Puzzle of the Month

February 2008: The race is on

2008-02-04T16:20:00.000-05:00

Alice and Bob ran a marathon (assumed to be exactly 26.2 miles long) with Alice running at a perfectly uniform eight-minute-per-mile pace, and Bob running in fits and starts, but taking exactly 8 minutes and 1 second to complete each mile interval (this refers to all intervals of the form (x, x+1), including, for example, the interval from 3.78 miles to 4.78 miles). Is it possible that Bob finished ahead of Alice?

November 2007: The lineup (solution)

2007-11-02T08:29:00.000-04:00

The sequence must terminate with a 0 or 9, and as we move from right to left, we always have a choice between the highest unused digit or the lowest--until we hit the left end where these two choices coincide. Thus, there are two choices at each of nine opportunities, i.e., 2⁹=512 ways.

November 2007: The lineup

2007-11-02T08:27:00.000-04:00

How many ways are there to write the numbers 0 through 9 in a row, such that each number other than the left-most is within one of some number to the left of it?

October 2007: On the right track (solution)

2007-10-09T10:49:00.000-04:00

The right-hand wheels travel 2π" farther than the left-hand wheels.

The two pairs of wheels travel the same distance along the straight tracks, so we may as well ignore them. We can take the curved tracks and reassemble them to form a circle. The circumference of the inside rail is 2πr and the circumference of the outside rail is 2π(r+1), so the difference is 2π". (If the track is not convex, then each right-hand curve cancels a left-hand curve, so we can toss them both out, leaving a single circle.)

Math fact: This applies much more generally. If a vehicle drives counterclockwise along any curvy route so that it loops back to the starting position without ever crossing its tracks (think of a Grand Prix race track), then the right wheels travel 2πb farther than the left wheels (where b is the distance between the wheels).

Correct solutions: Ritoban Basu Thakur

October 2007: On the right track

2007-10-03T21:51:00.000-04:00

Your young cousin has made an elaborate train track for his toy trains. He used track pieces of two varieties--ones that are straight and ones that are arcs of a circle. The gauge of the track (the distance between the rails) is 1". Suppose he drives his train around the track one time counterclockwise. How much farther did the right wheels travel than the left wheels? (You may assume that the track is convex, but in fact, you don't need this assumption--you just need there to be no crossovers.)

September 2007: You make the rules (solution)

2007-09-05T16:44:00.000-04:00

There are several solutions. Here's one. Divide the ruler as follows:

1"=1"
2"=1"+1"
3"=1"+1"+1"
4"=4"
5"=5"
6"=1"+5"
7"=1"+1"+5"
8"=1"+1"+1"+5"
9"=5"+4"
10"=1"+5"+4"
11"=1"+1"+5"+4"
12"=1"+1"+1"+5"+4"

(Other submitted solutions are 1-3-1-5-2, 2-5-1-3-1, 1-3-3-3-2, and 1-1-4-3-3)

Correct solutions: Ritoban Basu-Thakur, Kristin Jekielek, Richard Rast, Anna Marion, Danielle Newcomer ('07), Mike Scanish ('01)

September 2007: You make the rules

2007-09-05T16:19:00.001-04:00

The object of this month's puzzle is to design a ruler that can measure lengths of 1", 2", 3",..., 12". You are given an unmarked 12" ruler and you are allowed to make four marks that divide it into five segments (the marks must be perpendicular to the length of the ruler, as shown below). Describe how to construct such a ruler.

Here's an example to illustrate the idea. Suppose the problem was the same, but you had an unmarked 6" ruler that had to measure 1", 2", 3",..., 6". You can accomplish this by making two marks on the ruler leaving gaps of 1", 3", and 2":

It is obvious how to measure 1", 2", 3", and 6" lengths using this ruler, but also 4"=1"+3" and 5"=3"+2".

