts)	3rd (1 pt)	3rd (1 pt)	3rd (1 pt)	3rd (1 pt)	9 pts
Charlie	3rd (1 pt)	2nd (2 pts)	2nd (2 pts)	2nd (2 pts)	2nd (2 pts)	9 pts

April 29, 2024April 28, 2024 | Puzzles

“A Whisky Puzzle”

In 1895 a London shopkeeper attracted customers with a glass cask of whiskey — they were puzzled to find that no matter how much liquid they drew off, the level in the cask never dropped. The container could be viewed from any angle, and it stood well away from the wall. How was this possible?

A hidden pipe connected the bottom of the cask to a tank in another room. When a customer drew a glass of whisky, a confederate there would open a tap to replenish tank B, and the liquid, seeking its own level, would maintain the same height in the cask.

(James Scott, “Shopkeepers’ Advertising Novelties,” Strand, November 1895. See Desert Downpour.)

April 22, 2024April 21, 2024 | Puzzles · Science & Math

Constitutional Crisis

From Lee Sallows:

(Thanks, Lee!)

April 17, 2024April 16, 2024 | Puzzles · Science & Math

Black and White

crypton chess puzzle

An old puzzle by Paul Hoffman from Science Digest. Dr. Crypton is playing chess with his boss. Crypton has the white pieces. What move can he play that will not checkmate Black? There’s no funny business; the problem is just what it seems, except that Crypton has promised never to put a knight on any square adjacent to the black king, so 1. Ne6 doesn’t count as a solution.

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1. Rc5+ unpins the black rook on b4, permitting the reply 1. … Rxf4 and thus avoiding mate. From Dr. Crypton and His Problems: Mind Benders From Science Digest, 1982.

April 16, 2024July 17, 2024 | Puzzles

Empty Vessel

A puzzle by Soviet science writer Yakov Perelman: A bottle full of gasoline has a mass of 1,000 grams. The same bottle filled with acid has a mass of 1,600 grams. The acid is twice as dense as the gasoline. What’s the mass of the bottle?

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The mass of the bottle plus the mass of the gasoline is 1,000 grams. And the mass of the bottle plus twice the mass of the gasoline is 1,600 grams (because the acid is twice as dense as the gasoline). So the difference, 600 grams, is the mass of the bottle’s volume of gasoline. The mass of the bottle full of gasoline is 1,000 grams, so the mass of the bottle alone is 400 grams. (This checks out: It tells us that the mass of the acid is 1,600 – 400 = 1,200 grams, which is twice the mass of the gasoline.) From the March-April 1993 issue of Quantum (page 32, “Kaleidoscope”).

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April 9, 2024April 8, 2024 | Puzzles

Win Count

Imagine a game of tic-tac-toe (noughts and crosses) played in three dimensions in an 8×8×8 cube. A player wins by marking some straight line of eight cells through the large cube. How many such winning lines are there?

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This problem was proposed in the American Mathematical Monthly by A.L. Rubinoff of the University of Toronto in 1948 (Problem E773). It’s possible simply to count up the lines, but Leo Moser of the University of Manitoba hit on a simpler solution. Encase the 8×8×8 cube in a “shell” of unit thickness, and extend each winning line so that it pierces the shell at either end. Each winning line pierces two cells in the shell, and each cell in the shell is pierced by only one winning line. Thus each winning line corresponds to a unique pair of cells in the shell, and the number of winning lines is just half the number of cells in the shell: $\displaystyle \frac{10^{3} - 8^{3}}{2} = \frac{1000 - 512}{2} = 244.$ This solving strategy can be generalized to cubes of any size. (Via Ross Honsberger, Mathematical Morsels, 1978.)

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April 8, 2024April 7, 2024 | Puzzles

“Three Threes Are Ten”

This little trick often puzzles many:–

Place three matches, coins, or other articles on the table, and by picking each one up and placing it back three times, counting each time to finish with number 10, instead of 9. Pick up the first match and return it to the table saying 1; the same with the second and third, saying 2 and 3; repeat this counting 4; but the fifth match must be held in the hand, saying at the time it is picked up, 5; the other two are also picked up and held in hand, making 6 and 7; the three matches are then returned to the table as 8, 9, and 10. If done quickly few are able to see through it.

