| Section | TA | Studio details | Office hours |
|---|---|---|---|
| C01 | Logan Hart | MW 12:30pm-1:20pm, Skiles 154 | W 4:30pm-5:30pm, BlueJeans |
| C02 | Weiwei Zhang | MW 2:00pm-2:50pm, Guggenheim 244 | M 3:30pm-4:30pm, BlueJeans |
| C03 | Logan Hart | MW 2:00pm-2:50pm, Guggenheim 264 | W 4:30pm-5:30pm, BlueJeans |
The following is a tentative, nonbinding schedule. It will be updated as the term progresses.
The green checks ✓ and red crosses ✗ indicate whether or not the masking extra credit was earned in that lecture.
| Topics covered | Notes | Exercises | |
|---|---|---|---|
| Week 1 (8/23-8/27) |
T: Preliminaries, sections 1.1 and 1.2 ✓ R: Sections 1.3, 2.1, 2.2 ✓ |
Day 1 Day 2 |
1.2 (14, 19, 25, 26, 28) 1.3 (1-6, 13, 16, 21, 23, 32) 2.1 (7, 9, 12, 13, 17, 29, 35) 2.2 (5, 7, 9, 17, 31, 33) Extra Problems |
| Week 2 (8/30-9/3) |
T: Section 2.3 ✗ W: Quiz 1 (covers 8/23-8/30) R: Sections 2.4, 2.5 ✓ |
Day 3 Day 4 |
2.3 (1, 3, 5, 25) 1.1 (4, 12, 14) 2.4 (1, 3, 5, 7, 11, 15, 23) Extra Problems |
| Week 3 (9/6-9/10) |
M: Labor Day T: Section 2.5 ✓ W: Quiz 2 (covers 8/31-9/1) R: Sections 3.2, 3.3 (will also need 6.1 and 6.2) ✓ |
Day 5 Day 6 |
2.5 (5, 10-12) 3.2 (1-8) 6.1 (2, 3) 6.2 (8, 9) Extra Problems |
| Week 4 (9/13-9/17) |
T: Sections 3.3 and 6.3 ✓ W: Quiz 3 (covers 9/2-9/13) R: Sections 3.3, 3.4, and 6.4 ✓ |
Day 7 Day 8 |
3.3 (5, 6, 10, 11, 12, 15, 16, 17, 23) 3.4 (1, 3, 5, 6, 7, 22) 6.3 (4, 10, 16, 21) 6.4 (4, 6) Extra Problems |
| Week 5 (9/20-9/24) |
T: Sections 3.4, 6.4, and 3.5 ✓ W: Quiz 4 (covers 9/14-9/20) R: Shifted systems ✓ |
Day 9 Day 10 |
3.5 (2, 4, 6, 10, 11, 12) Extra Problems |
Week 6 (9/27-10/1) |
T: Sections 4.1 and 4.3 ✓ R: Sections 4.1 and 4.3 ✓ |
Day 11 Day 12 |
4.1 (1-7, 15, 22-26) 4.3 (5, 13, 21, 29, 37, 45, 47, 49, 51) Midterm 1 review: problems 1-14 here |
| Week 7 (10/4-10/8) |
T: Midterm 1 (covers 8/23-9/27) R: Section 4.4 ✓ |
Day 14 | 4.4 (3, 5, 6, 11, 17, 19, 20, 25) Extra Problems |
| Week 8 (10/11-10/15) |
M: fall break T: fall break R: Section 4.5 ✗ |
Day 15 | no textbook problems Extra Problems |
| Week 9 (10/18-10/22) |
T: Sections 4.5 and 4.6 ✓ W: Quiz 5 (covers 9/28-10/18) R: Sections 4.6 and 4.7 ✓ |
Day 16 Day 17 |
4.5 (1,3,5,7,9,17,19,21,29,37) 4.6 (9,11,17) 4.7 (3,5,7,9,11,19) Extra Problems |
| Week 10 (10/25-10/29) |
T: Sections 5.1, 5.2, and 5.3 ✗ W: Quiz 6 (covers 10/19-10/25) R: Sections 5.2, 5.3, 5.4✓ |
Day 18 Day 19 |
5.1 (13,15,17,21,23) 5.2 (5,7,9,13,19,25) 5.3 (9,13,17,21) 5.4 (1,5,9,13) Extra Problems |
| Week 11 (11/1-11/5) |
T: Sections 5.4 and 5.5 ✗ W: Quiz 7 (covers 10/26-11/1) R: Sections 5.5 and 5.6✓ |
Day 20 Day 21 |
5.4 (11,15,25) 5.5 (1,3,7,11,15,17,23) 5.6 (1,3,5,13,19) no extras this week |
| Week 12 (11/8-11/12) |
T: Sections 5.7 and 5.8 ✗ R: Sections 7.1 and 7.2✓ |
Day 22 Day 23 |
5.7 (1,3,9,11,17,19) 5.8 (2,3,5,7,9,11,17,21) Extra Problems Midterm 2 review: problems 15-35 here |
| Week 13 (11/15-11/19) |
T: Midterm 2 (covers 9/28-11/8) R: Section 7.3 ✗ |
Day 25 | 7.3 (1,2) no extras this week |
| Week 14 (11/22-11/26) |
M: Quiz 8 (covers 11/9-11/17) T: Sections 7.4 and 7.6 ✗ W-F: Thanksgiving break |
Day 26 | 7.4 (1,3) Extra Problems |
| Week 15 (11/29-12/3) |
T: Sections 8.1 and 8.2✓ W: Quiz 9 (covers 11/18-11/29) R: Section 8.3✓ |
Day 27 Day 28 |
8.1 (3,11) 8.2 (15,21) 8.3 (7,11) Extra Problems |
| Week 16 (12/6-12/10) |
M: Last studio (review) T: Last lecture (review)✓ R: Review session (optional, online-only) |
Day 29 | |
| Finals Week (12/13-12/17) |
R: Final Exam (8:00am-10:50am, cumulative) |
Just for fun, I'll post a weekly puzzle here. These are mostly stolen from Martin Gardner's book 'My Best Mathematical and Logic Puzzles,' but maybe updated a bit. The puzzles don't really have anything to do with the course, but feel free to come discuss them in office hours or on Discord!
| Puzzle | Solution (hover/tap to view) | |
|---|---|---|
| Week 1 (8/23-8/27) |