Summer 2007: Nurikabe (solution)

2007-04-25T13:42:00.000-04:00

Correct solutions: Alison Dethoff, Evan Templeton

Summer 2007: Nurikabe

2007-04-25T13:20:00.000-04:00

The idea of Nurikabe is to color each square in an n x n grid white or black according to the following rules:

Each numbered cell must be part of a connected white region (connected always means edge-to-edge, not corner-to-corner).
Each connected white region must contain exactly one number, and this value is the number of white cells in the region (including the numbered cell).
The black region must be connected.
There can be no 2 x 2 blocks of black cells.

Below is a 5 x 5 game and its solution.

This month's puzzle is to solve the following 9 x 9 Nurikabe puzzle.

April 2007: Kakuro (solution)

2007-04-03T10:52:00.000-04:00

Correct solutions: Sunil Baidar

April 2007: Kakuro

2007-04-03T10:08:00.000-04:00

Kakuro is a crossword-like puzzle with numbers instead of letters. The object of the game is to place the digits 1 through 9 in each of the open cells so that the sum of each "word" matches the clue and there is no duplication of digits in a given word. The clues are given above or to the left of the given word. In other words, numbers above the diagonal are clues for the word to the right and numbers below the diagonal are clues for the word below. An example, with solution, is shown below.This month's puzzle is to solve the following Kakuro puzzle.

Source

March 2007: Nine ways to eight (solution)

2007-03-04T20:44:00.000-05:00

2, 7, 1: 2+7-1=8 or 7+1²=8
2, 7, 2: 7+(2/2)=8
2, 7, 3: 7+3-2=8 or 2*(7-3)=8
2, 7, 4: 2^7-4=8
2, 7, 5: 7+(.2)*5=8 or 7+(.5)*2=8
2, 7, 6: 2*7-6=8
2, 7, 7: (7/.7)-2=8
2, 7, 8: 7+.2+.8=8 or 7.2+.8=8 or 7.8+.2=8
2, 7, 9: (7+9)/2=8

Correct solutions: Katie Shrader, Rob Peters ('02), Sunil Baidar, Christina Lakin, Travis Doll

March 2007: Nine ways to eight

2007-03-02T14:58:00.000-05:00

Create nine different mathematical expressions that equal 8. You must use the digits 2, 7 and one other. That other digit must be a 1 in the first expression, 2 in the next expression, and so on, up to 9. You can use a digit once and only once in each expression.

You may use the four arithmetic symbols: plus, minus, times, and divided by, as well as exponents and decimal points. You may use parentheses as you need them. For example: using the digits 2, 7, and 1 you can make the expression (2+7)-1= 8.

January/February 2007: The Sixteen Grid (solution)

2007-03-02T14:54:00.000-05:00

Correct solutions: Katie Shrader, Andrew Hulme

January/February 2007: The Sixteen Grid

2007-01-20T10:12:00.000-05:00

Each of the numbers 1, 2, …, 16 is used exactly once in the empty cells to form arithmetic expressions connected by symbols for the four basic operations. Each row (column) is an arithmetic expression, read and performed left to right (top to bottom), disregarding the usual order of operations, to yield the result at the right (bottom).

Hints

Before you begin, here’s a good clue,
Beware of the list, for one is untrue.

To get a start, with vim and vigor,
See that the top right number couldn't possibly be bigger.

For another corner, you must reject
Any and all numbers that are not perfect.

Stumped? Don't fall into depression:
The first three entries in one row are in geometric progression.

For salvation don't look to the gods or run for the border,
One row is consecutive odds (though not in order).

If you are really stuck, can linear algebra cure your anxiety?
The puzzle complete has determinant -15090

Now, buster, before you tire of reading these rhymes
No row or column has more than two primes.

When you solve this you'll feel like a hero,
Three numbers in one column, in one base, end in zero.

Puzzle was originally posted by the American Mathematical Society.

November 2006: Afghan Bands (solution)

2006-11-10T09:56:00.000-05:00

To create the belts for the Siamese twins, take a standard Möbius band (with one half twist), then, as before, cut down the length of the strip. But instead of cutting along the centerline, cut the strip in thirds (as you will see, this will only require one cut). The procedure is demonstrated in Experiment #3 in the following video.

