You have nine coins and a balance scale. One of the coins is lighter than the others. Is it possible to identify it in only two weighings?
Henrietta wants a yacht. Her parents think she’s too young. Like all rich people, they settle disagreements by playing competitive lawn tennis.
Henrietta must play three singles matches against her parents. If she wins two matches in a row she gets the yacht. Her mother is a better player than her father. Should she play mother-father-mother or father-mother-father?
An urn contains 75 white balls and 150 black ones. A pile of black ones is also available.
The following two-step operation occurs repeatedly. First we withdraw two balls at random from the urn, then:
- If both are black, we put one of them back in the urn and throw the other away.
- If one is black and the other white, we put the white one back and throw the black one away.
- If both are white, we throw both away and put a black ball from the pile into the urn.
Because the urn loses a ball at each step, eventually it will contain a single ball. What color is that ball?
Inventing the transformer wasn’t challenge enough for Ottó Bláthy — in his spare time he invented puzzles like this:
White to self-mate in 342 moves.
1.c4+ Kc1 2.Qd2+ Kb1 3.Qxd1+ Bc1 4.Qb3+ Bb2 5.Qd3+ Kc1 6.Qe3+ Kb1 7.Nd2+ Kc1 8.Nf1+ Kb1 9.Qe4+ Kc1 10.Qf4+ Kb1 11.Qf5+ Kc1 12.Qg5+ Kb1 13.Qg6+ Kc1 14.Bh6+ Rxh6 15.Qxh6+ Kb1 16.Qg6+ Kc1 17.Qg5+ Kb1 18.Qf5+ Kc1 19.Qf4+ Kb1 20.Qe4+ Kc1 21.Qe3+ Kb1 22.Nd2+ Kc1 23.Ne4+ Kb1 24.Qd3+ Kc1 25.Qd2+ Kb1 26.Qd1+ Bc1 27.Qd3+ Kb2 28.Qd4+ Kb1 29.Nc3+ Kb2 30.Nxb5+ Kb1 31.Nc3+ Kb2 32.Ne4+ Kb1 33.Qd3+ Kb2 34.Qc3+ Kb1 35.Qb3+ Bb2 36.Qd3+ Kc1 37.Qe3+ Kb1 38.Nd2+ Kc1 39.Nf1+ Kb1 40.Qd3+ Kc1 41.Qd2+ Kb1 42.Qd1+ Bc1 43.Qd3+ Kb2 44.Qd4+ Kb1 45.Qe4+ Kb2 46.Qe5+ Kb1 47.Qf5+ Kb2 48.Qf6+ Kb1 49.Qg6+ Kb2 50.Qg7+ Kb1 51.Qxb7+ Bb2 52.Qh7+ Kc1 53.Qh6+ Kb1 54.Qg6+ Kc1 55.Qg5+ Kb1 56.Qf5+ Kc1 57.Qf4+ Kb1 58.Qe4+ Kc1 59.Qe3+ Kb1 60.Qd3+ Kc1 61.Qd2+ Kb1 62.Qd1+ Bc1 63.g4 Kb2 64.Qd4+ Kb1 65.Qe4+ Kb2 66.Qe5+ Kb1 67.Qf5+ Kb2 68.Qf6+ Kb1 69.Qg6+ Kb2 70.Qg7+ Kb1 71.Qb7+ Bb2 72.Qh7+ Kc1 73.Qh6+ Kb1 74.Qg6+ Kc1 75.Qg5+ Kb1 76.Qf5+ Kc1 77.Qf4+ Kb1 78.Qe4+ Kc1 79.Qe3+ Kb1 80.Qd3+ Kc1 81.Qd2+ Kb1 82.Qd1+ Bc1 83.Kf8 Kb2 84.Qd4+ Kb1 85.Qe4+ Kb2 86.Qe5+ Kb1 87.Qf5+ Kb2 88.Qf6+ Kb1 89.Qg6+ Kb2 90.Qg7+ Kb1 91.Qb7+ Bb2 92.Qh7+ Kc1 93.Qh6+ Kb1 94.Qg6+ Kc1 95.Qg5+ Kb1 96.Qf5+ Kc1 97.Qf4+ Kb1 98.Qe4+ Kc1 99.Qe3+ Kb1 100.Qd3+ Kc1 101.Qd2+ Kb1 102.Qd1+ Bc1 103.Ke8 Kb2 104.Qd4+ Kb1 105.Qe4+ Kb2 106.Qe5+ Kb1 107.Qf5+ Kb2 108.Qf6+ Kb1 109.Qg6+ Kb2 110.Qg7+ Kb1 111.Qb7+ Bb2 112.Qh7+ Kc1 113.Qh6+ Kb1 114.Qg6+ Kc1 115.Qg5+ Kb1 116.Qf5+ Kc1 117.Qf4+ Kb1 118.Qe4+ Kc1 119.Qe3+ Kb1 120.Qd3+ Kc1 121.Qd2+ Kb1 122.Qd1+ Bc1 123.Kd8 Kb2 124.Qd4+ Kb1 125.Qe4+ Kb2 126.Qe5+ Kb1 127.Qf5+ Kb2 128.Qf6+ Kb1 129.Qg6+ Kb2 130.Qg7+ Kb1 131.Qb7+ Bb2 132.Qh7+ Kc1 133.Qh6+ Kb1 134.Qg6+ Kc1 135.Qg5+ Kb1 136.Qf5+ Kc1 137.Qf4+ Kb1 138.Qe4+ Kc1 139.Qe3+ Kb1 140.Qd3+ Kc1 141.Qd2+ Kb1 142.Qd1+ Bc1 143.Kc8 Kb2 144.Qd4+ Kb1 145.Qe4+ Kb2 146.Qe5+ Kb1 147.Qf5+ Kb2 148.Qf6+ Kb1 149.Qg6+ Kb2 150.Qg7+ Kb1 151.Qb7+ Bb2 152.Qh7+ Kc1 153.Qh6+ Kb1 154.Qg6+ Kc1 155.Qg5+ Kb1 156.Qf5+ Kc1 157.Qf4+ Kb1 158.Qe4+ Kc1 159.Qe3+ Kb1 160.Qd3+ Kc1 161.Qd2+ Kb1 162.Qd1+ Bc1 163.Kb8 Kb2 164.Qd4+ Kb1 165.Qe4+ Kb2 166.Qe5+ Kb1 167.Qf5+ Kb2 168.Qf6+ Kb1 169.Qg6+ Kb2 170.Qg7+ Kb1 171.Qb7+ Bb2 172.Qh7+ Kc1 173.Qh6+ Kb1 174.Qg6+ Kc1 175.Qg5+ Kb1 176.Qf5+ Kc1 177.Qf4+ Kb1 178.Qe4+ Kc1 179.Qe3+ Kb1 180.Qd3+ Kc1 181.Qd2+ Kb1 182.Qd1+ Bc1 183.Ka7 Kb2 184.Qd4+ Kb1 185.Qe4+ Kb2 186.Qe5+ Kb1 187.Qf5+ Kb2 188.Qf6+ Kb1 189.Qg6+ Kb2 190.Qg7+ Kb1 191.Qb7+ Bb2 192.Qh7+ Kc1 193.Qh6+ Kb1 194.Qg6+ Kc1 195.Qg5+ Kb1 196.Qf5+ Kc1 197.Qf4+ Kb1 198.Qe4+ Kc1 199.Qe3+ Kb1 200.Qd3+ Kc1 201.Qd2+ Kb1 202.Qd1+ Bc1 203.Kxa6 Kb2 204.Qd4+ Kb1 205.Qe4+ Kb2 206.Qe5+ Kb1 207.Qf5+ Kb2 208.Qf6+ Kb1 209.Qg6+ Kb2 210.Qg7+ Kb1 211.Qb7+ Bb2 212.Qh7+ Kc1 213.Qh6+ Kb1 214.Qg6+ Kc1 215.Qg5+ Kb1 216.Qf5+ Kc1 217.Qf4+ Kb1 218.Qe4+ Kc1 219.Qe3+ Kb1 220.Qd3+ Kc1 221.Qd2+ Kb1 222.Qd1+ Bc1 223.Ka7 Kb2 224.Qd4+ Kb1 225.Qe4+ Kb2 226.Qe5+ Kb1 227.Qf5+ Kb2 228.Qf6+ Kb1 229.Qg6+ Kb2 230.Qg7+ Kb1 231.Qb7+ Bb2 232.Qh7+ Kc1 233.Qh6+ Kb1 234.Qg6+ Kc1 235.Qg5+ Kb1 236.Qf5+ Kc1 237.Qf4+ Kb1 238.Qe4+ Kc1 239.Qe3+ Kb1 240.Qd3+ Kc1 241.Qd2+ Kb1 242.Qd1+ Bc1 243.Kb8 Kb2 244.Qd4+ Kb1 245.Qe4+ Kb2 246.Qe5+ Kb1 247.Qf5+ Kb2 248.Qf6+ Kb1 249.Qg6+ Kb2 250.Qg7+ Kb1 251.Qb7+ Bb2 252.Qh7+ Kc1 253.Qh6+ Kb1 254.Qg6+ Kc1 255.Qg5+ Kb1 256.Qf5+ Kc1 257.Qf4+ Kb1 258.Qe4+ Kc1 259.Qe3+ Kb1 260.Qd3+ Kc1 261.Qd2+ Kb1 262.Qd1+ Bc1 263.a6 Kb2 264.Qd4+ Kb1 265.Qe4+ Kb2 266.Qe5+ Kb1 267.Qf5+ Kb2 268.Qf6+ Kb1 269.Qg6+ Kb2 270.Qg7+ Kb1 271.Qb7+ Bb2 272.Qh7+ Kc1 273.Qh6+ Kb1 274.Qg6+ Kc1 275.Qg5+ Kb1 276.Qf5+ Kc1 277.Qf4+ Kb1 278.Qe4+ Kc1 279.Qe3+ Kb1 280.Qd3+ Kc1 281.Qd2+ Kb1 282.Qd1+ Bc1 283.a7 Kb2 284.Qd4+ Kb1 285.Qe4+ Kb2 286.Qe5+ Kb1 287.Qf5+ Kb2 288.Qf6+ Kb1 289.Qg6+ Kb2 290.Qg7+ Kb1 291.Qb7+ Bb2 292.Qh7+ Kc1 293.Qh6+ Kb1 294.Qg6+ Kc1 295.Qg5+ Kb1 296.Qf5+ Kc1 297.Qf4+ Kb1 298.Qe4+ Kc1 299.Qe3+ Kb1 300.Qd3+ Kc1 301.Qd2+ Kb1 302.Qd1+ Bc1 303.a8=Q Kb2 304.Qd4+ Kb1 305.Qb7+ Bb2 306.Qh7+ Kc1 307.Qh6+ Kb1 308.Qd1+ Bc1 309.Ka7 Kb2 310.Qf6+ Kb1 311.Qb3+ Bb2 