20 to 3

icosagon puzzle

Prove that, if we choose nine vertices of a regular icosagon, some three of these will form an isosceles triangle.

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Label the vertices, in order, A, B, C, D, A, B, C, D, etc. Now each label defines a regular pentagon. Of our nine chosen points, at least three must bear the same label (two of each label would account for only eight points). So we have three vertices of a regular pentagon. Since there are only two possible distances between a pentagon’s vertices, the distances between two pairs of these points must be equal. By Polish mathematician Paul Vaderlind.

December 17, 2013December 27, 2013 | Puzzles

Black and White

ferber chess problem

By Eugène Ferber. White to mate in two moves.

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1. Qh1! sets up a half-pin on the long white diagonal. No matter how Black plays, White can mate with the rook. From the Times, July 11, 1902.

December 13, 2013December 20, 2013 | Puzzles

The Froggy Problem

Speaking of Lewis Carroll — and further to Wednesday’s logic exercise — here’s the king of all Carroll’s logic problems. What’s the strongest conclusion that can be drawn from these premises?

When the day is fine, I tell Froggy “You’re quite the dandy, old chap!”
Whenever I let Froggy forget that 10 pounds he owes me, and he begins to strut about like a peacock, his mother declares “He shall not go out a-wooing!”
Now that Froggy’s hair is out of curl, he has put away his gorgeous waistcoat.
Whenever I go out on the roof to enjoy a quiet cigar, I’m sure to discover that my purse is empty.
When my tailor calls with his little bill, and I remind Froggy of that 10 pounds he owes me, he does not grin like a hyena.
When it is very hot, the thermometer is high.
When the day is fine, and I’m not in the humor for a cigar, and Froggy is grinning like a hyena, I never venture to hint that he’s quite the dandy.
When my tailor calls with his little bill and finds me with an empty pocket, I remind Froggy of that 10 pounds he owes me.
My railway shares are going up like anything!
When my purse is empty, and when, noticing that Froggy has got his gorgeous waistcoat on, I venture to remind him of that 10 pounds he owes me, things are apt to get rather warm.
Now that it looks like rain, and Froggy is grinning like a hyena, I can do without my cigar.
When the thermometer is high, you need not trouble yourself to take an umbrella.
When Froggy has his gorgeous waistcoat on, but is not strutting about like a peacock, I betake myself to a quiet cigar.
When I tell Froggy that he’s quite a dandy, he grins like a hyena.
When my purse is tolerably full, and Froggy’s hair is one mass of curls, and when he is not strutting about like a peacock, I go out on the roof.
When my railways shares are going up, and when it’s chilly and looks like rain, I have a quiet cigar.
When Froggy’s mother lets him go a-wooing, he seems nearly mad with joy, and puts on a waistcoat that is gorgeous beyond words.
When it is going to rain, and I am having a quiet cigar, and Froggy is not intending to go a-wooing, you had better take an umbrella.
When my railway shares are going up, and Froggy seems nearly mad with joy, that is the time my tailor always chooses for calling with his little bill.
When the day is cool and the thermometer low, and I say nothing to Froggy about his being quite the dandy, and there’s not the ghost of a grin on his face, I haven’t the heart for my cigar!

Unfortunately, Carroll died before he was able to publish the solution — but he warned that it contains “a beautiful ‘trap.'”

December 12, 2013December 11, 2013 | Puzzles

Made to Order

1. Find an expression for the number 1 that uses each of the digits 0-9 once.
2. Do the same for the number 100.
3. Write 31 using only the digit 3 five times.
4. Express 11 with three 2s.
5. Express 10 with two 2s.
6. Express 1 with three 8s.
7. Express 5 with two 2s.

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1 = 35/70 + 148/296 100 = 50 + 49 + 1/2 + 38/76 31 = 3³ + 3 + 3/3 11 = 22/2 10 = 2/.2 1 = 8^8-8 (Thanks, Ron.)

December 11, 2013December 20, 2013 | Puzzles

Animal Behavior

A logic exercise by Lewis Carroll. What conclusion can be drawn from these premises?

Animals are always mortally offended if I fail to notice them.
The only animals that belong to me are in that field.
No animal can guess a conundrum unless it has been properly trained in a Board-School.
None of the animals in that field are badgers.
When an animal is mortally offended, it rushes about wildly and howls.
I never notice any animal unless it belongs to me.
No animal that has been properly trained in a Board-School ever rushes about wildly and howls.

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No badger can guess a conundrum.

December 10, 2013December 20, 2013 | Puzzles

Query

How are the following numbers arranged?

0, 2, 3, 6, 7, 1, 9, 4, 5, 8

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In reverse alphabetical order.

December 9, 2013December 13, 2013 | Puzzles

Dog’s Mead

dog's mead puzzle

This puzzle was devised by W.T. Williams. The goal is to discover the age of Father Dunk’s mother-in-law (2 down), but the clues contain so much cross-reference and the grid so many interlocking solutions that practically the whole puzzle must be completed to find it.