No Magic At All: Mobius Strip - video powered by Metacafe

Solution to the bonus puzzle: To get the strip tied around the ring, put the strip through the ring and give the strip three half twists before taping. Then cut down the centerline.

Correct solutions from: Anne Maiale, Lisa Dubbs

November 2006: Afghan Bands

2006-11-01T13:35:00.000-05:00

The famous one-sided Mobius strip has appeared in the artwork of M. C. Escher and is the basis for the recycling symbol. It was also the inspiration for an old magic trick called the "Afghan bands" which dates back to at least 1882. A circus magician holds up three loops of fabric, which he explains are cloth belts. The problem, he laments, is that he needs belts for two clowns, the fat lady, and the Siamese twins. He takes the first band, rips it down the centerline and produces the belts for the two clowns. He rips the second band in the same way, but instead of two loops, he holds a single loop that has twice the circumference of the original---the belt for the fat lady. Finally, to get the belts for the twins he rips the third loop and obtains two belts that are linked together. The trick, as we see below, is that the loops have twists in them (zero, one, and two half twists, respectively). For maximum effect, the fabric or paper should be flexible and be much narrower than it is long so that the audience does not notice the twisting.It turns out that one of the Siamese twins has put on a lot of weight over the summer. This month's problem is to find a way to repeat the Afghan bands trick, but at the end obtain one large belt linked with one smaller belt. Your solution should describe how to perform the magic trick.

If you liked this puzzle, try this one. Run a strip of paper through a wedding ring and tape the ends together to form a twisted band. Using only scissors (and no tape) cut the band so that the resulting band is knotted around the wedding ring (see below).

Dickinson College students can submit answers to Dave Richeson or Barry Tesman . The list of solvers will be posted at the end of the month.

October 2006: The locker room (solution)

2006-10-08T16:36:00.000-04:00

The state of locker n is changed when the kth student passes through, for every divisor k of n. Since factors usually come in pairs {j, k} where j x k = n, the net effect of students j and k on this locker is nil. The exception is when n is a perfect square, in which case there is no other divisor to cancel the effect of the sqrt(n) student. Therefore, the lockers which are open at the end are exactly the perfect squares, 1, 4, 9, 16, 25, 36, 49, 64, 81, and 100.

Correct solutions: Evan Templeton, Matt Bachmann, Kaloyan Todorov, Theresa Sparacio ('04), Kristin Jekielek, Hiro Arai, Robert Pehlman, Jenny Witzeling, Dulguun Bayasgalan, Ritoban Basu-Thakur, Ritwik Niyogi, Sunil Baidar, Wade Robertson, Mike Scanish ('01), Lisa Dubbs, Amanda Janiec, Ben Raffeto

October 2006: The locker room

2006-10-05T16:20:00.000-04:00

Lockers numbered 1 to 100 stand in a row at the school gym. When the first student arrives, she opens all the lockers. The second student then goes through and recloses all the even-numbered lockers; the third student changes the state of every locker whose number is a multiple of 3.

This continues until 100 students have passed through. Which lockers are now open?

Dickinson College students can submit answers to Barry Tesman or Dave Richeson . The list of solvers will be posted at the end of the month.

September 2006: It's a colorful country (solution)

2006-08-23T13:27:00.000-04:00

There are several different answers to this month's puzzle. There are three problem states that make it impossible to color the country using only three colors: Nevada, Kentucky, and West Virginia. These three states have an odd number of neighbors. For instance, suppose we try to color Nevada and its five neighbors using only three colors. First, color Nevada blue. Then alternate colors around it: California red, Oregon yellow, Idaho red, Utah yellow. The problem is, Arizona borders Utah, California, and Nevada, so it can't be red, yellow, or blue. We need a fourth color. The same is true of the neighborhood around Kentucky and the neighborhood around West Virginia (notice that Kentucky and West Virginia are neighbors of each other).

So for the answer to the puzzle, one state has to be Nevada, California, Oregon, Idaho, Utah, or Arizona, and the other state has to be Kentucky, West Virginia, Ohio, or Virginia (the only states that border both Kentucky and West Virginia).

[We have not checked that all 24 pairs are attainable, but we have received at least 15 different solutions.]