Consider a cube whose sides are 3cm long. We can cut this cube into 27 cubes with 1cm sides with six cuts: two cuts parallel to the \( xy \)-plane, two parallel to the \( xz \)-plane, and two parallel to the \( yz \)-plane. (Think of a Rubik's cube.) Can you, perhaps by rearranging the pieces as you cut, achieve the same result with fewer cuts? If so, how? If not, why not? | The minimum number of cuts is six. Think about the 1x1x1 cube at the center of our 3x3x3 cube: it has six faces, none of which are initially exposed. Each face needs to be exposed by some cut, and no single cut can touch more than one face. So we need at least six cuts. Some of your classmates came up with other creative solutions. Discuss these on Discord! |
| Week 2 (8/30-9/3) |
Suppose we have 10 stacks of coins, each consisting of 10 coins. One of the stacks --- we don't know which --- consists entirely of counterfeit coins, but the rest are genuine. However, we do know (a) the mass of a genuine coin and (b) that the counterfeit coins are one gram heavier than the genuine coins. You may choose any collection of the coins to weigh on a kitchen scale. What is the smallest number of weighings needed to find the counterfeit coins? | Since we know the mass of a genuine coin, we can get away with just one weighing. First, number the stacks 1 through 10. Then place on the scale 1 coin from stack 1, 2 coins from stack 2, and so on, until we have 55 coins on the scale. The reading on the scale will exceed the mass of 55 genuine coins by some integer \( n\), and this integer tells us which stack is counterfeit. If, say, there are six extra grams, then that means we weighed six counterfeit coins, and thus the counterfeits came from stack six. |
| Week 3 (9/6-9/10) |
Every Saturday, Dusa takes a bus from Georgia Tech to either Ponce City Market or Krog Street Market for lunch. She arrives at her usual bus stop at some random time, and then takes whichever bus arrives first. Each market's bus arrives at Dusa's stop once per hour, on a fixed schedule. Assuming that the bus schedules are completely consistent, and Dusa's arrival times to the bus stop are completely, uniformly random, how is it that Dusa finds herself at Krog Street Market on 90% of her Saturdays? | The bus for Krog Street Market must arrive at Dusa's stop exactly six minutes before the Ponce City Market bus. This means that there are 54 minutes in each hour for which the KSM bus is next to arrive, but only 6 minutes in each hour for which the PCM bus is next. Since Dusa's arrival time is random, there's a 90% chance that she'll arrive after the PCM bus but before the KSM bus (rather than the other way around). |
| Week 4 (9/13-9/17) |
Suppose that an adventurer starts at her base camp and travels 10km due south. She then travels 10km due west, followed by 10km due north, only to find herself back at base camp. Find at least two locations on Earth where her base camp might be located. | The classic answer to this puzzle is that our adventurer starts at the North Pole. But there are infinitely many other points where her basepoint could be located --- all near the South Pole. There's a line of latitude roughly (but not exactly) 10km north of the South Pole which has a circumference of 10km. If our adventurer's basecamp is located 10km north of this latitude, then her journey will take her south to this latitude, then she'll make a full trip around the latitude, and then she'll head north to her base camp. In fact, there are infinitely many such lines -- for every positive integer \(n\), her base camp could be located 10km north of the line of latitude with circumference \(10/n\) km. |
| Week 5 (9/20-9/24) |
Say we have a rover on Mars, and this rover has an unlimited supply of batteries at its landing spot. We want to move the rover to a pickup spot 700 miles away, but it can only carry enough batteries to travel 420 miles, so we'll build up refueling stations along our route. We can deposit whatever quantity of batteries we want at any location, as long as we can either get back to our landing spot or make it to our pickup spot with the quantity left on board. What is the minimum quantity of batteries (measured in miles) needed to travel to our pickup spot? | The minimum supply of batteries we'll need is 1676 miles. On our first trip, we travel 60 miles away from the landing spot, deposit a 300 mile supply of batteries, and arrive back at the landing spot on empty. On the next trip, we travel to the first cache, refill, then travel 