By Josh Feldman

Every year, I give dozens of interviews for my alma mater, a way-too-tough-to-get-into university in Middlesex County, Massachusetts. My friends keep asking me why I bother giving interviews, as it’s a thankless job where way too many students (far more qualified than I ever was) get rejected every March. I tell everyone that there’s one big reason why I do it—so I can see what the cool kids are up to these days. Without any kids of my own, I view each interview as a big opportunity to learn about the goings-on in the high school world these days.

The most enjoyable part of the college interview is when I get to learn how students spend their free time, whether on the bus ride to school or those precious few minutes before class when there isn’t enough time to do homework, but a rare opportunity to socialize exists.

When I asked my dad about his high school days, he fondly recalls that everyone used to play the game “Eraser Wars,” which was simply everyone throwing erasers at each other. Back when I was in high school, the before-class game of choice was “Pencil Break,” where one person holds the ends of a pencil with their hands and their opponent would flick a pencil as hard as they could in an attempt to break the other person’s pencil. I never did well at Pencil Break—in fact, I lost the game pretty much every time I sat in for a match.

While it isn’t surprising that high schoolers still play games before class, it is good to know that the games of choice continue to evolve and become more sophisticated. For example, one student I recently interviewed said that he and his friends frequently play a game called “Immortal.” In the first round of Immortal, a student rolls two 2-sided dice (really just a coin), numbered 1 and 2. If snake eyes (a pair of ones) come up, the student dies and loses the game, but for any other roll the game continues. In round two, instead of using a two-sided dice, the students roll two 4-sided dice. Again, if student rolls snake eyes, they lose; if not, the game continues. In the third round, the student rolls two 6-sided dice; if they survive, they will roll two 8-sided dice in round four. The game continues with number of sides of the dice increasing by two each round until the student rolls snake eyes, or until the odds of rolling two ones becomes so remote that the person is declared “immortal.”

A second student I interviewed talked about a different game students played before classes started called “Pencil Toss.” It turns out that in this student’s homeroom class, the floor of the homeroom consists of long, parallel wooden boards that just happen to be the same length as that of a pencil. Hence students would throw a pencil in the air and see if the pencil would land solely on one board, or if the pencil would land on two different wooden boards. I didn’t ask if students bet money playing Pencil Toss or Immortal, but if your high schooler complains about getting their lunch money stolen, it’s quite possible that this is the true cause of the lost money.

Both the student who played ‘Immortal’ and the student who played ‘Pencil Toss’ asked me what the true mathematical odds of these games are. While I didn’t know the answers off the top of my head, I had a feeling my puzzle-solving friends could come to my rescue…

Given that all dice thrown in the game Immortal are indeed fair dice, what is the probability that the student playing the game will roll snake eyes and lose the game before becoming immortal?

Assume in the game Pencil Toss that the pencil’s width is infinitely small, the space between wooden boards is infinitely small, and that the pencil is indeed tossed fairly and randomly. What is the probability that the tossed pencil will land solely on one wooden board?

Solutions to previous puzzle:

A March Potpourri

**Problem 1. (Shah)**

** CAESARS = IDES**

In this puzzle, each letter represents a base-10 digit. No letter can represent more than one digit, and no digit can be assigned to more than one letter. There are no leading zeroes. There is only one solution.

**Answer**: CAESARS = 3045025 = 1745^{2} = IDES^{2}

**Solution**: (Following Promislow) By counting the
number of digits and keeping in mind that I and C can’t both be 1, we must have
14 < ID <
31. Because both numbers end in S, S = 1, 5, or 6 (but not 0 because that would
imply R = 0 also).

Case I: S = 5. When I posed this problem, I said only a few checks would be needed, so one might try this case first because it immediately implies R =2. This cuts the possibilities for ID down to 17, 18, 19, 30, or 31. If x = IDE then CAESA = x(x+1) and thus A = E(E+1) mod 10 and to avoid repeats, E must be 4, 7, or 9. That leaves us with 13 possibilities for IDE5 but two of these, 3175 and 3195 are too large. No more than 11 checking calculations will get IDES = 1745.

To show uniqueness, one needs to complete the checking and rule out the other cases S = 1 and S = 6.

~ ~ ~

In Nerdle, one needs to guess an equation. Equations are eight elements long with elements from the set {0 1 2 3 4 5 6 7 8 9 + − * / =}. Guesses must be valid base-10 equations, contain one and only one“=”, and can only have a number to the right of the “=”, not another calculation. Standard order of operations applies so * and / apply before + and −. For example, 3+2*5=13 is an acceptable guess. As in Wordle, after entering a guessed equation, a correct element in the right place turns green, while a correct element in the wrong place turns red.

**Problem 2. **Create two Nerdle equation guesses, which
combined use all the elements and will tell you exactly where “=” is located.

**Answer**: Everyone had a different answer; some went
fancy using the digits in order. Here is one pair that didn’t: 60/4-8=7 and
2*3+9=15. Notice that if the equal sign is not placed correctly in either of
these, then it must be in the fifth position.

**Problem 3**. (Polster & Ross) What number less than
10 is represented by 10.0102110122…? Continue the representation two more
places.

**Answer**: The number is p and the two additional places
are 22.

**Solution**: There were many hints in the original
description of all four problems that the expression is probably in base 3 and
realizing that -day occurs in March, the day before the Ides, leads to the
first part, which can be easily checked. For the second part, one needs to go
the other way, i.e., from an expression of in base 10 with great accuracy to an
expression in base 3. Methods for converting integers by iterating division by
3 are well known. To convert fractions, I iterated multiplication by 3.

~ ~ ~

The computer scientist and mathematician Donald Knuth
introduced the notation of an up-arrow, , to represent repeated exponentiation;
i.e., tetration. Thus 32 = 3^{3} = 27 and 33 = 3^{32 }= 3^{33}
= 3^{27 }= 7,625,597,484,987.

**Problem 4**. (Halmos) What is the smallest positive
integer *n* such that

9100 < 3*n*? Prove it.

**Answer**: 101

**Proof**: Clearly 3100* < *9100, so *n*
> 100.

By induction, we show (*) 9*n* < 3(*n*+1). For n=1

91 = 9 < 27 = 32 = 3(1+1)

(It is also clear that 92 = 9^{9} = 387,420,489 <
33.)

Now 9(*n*+1)* = *9^{9n}* = (*3^{2})^{9n}*
= *3^{2*9n}

Similarly, 9*n = *3* ^{x}* for some

*x*and by induction 9

*n*< 3(

*n*+1) = 3

*for some*

^{y}*y*>

*x.*But then2

***3

*3*

^{x}<

^{x}^{+1}

*<*3

*3(*

^{y}=*n*+1) thus

9(*n*+1)* = *3^{2*9n }*< *3^{3n+1}*
= *3(*n*+2)and the induction is proved.

Solvers

Good job, these turned out to be hard problems, in one way or another.

*Bob Byrne, Rui Guo, Clive Keatinge, Chris Norman, David
Promislow, Noam Segal, John Snyder. *