312.Qf5+ Kc1 313.Qf4+ Kb1 314.Qd1+ Bc1 315.Ka6 Kb2 316.Qe5+ Kb1 317.Qb3+ Bb2 318.Qe4+ Kc1 319.Qee3+ Kb1 320.Qd1+ Bc1 321.Ka5 Kb2 322.Qed4+ Kb1 323.Kb4 Qg1 324.Qb3+ Bb2 325.Qdd3+ Kc1 326.Qe3+ Kb1 327.Nd2+ Kc1 328.Ne4+ Kb1 329.Qed3+ Kc1 330.Qd2+ Kb1 331.Qdd1+ Qxd1 332.Qxd1+ Bc1 333.Qd3+ Kb2 334.Qd4+ Kb1 335.Nc3+ Kb2 336.Nb5+ Kb1 337.Qd3+ Kb2 338.Qc3+ Kb1 339.Qb3+ Bb2 340.Qd1+ Bc1 341.Nxa3+ Kb2 342.Qd2+ Bxd2#
This was typical for Bláthy — in one volume of his compositions, the shortest problem required mate in 30 moves.
Two lines that intersect at A are tangent to a circle at B and C. AB and AC are both 10. If a third tangent (green) touches the circle somewhere between B and C, what is the perimeter of the triangle formed by the three lines?
ABC is a right triangle, and D is the midpoint of its hypotenuse.
Prove that BD is half the length of AC.
What is the product of this series?
(x – a) (x – b) (x – c) … (x – z)
This yields to an insight, so I’ll withhold the answer.
From J. Newton Friend, Numbers: Fun & Facts, 1954:
A boy jumped onto one end of a piece of tree trunk lying on the top of a hill. Now the log happened to be exactly 13 feet long, an unlucky omen for the youth, and the impact caused it to begin rolling down the hill. As it rolled, he managed to keep himself upright on top and slowly walked across the log to the other end which he reached just as the log came to rest at the bottom of the hill, 84 feet from where it began to roll.
The log was 2 feet in diameter. How far did the boy actually travel and how far would he have travelled had the log been 3 feet in diameter?
From a 1987 Hungarian math contest for 11-year-olds:
How can a 3 × 3 × 3 cube be divided into 20 cubes (not necessarily the same size)?
“I used to flatter myself that I would immediately be able to see through any problem that might be asked of an 11-year-old,” writes University of Waterloo mathematician Ross Honsberger. “I don’t take anything for granted anymore!”
Point A is the center of one square and the vertex of another. The side of each square is 4 inches. What is the area of the shaded region?
Here’s a long corridor with a moving walkway. Let’s race to the far end and back. We’ll both run at the same speed, but you run on the floor and I’ll run on the walkway, going “downstream” to the far end and “upstream” back to this point. Who will win?
This puzzle, by Les Marvin and Sherry Nolan, appeared in the Journal of Recreational Mathematics in 1977. “White to play in the adjoining diagram. If both players play optimally, will White win, lose, or draw?”