84 more miles before depositing 252 miles' worth of batteries. We then head back to the first cache, refuel, and head home. We now have 180 miles' worth of batteries at cache 1, 60 miles from the landing spot, and 252 miles' worth at cache 2, 84 miles past that. We do one more setup run, depositing 140 miles' worth of batteries at a spot that's 140 miles past our second cache. After this run, we have a 60 mile supply at cache 1, an 84 mile supply at cache 2, and a 140 mile supply at cache 3. For our final trip, we leave the landing spot with 416 miles' worth of batteries, refuel at each cache, and make it to the pickup spot on empty.
In general, if our rover has a capacity of \(c\) miles' worth of batteries, then we can travel a maximum distance of \(c(1+\frac{1}{3}+\frac{1}{5} + \cdots + \frac{1}{2m+1})\) after \(m\) setup trips. Let's discuss on Discord or in office hours why this tells us that 1676 is the minimum above! |
| Week 6 (9/27-10/1) |
Consider three boxes, each of which contains two marbles. One box contains two black marbles, one box contains two white marbles, and one box contains one marble of each color. The boxes have been labeled BB, WW, and BW, but these labels are wrong. You are allowed to draw one marble at a time, examine its color, and then return the marble to its box. What is the minimum number of drawings needed to determine the contents of all three boxes? | One draw will suffice. Pull a marble from the box labeled BW. If it's black, then the box must be BB --- it can't be BW, by the label, and it can't be WW, since it has a black marble. But then the box labeled WW must be BW, since it's neither BW nor BB. Finally, the box labeled BB must be WW. The analysis is similar if we pull a white marble from the box labeled BW. |
| Week 7 (10/4-10/8) |
Can you choose five points on Earth so that no hemisphere contains four of them? Why or why not? (Here, the equator counts as part of both the northern and southern hemispheres, and we can think about hemispheres other than these two.) | Given any five points on Earth, we can choose a hemisphere which contains four of them. To see this, pick two points and consider the great circle that passes through them. This splits Earth into two hemispheres, and the great circle we've chosen is a part of both hemispheres. The remaining three points must be split between these two hemispheres, so at least one of the hemispheres contains at least two points (in addition to the two on our great circle). Thus this hemisphere contains four of the original five points. |
| Week 8 (10/11-10/15) |
Suppose we have two pieces of flammable string and some lighters. We know that each string, if lit at one end, would take an hour to burn; but the burning rate is not consistent. Half the string might be gone in 10 minutes, with the rest of the string taking 50 minutes to burn; we just don't know. Given this scenario, how might we measure 15 minutes of time? | One way to do this is as follows: we can simultaneously light both ends of string A and one end of string B. Whenever string A has burned completely, we light the other end of string B. This will necessarily occur 30 minutes after we started, meaning that half of string B remains. This marks the beginning of our 15 minutes, and the end occurs when string B has burned completely. |
| Week 9 (10/18-10/22) |
Georgia Tech, UGA, and Clemson University recently held an 'academic track meet,' which Georgia Tech (naturally) won with 22 points, while UGA and Clemson each scored 9. In each event, the third place school earned some positive number of points, the second place school earned more, and the first place school earned most, all point totals being integers. Given that UGA won the Herschel Walker trivia contest, how did the scoring work? | I only know how to solve this in an ad hoc manner. Let \(p=p_1+p_2+p_3\) be the total number of points awarded in each event, and \(E\) the number of events. Then \(pE=22+9+9=40\), so both \(p\) and \(E\) are factors of 40. Next, let's think about \(p_1\), which must be at least 3, but no more than 9. In fact, \(p_1\neq 9\), since \(p_1=9\) leads to obvious nonsense. We also can't have \(p_1=8\), since UGA's score would then tell us that there were at most two events, and the max score possible would be 16, below GT's 22. We can similarly rule out \(p_1=7\) and, with slightly more work, \(p_1=6\). Next, \(p_1\neq 3\), since \(p_1=3\) would mean \(p=6\), which is not a factor of 40. Similarly, \(p_1\neq 4\). So at last we see that \(p_1=5\). The max score possible is then \(p_1E\), so \(E\geq 5\), meaning that \(p\leq 8\). We finally see that \(p_1=5\), \(p_2=2\), and \(p_3=1\). |