— John Scott, The Puzzle King, 1899

04/20/2024 Reader Vladamir Tsepis adds, “This reminds me of the way to convince children you have 11 fingers. Start by showing your left hand splayed, curl down the thumb and index finger counting ‘one, two…’, then of the remaining say ‘let’s skip these three’. Move to your right hand, bend each finger in turn as you count ‘four, five, six, seven, eight…’. Return to the left hand counting off the three we skipped ‘nine, ten, eleven.'”

April 5, 2024April 20, 2024 | Puzzles

Memorial

From The Book of 500 Curious Puzzles, 1859:

Following is the epitaph of Ellinor Bachellor, an old pie woman. How should we read it?

Bene A. Thin Thed Ustt HEMO. Uld yo
L.D.C. RUSTO! Fnel L.B.
Ach El Lor. Lat. ELY,
Wa. S. shove N. W. How — Ass! kill’d I. N. T. H.
Ear T. Sofp, I, Escu Star.
D. San D T Art. San D K. N E. W. E
Ver — Yus E. — Oft He ove N, W. Hens He
‘Dli V’DL. on geno
Ug H S hem A.D.E. he R. la Stp. Uf — fap
Uf. F. B Y he. R hu
S. Ban D. M.
Uch pra is ‘D. No. Wheres Hedot
HL. i. e. Tom. A kead I.R.T.P. Yein hop Esthathe
R. C. RUSTWI,
L L B. Era is ‘–D!

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Beneath in the dust The mouldy old crust Of Nell Bachellor lately was shoven: Who was skilled in the arts Of pies, custards, and tarts, And knew every use of the oven. When she’d liv’d long enough, She made her last puff, A puff by her husband much prais’d: Now here she doth lie, To make a dirt pie, In hopes that her crust will be rais’d.

April 4, 2024April 3, 2024 | Puzzles

Counting Up

A problem from Daniel J. Velleman and Stan Wagon’s excellent 2020 book Bicycle or Unicycle?: A Collection of Intriguing Mathematical Puzzles:

A square grid measures 999×999. Each square is either black or white. Each black square that’s not on the border of the grid has exactly five white squares among its eight immediate neighbors (those that adjoin it horizontally, vertically, or diagonally). Each white square that’s not on the border has exactly four black squares among its immediate neighbors. Of the 999 × 999 = 998001 squares in the grid, how many are black and how many white?

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999 is 3 × 333, so we can divide the grid into 333² 3×3 subgrids. Each of those subgrids consists of a central square surrounded by eight neighbors. According to our information, if a subgrid’s central square is black then exactly five of its neighbors are white, and thus the subgrid contains four black and five white squares. If the central square is white, then the subgrid likewise has four black and five white squares. So the entire grid has 333² × 4 = 443556 black and 333² × 5 = 554445 white squares.

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March 26, 2024March 25, 2024 | Puzzles

Specialist

A puzzle by Soviet science writer Yakov Perelman: Six carpenters and a cabinetmaker were hired to do a job. Each carpenter was paid 20 rubles, and the cabinetmaker was paid 3 rubles more than the average wage of the whole group. How much did the cabinetmaker make?

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We’re told that the cabinetmaker made 3 rubles more than the average wage of the seven together. So if we relieve him of 3 rubles he’ll be left with the average wage. What is that wage? Distribute the 3 rubles we’ve just taken from the cabinetmaker equally among the other six workers. Each will get half a ruble. So the average wage is 20.5 rubles, and the cabinetmaker got 23.5 rubles. From Perelman’s 1924 book For Young Mathematicians, via the March-April 1993 issue of Quantum (“Kaleidoscope,” page 32).

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March 25, 2024March 24, 2024 | Puzzles