Denote by S the sum of all the numbers that remain on the blackboard. At the start this sum is S = 1 + 2 + … + 2n = n(2n + 1), which is odd. Each step reduces S by 2 min(a, b), which is an even number. Thus the parity of S never changes. Since it started odd, it will end odd.

From Arthur Engel’s Problem-Solving Strategies, 2003.

09/01/2024 UPDATE: Readers Seth Cohen and Robert Filman point out a simpler solution. In Seth’s words:

Call the number of odd numbers on the board n. At the start, n is odd.

• If a and b are even, so is a–b, so you don’t lose or gain any odd numbers, so n is still odd.
• If a and b are odd, a–b is even. So you lose two odd numbers and gain none, and n remains odd.
• If a and b are one odd and one even, a–b is odd. You lose an odd number and gain an odd number. Thus, n stays odd.

No matter what you do, n is always odd, so once you’re down to one number, it’ll be odd.

Thanks to both.

The water in a river doesn’t move with a single velocity — friction with the air and the channel slows some of the stream. Water moves fastest just below the surface in midstream:

As a raft is laden its draft increases, submerging it into the faster layers, which carry it along more quickly.

From Physical Paradoxes and Sophisms, 1987.

This solution, by Titu Zvonaru, is given in the February 2008 issue of Crux Mathematicorum:

Suppose that the most unfortunate player beat 3 others. Label the players so that this player is P₁ and that the players he beat follow him immediately in the lineup, here as P₂, P₃, and P₄. Since the minimum number of wins is 3, there must be some player P_k such that player P₂ beat player P_k and k is greater than 4 (otherwise player P₂ could have beaten only players P₃ and P₄, which would give him the minimum number of wins, contrary to our supposition). This means that player P_k must have beaten player P₁, and we can declare that A = P₁, B = P₂, and C = P_k.

The same argument holds for any minimum number of wins. (It wouldn’t work if the minimum were zero, but we’ve been told that’s not the case.)

09/02/2024 Reader Robert Filman offers an inductive proof:

The theorem is obviously true for n = 1, 2 and 3. (n=1 is not a possible situation, because everyone won a game, and with n=1, there are no games; with n=2, there’s only one game, so the loser of that game had no one to beat. With n=3, the conditions require A>B, B>C, and C>A.)

Assume hypothesis is true for 1, 2, 3, 4, … n – 1 players, and prove it true for n players.

Pick any player, A. Divide the players into the ones that beat A (set W) and the ones that lost to A (set L). L is not empty, because A beat someone.
If any player x in L beat any player y in W, we have a triangle <a, x, y>
If not, consider the set L. No one in L beat anyone outside of L. Every player in L’s win came against another player in L. But the size of L is at most n – 1. So by induction, the theorem is true for L.
Hence, the theorem is always true.

(Thanks, Robert.)

Solution:

This problem has 4 solutions and answers: 5-13, 5-22, 18-13, 18-22. Only the first solution is fully drawn on the solution illustration, but all four are explained below.

Explanation:

First, let’s re-draw the situation to be less messy! Once we’ve organized the leaves a bit, we see that there are 4 distinct groups of leaves which can only be accessed via one leaf: Group 1 is only accessible via leaf 23, Group 2 via leaf 3, Group 3 via leaf 10, and Group 4 via leaf 11.

Here we see the problem — once Arabella has entered a group of leaves, she cannot get out again without crossing the ‘access leaf’ again. For Group 1 this is okay, since Arabella can start on leaf 1, move to leaf 6, then leaf 14, then exit at leaf 23, and doesn’t need to go back in. Similarly, Group 4 contains the last leaf, so Arabella can enter the group from leaf 11, then go to leaf 2, leaf 8, leaf 21, leaf 15, and exit at leaf 26.

But for Group 2 and Group 3, Arabella must enter and exit the group. This is where the extra web comes in handy! If we connect a leaf in Group 2 and a leaf in Group 3, then Arabella can go from Group 1 to Group 2 using leaf 23, then Group 2 to Group 3 using the new web, and finally from Group 3 to Group 4 using leaf 7.