The year is 1939. There are 20 shillings in a pound; 4840 square yards in an acre; a quarter of an acre in a rood; and 1760 yards in a mile.

ACROSS

1. Area in square yards of Dog’s Mead
5. Age of Martha, Father Dunk’s aunt
6. Difference in yards between length and breadth of Dog’s Mead
7. Number of roods in Dog’s Mead times 8 down
8. The year the Dunks acquired Dog’s Mead
10. Father Dunk’s age
11. Year of Mary’s birth
14. Perimeter in yards of Dog’s Mead
15. Cube of Father Dunk’s walking speed in miles per hour
16. 15 across minus 9 down

DOWN

1. Value in shillings per rood of Dog’s Mead
2. Square of the age of father Dunk’s mother-in-law
3. Age of Mary, father Dunk’s other daughter
4. Value in pounds of Dog’s Mead
6. Age of Ted, father Dunk’s son, who will be twice the age of his sister Mary in 1945
7. Square of the breadth of Dog’s Mead
8. Time in minutes it takes Father Dunk to walk 4/3 times round Dog’s Mead
9. The number which, when multiplied by 10 across, gives 10 down
10. See 9 down
12. Addition of the digits of 10 down plus 1
13. Number of years Dog’s Mead has been in the Dunk family

Hint: Start with 15 across, and keep your wits about you — most of the clues require some sort of insight or intelligent narrowing of the possible solutions; there’s very little mechanical plugging of numbers. With enough careful, dogged reasoning, it’s possible to complete the entire grid, but it’s stupendously hard.

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Chris Adams sent me the guided solution below. Throughout, L is the length, B the breadth, and all units are in yards/square yards unless otherwise noted. Both 8 and 11 Across are years before 1939; since both are four digits, both must begin with a 1. The only two-digit cubes are 27 and 64, so 15 Across must be one of these numbers, and hence Father Dunk’s walking speed in miles per hour is either three or four. Let S denote his speed in miles per hour; then 88S/3 is his speed in yards per minute. And thus the time it takes him to walk the perimeter is 2(L+B)/(88S/3) = 3(L+B)/(44S), and so 4/3 this time (8 down) is (L+B)/(11S), and this is between 10 and 19 inclusive. Now, the perimeter is less than 1000 (since 14 Across is a three-digit number), so L+B, which is one half the perimeter, is less than 500. So L+B is equal to 11 times S times some number between 10 and 19, and this is lower than 500. Regardless of the value of S, 16 to 19 give us a value greater than 500, so the time is between 10 and 15 minutes. Look at 9 Down, 10 Across, and 10 Down. Since 9 Down ends in a 1, and 9 Down times 10 Across equals 10 Down, this means that 10 Across and 10 Down end in the same number. This also means that 8 Down ends with the same number that 15 Across begins with. And 15 Across begins with either 2 or 6, while 8 Down ends with 0, 1, 2, 3, 4, or 5. Hence that number must be a 2, so it takes 12 minutes, and the speed is 3 mph, which yields 27 as a cube. We know the speed and the time it takes to go 4/3 of the way around the perimeter, so we can solve for the perimeter and get that the perimeter is 792 yards. Alternatively, one can get that from L+B = 11×3×12, and P=2(L+B). Now, 12 Down is the sum of the digits in 10 Down, plus 1. This must be 19, as any other two-digit number ending in 9 is greater than any possible sum of the digits. Hence 10 Down must be 792, 882, or 972, making 10 Across 72, 82, or 92, respectively. Only in the first case does 10 Across divide 10 Down, since 792 equals 72×11. Hence 10 Down is 792, 10 Across is 72, and 9 Down is 11. This also makes 16 Across 16. 10 Down is a square and ends in 76, so B ends in either a 4 or a 6. Also, since L-B is less than 100, and L+B equals 396, B must be between 150 and 190, so there are eight possibilities for B. Of these, only 174 and 176 end in 76. However, 174 ends in 276, meaning Mary was born in the 1200s, which is plainly false (her father is only 72). Hence 76, which ends in 976, is the correct answer. The breadth B is 176, so the length L is 220, and this makes the area (1 Across) 38720. It also makes the number of roods 32, so 7 Across is 384. Also, 6 Across is 44, and so 6 Down is 48, which is Ted’s age. In six years, he’ll be 54, which is twice Mary’s age then. So she’ll be 27 then, which makes her 21 (3 Down) now. It also means she was born in 1918 (11 Across). Putting 8 Across and 13 Down together, we can fill in the blanks and show that the family acquired the property in 1110, so they’ve owned it for 829 years. 2 Down is a square number ending in 1, so the original number ends in 1 or 9. It is also in the 7000s, and since 80 squared is 6400 and 90 squared is 8100, this makes Father Dunk’s mother-in-law 81 or 89. 81 doesn’t work, but 89 squared is 7921, and so the mother-in-law is 89. 1 Down and 4 Down are still open, but we know that 1 Down is 5/8 (or 20/32, since there are 32 roods) of 4 Down. And 1 Down is between 300 and 400, so 4 Down must be between 480 and 640. Since 4 Down ends in 44, the only choice is 544, which makes 1 Down 340. This puzzle inspired an even harder one by CUNY math professor William Sit.