I don’t believe JRM ever published the solution. My stab: Either king is vulnerable to a check from the bishop file, and White will win a straight race. So I think Black must play defense. But if White attacks c7 with both knights and Black defends it doubly, then White can simply trade off all four knights (1. Nc7+ Nxc7 2. Nxc7+ Nxc7 bxc7) and the pawn will queen. So I think White wins.
This isn’t a very “mathematical” solution, but I can’t find a reliable alternative involving the parity of the knights’ moves, which seems to be what’s expected. Any ideas?
06/06/2014 UPDATE: A reader ran this position through a couple of strong chess engines and finds that it’s likely a draw — here’s one example:
[FEN “k6n/Pp4n1/1P6/8/8/6p1/1N4Pp/N6K w – – 0 1″]
1.Nd1 Nf7 2.Nc2 Ne5 3.Nce3 Nd7 4.Nd5 Nxb6 5.Nxb6+ Kxa7 6.Nc8+ Ka6 7.Ne3 b5 8.Nd6 b4 9.Ne4 Nh5 10.Nc2 Kb5 11.Nxb4 Kxb4 12.Nxg3 Nxg3+ 13.Kxh2 Nf1+ 14.Kh3 Ne3 15.g4 Kb3 16.g5 Nd5 17.g6 Nf4+ 18.Kg3 Nxg6
There doesn’t seem to be a sure way for either side to reach a win. I suspect that Marvin and Nolan thought otherwise, but they were writing in 1977, without the benefit of computer analysis. Without a published solution, we can’t be sure.
Here is a curious problem. We may safely assume that you had two parents; each of your parents had two parents, so that you had four grandparents. Arguing along similar lines you must have had eight great grandparents and so on. Assuming an average of three generations per century the number of your ancestors since the Christian Era began must have been nearly 1 trillion–
1,000,000,000,000,000,000 or 1018
This is vastly more people than have ever lived on the Earth. What can we do about it?
— J. Newton Friend, Numbers: Fun & Facts, 1954
This one is slippery, so watch it closely.
A poor old lady, with little money and plenty of time, sat quietly one day trying to devise a plan for making a little change. She finally came up with a very clever idea. Taking an old necklace, which she knew was worth only $4, she went to a pawnshop and pawned it for $3. Then, on a street corner, she started a friendly acquaintance with a young man, finally persuading him to buy the pawnticket for only $2. Now, she had $5 altogether and thus had made $1 profit. The pawnbroker wasn’t out any money since he paid only $3 for a $4 item, and the young man paid only $2 to get the $4 necklace. Who lost?
— Raymond F. Lausmann, Fun With Figures, 1965
From Lewis Carroll:
I don’t know if you are fond of puzzles, or not. If you are, try this. … A gentleman (a nobleman let us say, to make it more interesting) had a sitting-room with only one window in it–a square window, 3 feet high and 3 feet wide. Now he had weak eyes, and the window gave too much light, so (don’t you like ‘so’ in a story?) he sent for the builder, and told him to alter it, so as only to give half the light. Only, he was to keep it square–he was to keep it 3 feet high–and he was to keep it 3 feet wide. How did he do it? Remember, he wasn’t allowed to use curtains, or shutters, or coloured glass, or anything of that sort.
A printer prints a sentence in a monospaced font. It inserts a space after the concluding period and then prints the same sentence again. It continues in this way until it has filled the page, running the sentences together into one long paragraph. The sentence is shorter than a full line, and no words are hyphenated. Prove that the finished page will always include a full column of blank spaces.
A groaner from Clark Kinnaird’s Encyclopedia of Puzzles and Pastimes (1946):
“A farmer had 3 3/7 haystacks in one field and 5 4/9 haystacks in another field. He put them all together. How many did he have then?”
I’ll withhold the answer.
This is a story of four brothers. Billy owed a dollar to Jerry. Jerry owed a dollar to Tommy, and Tommy owed a dollar to Billy. The three of them met one day at a family picnic. Being brothers and good friends, none wished to hound the other about his debt. Vincent, the fourth brother, arrived at the picnic with some beer. While he was busily unloading the truck, Billy walked over, unnoticed, and quietly asked Vincent for a loan of a dollar, which Vincent gladly gave to him. Billy then ambled over to Jerry and paid him the dollar he owed him; then Jerry paid Tommy the dollar he owed to him; Tommy then went over to Billy and paid him the dollar he owed him. Billy then walked back to Vincent and paid him back his dollar. All old debts were paid. Simple, isn’t it?
— Raymond F. Lausmann, Fun With Figures, 1965
A tangram paradox from Sam Loyd’s Eighth Book of Tan (1903). Each of these cups was composed using the same seven geometric shapes. But the first cup is whole, and the others contain vacancies of different sizes.
“Of course it is a fallacy, a paradox, or an optical illusion, for you will say the feat is impossible!” But how is it done?