| Week 10 (10/25-10/29) |
Long ago, mathematicians used to gather in large numbers at conferences. Even worse, they engaged in a ritual we have long since forgotten: shaking hands. Prove that all of these conferences had the following property: the number of mathematicians who shook hands an odd number of times is even. What can you say about the number of mathematicians who shook hands an even number of times? (A handshake requires two people, and the same pair can shake hands more than once.) | There are a few ways to prove this, but here's my favorite. At the start of the conference, before any handshakes have occurred, zero mathematicians have participated in an odd number of handshakes. Since zero is an even number, the property is satisfied. Next, notice that a handshake can't cause the property to become false: if the mathematicians participating in the handshake have the same parity (i.e., if one has shaken an odd number of hands, then so has the other), then they join or leave the odd mathematicians together, not changing the parity of the odd mathematicians. If one mathematician has shaken an odd number of hands while the other has shaken an even number, they swap places: the odd mathematician becomes even, and the even becomes odd. We can't say anything about the parity of the even mathematicians, except that it matches the parity of the conference overall. |
| Week 11 (11/1-11/5) |
Georgia Tech's School of Mathematics has a strange way of assigning teaching duties. The chair of the school puts 5 mathematicians into a single file line facing a wall, so that each mathematician can only see the heads of those people in front of them. The chair then places a gold or white hat on each mathematician's head; no mathematician can see their own hat, but they can see the hats in front of them. Starting from the back of the line, the chair asks each mathematician the color of their hat; if they guess correctly, they get to teach 2552. Knowing nothing of how the chair might choose their hats, how can the 5 mathematicians conspire to ensure that at least 4 of them get to teach 2552? | The mathematician at the back of the line will have to take their chances, but everyone else can be assured of teaching 2552. When asked for their hat color, the mathematician at the back can call out white if they see an odd number of white hats ahead of them, and call out gold if they see an even number of white hats. The next mathematician, able to count the number of white hats in front of them, can use this to deduce their own hat color. After that, everyone can deduce their hat color based on the parity given by the first mathematician and the fact that each of the colors called after that were correct. |
| Week 12 (11/8-11/12) |
Silas Marner has 60 gold coins and 6 safes. He wants to distribute his coins among the safes in such a way that, for any integer \(n\) between 1 and 60, he can empty some collection of safes onto his desk to view exactly \(n\) coins. How should he distribute the coins? | Silas can put 1, 2, 4, 8, 16, and 29 coins in his respective safes. For \(n\) between 1 and 31, the binary expression of \(n\) tells him which safes to open. For instance, 25 has the binary representation 11001, so Silas will open the first, fourth, and fifth safes. For numbers bigger than 31, he'll always open the last safe, and then the binary representation of \(n-29\) will tell him which other safes to open. |
| Week 13 (11/15-11/19) |