This month's solvers: Evan Templeton, Kristin Jekielek, James Doyle, Helen Delano, Hiro Arai, Lisa Dubbs, Dulguun Bayasgalan, Jim Matthews, Judi Matthews, Susan Matthews, and Judi Matthew's high school math classes

September 2006: It's a colorful country

2006-08-21T22:33:00.000-04:00

In 1852 Francis Guthrie noticed that it was possible to color all of the counties in England using only four colors (with bordering counties having different colors), and he conjectured that every map could be colored using four or fewer colors. For over a century the so-called four color problem was one of the most popular and elusive unsolved problems in mathematics. It did not become a theorem until 1976 when Kenneth Appel and Wolfgang Haken proved it using a computer to check thousands of special cases. Even now there is no pencil-and-paper proof of the theorem.

By the four color theorem we know that it is possible to color the continental United States using four colors. In fact, it can be colored using four colors so that one color is used for only two states. What are those two states?

Here is a blank map of the U.S.

Dickinson College students can submit answers to Dave Richeson or Barry Tesman. You do not have to submit your coloring, only the names of the two states (there is more than one answer). The list of solvers will be posted at the end of the month.

June 2006: Shaking hands (solution)

2006-06-05T15:33:00.000-04:00

This is the solution to the Shaking Hands puzzle.

Mr. Dickinson shook hands with 4 people.

Suppose the other four couples are Mr. and Mrs. A, Mr. and Mrs. B, Mr. and Mrs. C, and Mr. and Mrs. D. We know Mr. Dickinson did not shake hands with 8 people, because if he did, then everyone else would have shaken hands with at least one person, and we know that someone shook zero hands. Say that it was Mr. A who shook 8 hands. Then Mrs. A must have shaken 0 (in the graph below, lines indicate hand shakes). By a similar logic, Mr. Dickinson couldn't have shaken 7 hands, because then no one could have shaken 1 hand. Say Mr. B shook 7 hands. Then Mrs. B shook 1 hand. Arguing in the same way, Mr. Dickinson couldn't have shaken 6 hands, so someone else did---Mr. C, say. Then Mrs. C shook 2 hands. And again, someone else shook 5 hands, Mr. D, and his wife, Mrs. D, shook 3 hands. Thus, Mr. Dickinson shook 4 hands.

This month's winners: Evan Templeton, Ryan Deeds, Jared Lease, Shashwat Acharya, Hiro Arai

June 2006: Shaking hands

2006-06-05T15:30:00.000-04:00

Mr. and Mrs. Dickinson invited four couples to their house for a dinner party. Some of the guests knew each other, and others did not. Those who did not know each other shook hands upon meeting. Mrs. Dickinson happened to notice something curious: each person (not including herself) shook hands with a different number of people (one person shook no hands, one person shook one hand, one person shook two hands, etc.). How many hands did Mr. Dickinson shake?

Hint: Mr. Dickinson did not shake eight hands, for if he did, then no one could have shaken zero hands.

Dickinson College students can submit their answers to Dave Richeson. The list of solvers will be posted at the end of June.

From: "Effective Problem Solving, 2^nd Edition," Marvin Levine, Prentice-Hall, 1994

May 2006: Tait's 8 Coin Puzzle (solution)

2006-06-05T14:49:00.000-04:00

This is the solution to Tait's 8 Coin Puzzle:

This month's winners are: Evan Templeton, Jon Rogers, Jared Lease, Gordon Clark, Shashwat Acharya

May 2006: Tait's 8 Coin Puzzle

2006-05-03T13:58:00.000-04:00

In 1884 P. G. Tait presented the following puzzle (which he encountered while riding on a train). Start with eight adjacent coins alternating heads and tails. By moving four pairs of neighboring coins (and without changing their order), rearrange the coins to obtain four tails followed by four heads with no gaps between any of the coins.

For example, here is a solution for six coins using four moves.

Your solution should list the orders of the coins (including the gaps) at each stage.