However, not just any leaves from Group 2 and Group 3 will work. Arabella must cover all of Group 2 before moving to Group 3, and cover all of Group 3 before going to Group 4. This still allows for a few different combinations: from Group 2 we can use leaf 5 or leaf 18; from Group 3 we can use leaf 13 or leaf 22.

Here are the different paths Arabella could take for each combination (there are minor variations in all of the paths, such as switching the order of leaves 6 and 14, which are unaffected by the solution choice, but we will not list those):

• Leaves 5 & 13 — 1, 6, 14, 23, 3, 18, 25, 12, 20, 17, 9, 5, 13, 16, 19, 24, 4, 22, 10, 7, 11, 2, 8, 21, 15, 26
• Leaves 5 & 22 — 1, 6, 14, 23, 3, 18, 25, 12, 20, 17, 9, 5, 22, 4, 24, 19, 16, 13, 10, 7, 11, 2, 8, 21, 15, 26

It’s natural to suppose that we’re looking at this from Black’s side of the board, but that can’t be the case — if it were, Black could not have swapped king and queen without moving a pawn. So we’re looking at the position from White’s point of view, and that means that Black’s pawns are moving down, not up, the board. So 1. Nd3# immediately ends the game:

96 × 7² = 4704

(Leonard J. Gordon, “Literary Cryptarithmetic by Computer,” Word Ways 23:2 [May 1990], 67-70.)

07/17/2024 UPDATE: Reader Dave Lawrence points out that other solutions are possible:

76 × 8² = 4864
77 × 7² = 3773
86 × 9² = 6966

(Thanks, Dave.)

07/18/2024 Wait, I jumped the gun — these alternatives lack the required one-to-one correspondence between letters and numbers. So I believe LeVasseur’s solution is unique. Thanks to everyone who pointed this out.

If Bob’s number isn’t a divisor of 50, then 50 must be the sum and Alice will know her number. If Bob’s number is 50, then 50 must be the product and again Alice should know her number. She doesn’t, so Bob’s number must be a proper divisor of 50, that is, 1, 2, 5, 10, or 25.

By the same reasoning, if Alice’s number isn’t a proper divisor of 50 then Bob should know his number. Also, after Alice’s utterance Bob can infer that his own number is a proper divisor of 50, following her thinking above. So each number is a proper divisor of 50.

Now, if 50 is the sum of the two numbers, then both numbers are 25 (as 25 + 25 is the only way to combine two of the candidate divisors to get 50). This means that if Alice’s number isn’t 25, then Bob can conclude that 50 is a product and can infer his own number. The fact that he can’t do this shows that Alice’s number must be 25 (and Bob’s number is either 2 or 25).

(Via Matvey Borodin et al., “It’s Common Knowledge,” Recreational Mathematics Magazine 6:12 [December 2019], 9-32.)

“You can bet your last dollar that you will see Garfield defeated by Hancock.”

07/09/2024 UPDATE: Oh, probably it’s “last bottom dollar” — I should have seen that. (Thanks, John.)

Divide the coins into nine piles — call them A₁, B₁, C₁, A₂, B₂, C₂, A₃, B₃, and C₃, with sizes 24, 1, 2, 24, 1, 2, 23, 2, 1, respectively. Now use the scale to show that A₁ + B₁, A₂ + B₂, and A₃ + B₃ balance one another, as do B₁ + C₁, B₂ + C₂, and B₃ + C₃. This shows the judge that there are three different ways in which the fake coins might be distributed:

• one fake coin in each of A₁, A₂, A₃ (sizes 24, 24, 23)
• one fake coin in each of B₁, B₂, B₃ (sizes 1, 1, 2)
• one fake coin in each of C₁, C₂, C₃ (sizes 2, 2, 1)

Each possibility involves 3 fake coins, not 2, and we’ve managed to convey that fact without revealing the character of any particular coin.

(Nicholas Diaco and Tanya Khovanova, “Privacy and Counterfeit Coins,” Mathematical Intelligencer 38:3 [September 2016], 6-10.)