At the Atlanta Braves' recent championship parade, their mascot, Blooper, walked alongside the main parade float, which was 30 feet long. Starting at the back of the float, Blooper would walk forward until they reached the front of the (moving) float, then turn around and walk to the back of the float. When Blooper reached the back of the float, it had moved forward 30 feet. Assuming all speeds are constant, how far did Blooper walk over the 2 mile parade route? | We can denote by \(f\) and \(B\) the speeds, in feet per second, of the float and Blooper, respectively. As Blooper walks from the rear of the float to the front, their relative speed is \(B-f\), and thus this walk takes \(30/(B-f)\) seconds. Similarly, it takes Blooper \(30/(B+f)\) seconds to return to the rear of the float. In that same time, the float has moved 30 feet, and thus \(30/f\) seconds have elapsed. By equating this to the total time of Blooper's back-to-front and front-to-back trips, we get an equation relating \(B\) to \(f\). This can be solved to find that \(B=(1+\sqrt{2})f\). It follows that Blooper will walk \((1+\sqrt{2})2\approx 4.83\) miles over the float's 2 mile journey. |
| Week 14 (11/22-11/26) |
We're going to place 100 ants at distinct points on a log with a length of 1 meter. You can place the ants in any straight-line configuration you want (as long as they're not on top of each other) and each ant can face either left or right. When you tell them to start (they're very well-trained ants), all of the ants will begin walking in the direction you've chosen at a speed of 1 meter per minute. Whenever two ants collide, they both turn around and keep walking at the same speed. If an ant runs into the end of the log, it falls off. What is the greatest amount of time that you can keep at least one ant on the log? | No matter how they're configured, all of the ants will be gone from the log after one minute (and possibly sooner). The key insight is to think not of the ants themselves, but of their paths. When two ants collide, they each pick up the path that was being followed by the other. Said another way, if we don't care about the identities of the individual ants, ants which bounce off of each other are the same as ants which pass through each other. Since the ants move at 1 meter per minute, it's impossible to start a path which is more than a minute away from running off the log. |
| Week 15 (11/29-12/3) |
There are four instructors for 2552 next semester, and some number of students signed up to take the course. The instructors planned to meet to divide the students among their sections, but, realizing she couldn't make the meeting, Dr. Brown logged into the registration system early. She noticed that dividing the number of students by 4 left a remainder of 1, so she chose a student at random to drop from the course for "capacity issues." She then randomly assigned one quarter of the remaining students to her section. The next day, Dr. Green realized they wouldn't be able to make the meeting either. Not knowing that Dr. Brown had already assigned herself some students, Dr. Green logged in and did the same thing: they kicked out one student to leave a multiple of 4, and then assigned one quarter of the remaining students to themselves. Naturally, Dr. Black and Dr. White went through the same process, each booting a student and taking a quarter of the remaining students. Weeks later, the registrar noticed that the number of unassigned students was one greater than a multiple of 4. So he, too, dropped a student and assigned the remaining students evenly among the four sections. What is the smallest number of students who could have originally signed up for 2552? | Say there are \(N\) students who signed up to take 2552. After Dr. Brown drops a student and claims some for her own class, there will be \(\frac{3}{4}(N-1) = \frac{3}{4}(N+3)-3\) students left to be claimed. Dr. Green's actions will turn this into \(\left(\frac{3}{4}\right)^2(N+3)-3\). The other instructors will impact the list in a similar way, leaving the registrar with \(\left(\frac{3}{4}\right)^4(N+3)-3\) students on the list. We know that this number must have the form \(4M+1\), for some \(M\geq 1\), so \(\left(\frac{3}{4}\right)^4(N+3)-4=4M\). Since \(4\) and \(4M\) are both multiples of 4, we also need \(\left(\frac{3}{4}\right)^4(N+3)\) to be a multiple of 4. The smallest value of \(N\) for which this will be true occurs when \(N+3=4^5\), or \(N=1021\). |