Test your math strength against former pro-football player, John Urschel
Do you have what it takes to go head-to-head with former pro-football player turned professional mathematician, John Urschel? Now is your chance to flex your mental muscles and find out.
Starting today, John Urschel (@JohnCUrschel) will share a math problem on Twitter every week for the next six weeks. You and your students will have about 48 hours to solve the problem before we (@TICalculators) share the correct answer, every Wednesday, on our BulleTIn Board and Twitter.
You can play as a class or individually, just be sure to get in the game. We will be surprising some of you who answer correctly with prizes along the way. To answer the challenge, simply reply to John’s weekly math challenge on Twitter.
The Monday math challenges will tackle subjects from chess to cryptography -- things students are naturally interested in – and force students to explore the math behind them. John, who is currently pursuing his PhD in math at MIT, created each of the questions himself and wants to see students exercise their brains as they kick-off a new school year. John’s first problem looks for mathematics in one of his all-time favorite hobbies, chess. Let’s start the game clock!
John Urschel, MIT mathematics PhD student and former NFL player
There are 88 keys on a piano, with the lowest note being A0, with a frequency of 27.5 Hertz. Two consecutive keys differ in frequency by a factor of 12√2. The note A1 is the lowest piano note with an integer frequency (55 Hz). Does there exist a piano note with integer frequency other than octave shifts of A1?
Problem 4: Solultion
Because A1 is the lowest piano note with an integer frequency, it suffices to consider only notes higher than A1. If a note is k keys higher than A1, then its frequency is given by 55 X 2k/12. If the note is not an A note, then k can not be a multiple of 12. The question is if there exists a k which is not a multiple of 12 such that 55 X 2k/12 is an integer. It suffices to show that 2k/12 is irrational for all k not a multiple of 12.
We will prove something even stronger, that any fractional power of a prime is irrational. We will prove this by contradiction. Let p be a prime and x a positive fraction. x is not a whole number, and so can be written as x = a/b where a and b (b ≠ 1) are relatively prime natural numbers. Suppose px is rational. Then px =pa/b = q for some rational number q. Then pa = qb, and q must be a power of p, q = pc (because p is prime). But this implies that a = bc, which contradicts the assumption that a and b are mutually prime.
Therefore, all piano notes with integer frequency must be an A note in some octave.
Check back on 10/22 for the next problem.
Last week's challenge:
Here’s a Valentine’s Day cryptoarithmetic puzzle. Is it possible to replace each unique letter in this equation with a unique digit (0 to 9) so that it holds? Note that once a letter is given a digit, that is its digit throughout the equation (if you set O to be 3 in YOU, it is also 3 in LOVE).
Problem 3 Solution
Sadly, we just can’t make this equation work. Because U + E has E in the ones place, we know U is zero. Y, L and O have to be nine, one, and zero, otherwise we can’t get a four digit number from a three and two digit one. But the problem is that O+M has to be at least eleven, which isn’t possible if O is zero. This equation just isn’t meant to be.
However, it turns out that just by adding in math, we can make this equation work, and be indivisible (YOU, ME, MATH, and LOVE are all prime numbers). This one is not for the faint of heart, but if you enjoyed this challenge, you might want to give this one a try. The solution is unique.
Suppose we have the following model for information (e.g. photo, video, post) transfer (i.e. spread) in a grid: each square is colored either green (if it knows a given piece of information) or red (if it does not). A red square learns this information (becomes green) if at least two of its neighbors (sharing an edge) already knows it. For example, in this situation the bottom middle square becomes green:
On a 10 X 10 grid, what is the minimum number of initial green squares needed for the entire grid to become green?
Problem 2 Solution:
Making either long diagonal exclusively green (10 initial green squares) is enough to make the entire grid green (after k steps, all squares within distance k of the diagonal becomes green). The hard part is showing that you need at least 10 squares. One way to do this is by using something called a “monotonicity" argument. Note that the total perimeter of the green squares never increases, but is 40 if the entire grid is filled.
On a chessboard, it takes a king, on the h4 square, six moves to reach the b4 square. How many different, six move routes could the king take?
Problem 1 Solution:
The number of six move routes is actually equivalent to the number of strings consisting of -1, 0, +1 of length six that add up to zero (the path h4-g3-f3-e2-d3-c4-b4 would correspond to -1, 0,-1,+1, +1, 0):
What if you had the setup below? What move would you play as white?
The chess puzzle is a famous one, composed by Grigoriev. An attack on the b7 pawn loses:
1. Kg5 Kc2 2. Kf4 Kd3 3. Ke5 Kc4 4. Kd6 Kb5 5. Kc7 Ka 6.
Of the 141 routes the white king has to get to b4, only one holds the draw:
1. Kg3! Kc2 2. Kf2! Kd3 3. Ke1! Kc4 4. Kd2 Kb5 5. Kc3 Kxb6 6. Kb4.