May winners: Evan Templeton, Jon Rogers, Jared Lease, Gordon Clark, Shashwat Acharya

P.G. Tait, "Listing's Topologie," Philosophical Magazine, January 1884

[See the solution to last month's puzzle, Queen Domination]

April 2006: Queen domination (solution)

2006-05-03T13:56:00.000-04:00

There are many solutions to this puzzle, but they are all variations of the following solution: a3, e4, b8, d11.

April 2006: Queen domination

2006-04-14T09:58:00.000-04:00

A chessboard is dominated if every square either has a piece on it or is attacked by a piece. An 8x8 board can be dominated with 12 Knights as shown below.

Can you dominate a 5x12 board with 4 Queens?

Your final solution should be the chessboard with 4 Queens drawn in. Or, submit the positions of the 4 Queens based on the labeling of the board above.

You can download a printable version here: (pdf)

[If you solve this one, try dominating the 5x12 board with 10 Knights; a much harder problem.]

March 2006: A Lewis Carroll puzzle (solution)

2006-04-14T09:57:00.000-04:00

This is the solution to last month's puzzle, A Lewis Carroll puzzle.

The only animals in this house are cats.
Every animal is suitable for a pet, that loves to gaze at the moon.
When I detest an animal, I avoid it.
No animals are carnivorous, unless they prowl at night.
No cat fails to kill mice.
No animals ever take to me, except what are in this house.
Kangaroos are not suitable for pets.
None but carnivora kill mice.
I detest animals that do not take to me.
Animals, that prowl at night, always love to gaze at the moon.

Logical conclusion: I avoid kangaroos.

Justification:

H: it is in this house
F: it is a cat
G: it loves to gaze at the moon
D: I detest it
A: I avoid it
P: it is suitable for a pet
C: it is a carnivore
N: it prowls at night
T: it takes to me
K: it is a kangaroo
M: it eats mice

The riddle breaks down as follows (note: each sentence can be replaced by its contrapositive).
H --> F
G --> P
D --> A
~N --> ~C
F --> M
~H --> ~T
K --> ~P
M --> C
~T --> D
N --> G

From which we conclude:
~A --> ~D --> T --> H --> F --> M --> C --> N --> G --> P --> ~K

Possible solutions:
If I do not avoid it, then it is not a kangaroo.
If it is a kangaroo, then I avoid it.
I avoid kangaroos.
Etc.

March 2006: A Lewis Carroll puzzle

2006-04-14T09:54:00.000-04:00

In addition to being the author of classic children's books such as Alice in Wonderland, Lewis Carroll was an accomplished logician. Below is one of Lewis Carroll's famous logic puzzles. Use all ten statements of the riddle to draw the logical conclusion.

The only animals in this house are cats.
Every animal is suitable for a pet, that loves to gaze at the moon.
When I detest an animal, I avoid it.
No animals are carnivorous, unless they prowl at night.
No cat fails to kill mice.
No animals ever take to me, except what are in this house.
Kangaroos are not suitable for pets.
None but carnivora kill mice.
I detest animals that do not take to me.
Animals, that prowl at night, always love to gaze at the moon.

Hint: turn this into a puzzle of symbolic logic. For example, here is how you could solve another of Carroll's puzzles:

Things sold in the street are of no great value.
Nothing but rubbish can be had for a song.
Eggs of the Great Auk are very valuable.
It is only what is sold in the street that is really rubbish.

S: it is sold in the streets
V: it is valuable
R: it is rubbish
I: it is inexpensive
E: it is the egg of the Great Auk

Symbolically the puzzle becomes:

S --> ~V (or, using the contrapositive, V --> ~S)
I --> R (or ~R --> ~I)
E --> V (or ~V --> ~E)
R -->S (~S --> ~R)

Stringing these together we obtain: I --> R --> S --> ~V --> ~E

Thus the conclusion is I --> ~E or E --> ~I which can be written as:
If it is inexpensive then it is not the egg of a Great Auk.
If it is the egg of the Great Auk then it is not inexpensive.
Nothing inexpensive can be the egg of the Great Auk.
No egg of the Great Auk is inexpensive.
An egg of the Great Auk cannot be had for a song.
Etc.