A puzzle by Yoshinao Katagiri: A boy and a girl played rock paper scissors 10 times. Altogether the boy played rock three times, scissors six times, and paper once, and the girl played rock twice, scissors four times, and paper four times (though, in each case, the order of these plays is unknown). There were no ties. Who won?
This solution is by Gustavo Krimker of Universidad CAECE, Buenos Aires.
The distance between the first and fifth points is 19. That means that 19 is also the sum of each of three pairs of intermediate distances:
(the distance from the first point to the second point) + (the distance from the second point to the fifth point) = 19
(the distance from the first point to the third point) + (the distance from the third point to the fifth point) = 19
(the distance from the first point to the fourth point) + (the distance from the fourth point to the fifth point) = 19
Looking over the list of distances we’ve been given, the only pairs that add to 19 are (2, 17) and (4, 15). So those pairs correspond to two of the bulleted sums above. What’s the third pair? It must be (7, k) or (8, k). (It can’t be (5, k) or (k, 13) because we know that k falls between 8 and 13 and so those pairs can’t be made to total 19.) And if the third case is (7, k) or (8, k), then k must be 11 or 12.
Now, 17 is the second-greatest distance on the list; that means it’s the distance between one of the end points in the row of five and the penultimate point on the opposite side. Either way, that means that 17 is the sum of two pairs of intermediate distances, using the same reasoning as above. One of those pairs is (4, 13), and the other must be (5, k), (7, k), or (8, k). That means that k must be 12, 10, or 9. The only one of these that matches a candidate above is 12.
07/20/2021 UPDATE: This is cooked. 1. Qd6+ works just as well and is simpler. (Thanks, Travis.)
07/20/2021 UPDATE: Blimey, doubly cooked! 1. Rc3 works too. (Thanks, Chris.) And the pawn seems to have no role in any of this, not even in Carney’s intended solution. Not sure how Wadworth can have chosen this without noticing the errors — but I guess I just did the same thing!
You’re in a dark room. The only light comes from an old LED digital alarm clock with four seven-segment displays. The time is displayed in 24-hour format, HH:MM (no seconds), and the leading digit is blank if not used. How much time passes between the room’s darkest state and its lightest?
07/16/2021 UPDATE: Properly speaking the interval is one minute less than that, because it extends from the last moment of 1:11 to the first of 20:08. Also, given my wording, the answer might equally be 5 hours 3 minutes, as that’s the interval “between” 20:08 and 1:11. (Thanks, Michael and Joe.)
A problem from the January 1990 issue of Quantum: Forty-one rooks are placed on a 10 × 10 chessboard. Prove that some five of them don’t attack one another. (Two rooks attack one another if they occupy the same row or column.)
Roll the board into a cylinder and paint each of its diagonals a different color. These 10 regions contain 41 = 4 × 10 + 1 rooks, so at least one region must contain 5 rooks. And rooks on the same diagonal don’t attack one another.
(Alexander Soifer and Edward Lozansky, “Pigeons in Every Pigeonhole,” Quantum 1:1 [January 1990], 25-28, 32.)
Steve, Tony, and Bruce have a plate of 1,000 cookies to share. They decide to share them in the following way: beginning with Steve, each of them in turn takes as many cookies as he likes (they must take an integer amount, greater than or equal to 1), and then passes the plate clockwise (with Tony sitting to Steve’s left, and Bruce sitting to Tony’s left). Nobody wants to feel like he hogged too many cookies, so they all want to avoid being the player at the end who has taken the most cookies. Additionally, nobody wants to feel cheated by finishing with the fewest cookies. Finally, given that the previous two conditions are definitely met, or definitely cannot be met, each player would like to maximize the number of cookies he eats. The players’ objectives can be summarized as follows:
Have one player who has eaten more cookies than you, and one player who has eaten fewer cookies than you.
Eat as many cookies as possible.
Objective #1 takes infinite priority over Objective #2. Assuming that all players are perfectly rational, that they are all aware of each other’s rationality and objectives, and that they cannot communicate with each other in any way, how many cookies should Steve take to ensure he meets both objectives and how many cookies will Tony and Bruce take if Steve takes the winning amount?
First, suppose Steve takes 335 cookies. We will argue that Tony can then win by taking 334 cookies. If Tony does take 334, Bruce will be left with a plate of 331 cookies to take from. No matter how Bruce plays from here, he cannot meet objective #1, because he will need at least 335 cookies to do so, and has only 331 to take from. Therefore, Bruce will maximize the number of cookies he eats, and take the remaining 331. Tony finishes with 334, the middle amount, so he meets objective #1, justifying his decision to take 334 cookies. (Note that if Tony had taken more than 334, Bruce would still have taken all the remaining cookies, but Tony would not have met objective #1, so taking 334 is in fact Tony’s best play.)
Therefore, Steve will fail objective #1 if he takes 335 cookies. Furthermore, we can repeat the argument above if Steve takes more than 335 cookies, by always having Tony take 1 fewer cookie than Steve (or all remaining cookies, if this amount is less than the number Steve took).
Now suppose instead that Steve takes 334 cookies. We will show that Tony cannot take any amount of cookies resulting in him (Tony) meeting objective #1. First, note that if Tony takes 333 or more cookies, Bruce will be unable to meet objective #1 (he will need at least 334, but there are at most 333 remaining), and so will take all remaining cookies, resulting in a “loss” for Tony.
Now suppose instead that Tony takes x cookies, where x ≤ 332. Then there will be 666 – x cookies remaining. Suppose Bruce takes 333 cookies. This leaves 333 – x cookies on the plate. Even if Steve takes only 1 cookie each turn from now on, Bruce can ensure that he will have fewer cookies than Steve by also taking 1. Suppose Steve takes s cookies. Then on Tony’s next turn, he has 333 – x – s cookies to take from. Even if Tony takes all remaining cookies, total over his two turns, he will have taken 333 – x – s + x = 333 – s, which is at most 332 (since s ≥ 1), so Tony will definitely finish with fewer cookies than Bruce. Thus, if Tony takes 332 or fewer cookies, Bruce can win by taking 333 cookies, since this guarantees he will finish with less than Steve and more than Tony.
Therefore, if Steve takes 334 cookies, Tony cannot meet objective #1, and so will take as many cookies as possible, leaving none for Bruce. This leaves Steve with the middle amount of cookies, so taking 334 cookies is in fact a winning strategy for Steve. As shown above, Steve cannot do any better than this.
I just ran across this anecdote by Jason Rosenhouse in Notices of the American Mathematical Society. In a middle-school algebra class Rosenhouse’s brother was given this problem:
There are some horses and chickens in a barn, fifty animals in all. Horses have four legs while chickens have two. If there are 130 legs in the barn, then how many horses and how many chickens are there?
The normal solution is straightforward, but Rosenhouse’s brother found an alternative that’s even easier: “You just tell the horses to stand on their hind legs. Now there are fifty animals each with two legs on the ground, accounting for one hundred legs. That means there are thirty legs in the air. Since every horse has two legs in the air, we find that there are fifteen horses, and therefore thirty-five chickens.”
(Jason Rosenhouse, “Book Review: Bicycle or Unicycle?: A Collection of Intriguing Mathematical Puzzles,” Notices of the American Mathematical Society, 67:9 [October 2020], 1382-1385.)