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dog's mead solution

Chris Adams sent me the guided solution below. Throughout, L is the length, B the breadth, and all units are in yards/square yards unless otherwise noted.

Both 8 and 11 Across are years before 1939; since both are four digits, both must begin with a 1.
The only two-digit cubes are 27 and 64, so 15 Across must be one of these numbers, and hence Father Dunk’s walking speed in miles per hour is either three or four. Let S denote his speed in miles per hour; then 88S/3 is his speed in yards per minute. And thus the time it takes him to walk the perimeter is 2(L+B)/(88S/3) = 3(L+B)/(44S), and so 4/3 this time (8 down) is (L+B)/(11S), and this is between 10 and 19 inclusive.
Now, the perimeter is less than 1000 (since 14 Across is a three-digit number), so L+B, which is one half the perimeter, is less than 500. So L+B is equal to 11 times S times some number between 10 and 19, and this is lower than 500. Regardless of the value of S, 16 to 19 give us a value greater than 500, so the time is between 10 and 15 minutes.
Look at 9 Down, 10 Across, and 10 Down. Since 9 Down ends in a 1, and 9 Down times 10 Across equals 10 Down, this means that 10 Across and 10 Down end in the same number. This also means that 8 Down ends with the same number that 15 Across begins with. And 15 Across begins with either 2 or 6, while 8 Down ends with 0, 1, 2, 3, 4, or 5. Hence that number must be a 2, so it takes 12 minutes, and the speed is 3 mph, which yields 27 as a cube.
We know the speed and the time it takes to go 4/3 of the way around the perimeter, so we can solve for the perimeter and get that the perimeter is 792 yards. Alternatively, one can get that from L+B = 11×3×12, and P=2(L+B).
Now, 12 Down is the sum of the digits in 10 Down, plus 1. This must be 19, as any other two-digit number ending in 9 is greater than any possible sum of the digits. Hence 10 Down must be 792, 882, or 972, making 10 Across 72, 82, or 92, respectively. Only in the first case does 10 Across divide 10 Down, since 792 equals 72×11. Hence 10 Down is 792, 10 Across is 72, and 9 Down is 11. This also makes 16 Across 16.
10 Down is a square and ends in 76, so B ends in either a 4 or a 6. Also, since L-B is less than 100, and L+B equals 396, B must be between 150 and 190, so there are eight possibilities for B. Of these, only 174 and 176 end in 76. However, 174 ends in 276, meaning Mary was born in the 1200s, which is plainly false (her father is only 72). Hence 76, which ends in 976, is the correct answer.
The breadth B is 176, so the length L is 220, and this makes the area (1 Across) 38720. It also makes the number of roods 32, so 7 Across is 384. Also, 6 Across is 44, and so 6 Down is 48, which is Ted’s age. In six years, he’ll be 54, which is twice Mary’s age then. So she’ll be 27 then, which makes her 21 (3 Down) now. It also means she was born in 1918 (11 Across).
Putting 8 Across and 13 Down together, we can fill in the blanks and show that the family acquired the property in 1110, so they’ve owned it for 829 years.
2 Down is a square number ending in 1, so the original number ends in 1 or 9. It is also in the 7000s, and since 80 squared is 6400 and 90 squared is 8100, this makes Father Dunk’s mother-in-law 81 or 89. 81 doesn’t work, but 89 squared is 7921, and so the mother-in-law is 89.
1 Down and 4 Down are still open, but we know that 1 Down is 5/8 (or 20/32, since there are 32 roods) of 4 Down. And 1 Down is between 300 and 400, so 4 Down must be between 480 and 640. Since 4 Down ends in 44, the only choice is 544, which makes 1 Down 340.

This puzzle inspired an even harder one by CUNY math professor William Sit.

December 7, 2013December 13, 2013 | Puzzles

Black and White

carpenter chess problem

By Paud D. Carpenter. White to mate in two moves.

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1. Kh3! Kf4 2. Rf6#. From The Tablet, 1920.

December 6, 2013December 13, 2013 | Puzzles

Emptied Nest

Solve for x:

emptied nest problem

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Square both sides: Now we can substitute the original equation, giving x + 2 = 4 and x = 2.

December 5, 2013December 13, 2013 | Puzzles

Black and White

palitzsch chess problem

By Friedrich Martin Palitzsch. White to mate in two moves.

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1. Re1! If Black takes the rook then 2. Bc3 is mate; if he tries to flee then the queen finishes him from e2 or d1. From Deutsches Wochenschach, July 22, 1917.

December 1, 2013December 6, 2013 | Puzzles