The Sum of our Knowledge

Many times, we use the same old icebreakers at the beginning of a class:  go around the room, introduce yourself, your major, a hobby or something interesting about yourself . What if instead, we could get our students working together and engaging in mathematical problem-solving from the get-go? Today’s post is the first in a four-part series where we'll present several ideas and strategies to accomplish this, most of which are appropriate for any level of mathematics course. Review Dominos We have found that this activity works great in many situations-- Our favorites uses: An icebreaker on the first day of class, while also reviewing material from a previous class. Low stakes group work to practice basic computation relating to newly learned material on an average class day A group warm-up, reviewing material from a previous day that is crucial to know before beginning class. A small-group project before an exam review. This activity requires a small amount of preparation up fro...

This problem, along with a couple of others, was inspired by the work of John H. Conway* and appeared in  the October 15, 2020 edition of Quanta Magazine . Let a, b, c, d, e, f, g, h, i, j be the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 in some order. Each digit appears exactly once. (For those that want a more precise mathematical phrasing, there is a one-to-one correspondence between the set {a, b, c, d, e, f, g, h, i, j} and the set {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}, or better yet, there is a bijection between the two sets.)  We form other integers by concatenating digits, so for example, 'ab' is a two-digit number where 'a' is the tens digit and 'b' is the ones digit. If abcdefghij is a 10-digit number with the following properties: a is divisible by 1 (OK, that much is obvious) ab is divisible by 2 (So, b must be even) abc is divisible by 3 (Do you know the test for divisibility by 3?) abcd is divisible by 4 abcde is divisible by 5 abcdef is divisible by 6 abcdefg ...

COME ON DOWWWWNNNN!!!!!!   You’ve been chosen to be a contestant on Let’s Make a Deal!   You’re now standing in front of three doors -- behind one of these doors is a car, and behind each of the other two is a goat. Monty, the host, says, “Pick a door! If it has the car behind it, you’ll drive it home today!”  You choose door #1, and feel really good about your choice.  The host then opens door #3,  which has a goat behind it, and asks “Do you want to switch doors?”  Question: What is the probability that you will win the car now that you know there was a goat behind door #3? Should you switch? Assumptions: Before the show, the car was equally likely to be placed behind any of the three doors. Once you make your selection of door #1, Monty will choose to reveal what is behind one of the other two doors, but will definitely not show you where the car is. Update: 11/19/20: read below for the solution.

Here's a puzzle that was first passed on to me by Al Lippert, who spent years providing the "Problem of the Day" at MathPath .  What goes after the last equals sign? Hints: The answer we're looking for isn't 80.  And the equalities are all true! Update: 11/6/20: read below for the solution.