Math Jams

## Who Wants to Be a Mathematician, Round 2

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AoPS instructor David Patrick will discuss the problems on Round 2 of the 2016-2017 Who Wants to Be a Mathematician national contest. We will also be joined by Mike Breen and Bill Butterworth, the creators of the game. Mike is also the host of the national finals, to be held in Atlanta in January 2017.

#### Facilitator: Dave Patrick

DPatrick 2016-11-09 19:30:44
Welcome to the 2016-17 Who Wants to Be a Mathematician Round 2 Math Jam!
DPatrick 2016-11-09 19:30:55
I'm Dave Patrick, and I'll be leading our discussion tonight. Many of you know me from around AoPS: I've taught dozens of AoPS classes over the past 12 years, and I've written or co-written a few of our textbooks. I also once was a contestant on ABC's Who Wants to Be a Millionaire back before I started working at AoPS, when Regis was still the host. (No, I didn't win the million bucks, but I did win enough to buy a new car.)
DPatrick 2016-11-09 19:31:17
Before we get started I would like to take a moment to explain our virtual classroom procedures to those who have not previously participated in a Math Jam or one of our online classes.
DPatrick 2016-11-09 19:31:31
The classroom is moderated, meaning that students can type into the classroom, but these comments will not go directly into the room. These comments go to the instructors, who may choose to share your comments with the room.
DPatrick 2016-11-09 19:31:41
This helps keep the session organized and on track. This also means that only well-written comments will be dropped into the classroom, so please take time writing responses that are complete and easy to read.
DPatrick 2016-11-09 19:31:57
Also, we won't be going through all the math quite as thoroughly as we do in our classes -- I can't teach all the material for every problem as we go. Another difference between tonight and our regular online classes is that it is very unlikely that we'll be able to answer every single question you ask. We usually do in our classes, but we have a large number of students tonight! So, please go ahead and ask questions, but also please understand if we aren't able to answer them all!
DPatrick 2016-11-09 19:32:13
Assisting with the Math Jam tonight is Josh Morris (QuantumCat).
QuantumCat 2016-11-09 19:32:24
Hey everyone!
DPatrick 2016-11-09 19:32:26
Josh, after getting his Bachelor's and Master's in Physics from UT El Paso and teaching high school physics, decided he loved science and working with students so much he went back to school to work pursue a career in medicine. While he's applying to medical school and going through the interview process, he's helping out here at Art of Problem Solving and keeping his math skills sharp. In his spare time, he loves cooking, running, reading and is (still) slowly teaching himself how to play the piano.
DPatrick 2016-11-09 19:32:41
Josh can chat with you if you are having trouble, but again, due to the size of tonight's session, we may not be able to get to every question.
DPatrick 2016-11-09 19:32:52
Also joining us tonight are the co-creators of WWTBAM, Mike Breen (mikebreen) and Bill Butterworth (TPiR).
mikebreen 2016-11-09 19:33:05
Hello, everyone!
TPiR 2016-11-09 19:33:16
Hello all. Good to be here.
DPatrick 2016-11-09 19:33:18
Mike taught at Alfred University and Tennessee Technological University before becoming AMS Public Awareness Officer in 2000. He and Bill began Who Wants to Be a Mathematician for the American Mathematical Society in 2001. The first national game was in 2010. Mike has been on Jeopardy! and Wheel of Fortune (if you want to know if he won lots of money on either show, note that he is still working for a living) and may be the only person ever to cut his hand on the wheel. Who Wants to Be a Mathematician has so far been much safer.
DPatrick 2016-11-09 19:33:41
Bill earned an undergraduate degree in mathematics from Santa Clara University and a Ph.D. from Northwestern University, and is currently an associate professor and associate chair of mathematics at DePaul University. He shares a life-long interest in game shows with colleague Mike Breen, with whom he works as the not-so-lovely assistant on the mathematics game show Who Wants to Be a Mathematician. In addition to authoring articles and presenting talks related to game-show mathematics, Bill served as mathematics consultant to the CBS television show The Price is Right from 1997 to 2009. (Hence, his username.)
DPatrick 2016-11-09 19:34:08
As you can see, we have a lot of game show background here tonight!
mathpractice2011 2016-11-09 19:34:11
I have been on wheel of fortune too
mikebreen 2016-11-09 19:34:17
Bill is in the Cubs' afterglow
DPatrick 2016-11-09 19:34:28
Indeed, as is levans (our webmaster)!
DPatrick 2016-11-09 19:34:45
Who Wants to Be a Mathematician is run by the American Mathematical Society (AMS). The AMS promotes mathematical research, fosters excellence in mathematics education, and increases the awareness of the value of mathematics to society.
DPatrick 2016-11-09 19:35:02
Round 2 of the national contest consisted of 10 questions, with a 15-minute time limit. So the problems are quick: you have an average of 1.5 minutes per question. (Compare that to the AMC 10/12 which has an average of 3 minutes per question, or the AIME which has an average of 12 minutes per question.)
DPatrick 2016-11-09 19:35:42
We'll take a bit longer than 15 minutes tonight, because we'll stop along the way to discuss each question. Please also remember that the purpose of this Math Jam is to work through the solutions to the problems, and not to merely present the answers. "Working through the solutions" often includes discussing problem-solving tactics. So please, when a question is posted, do not simply respond with the final answer. That's not why we're here. We're going to work through the problems step-by-step.
DPatrick 2016-11-09 19:36:02
Several of the questions have interesting sidetracks, so we'll also stop and view some of the scenery along the way.
mathpractice2011 2016-11-09 19:36:07
How long is this session
DPatrick 2016-11-09 19:36:10
DPatrick 2016-11-09 19:36:29
Anyway, let's get started.
DPatrick 2016-11-09 19:36:33
1. What is the sum of the roots of the polynomial $x^2 - 12x + 4$?
declanhiller 2016-11-09 19:37:24
would we use vieta's theorem?
Jorvis 2016-11-09 19:37:24
Use vieta's formula -> -(-12) = 12
mc0131 2016-11-09 19:37:24
Vietta's
EpicRuler101 2016-11-09 19:37:24
vietas
chocolatemushrooms 2016-11-09 19:37:24
12 by vieta's
zetaintegrator 2016-11-09 19:37:24
vieta's rule
NewtonFTW 2016-11-09 19:37:24
The sum of the roots of any quadratic equation are -b/a
DPatrick 2016-11-09 19:37:41
Right! The fastest solution is to use one of Vieta's Formulas, which tell us that the sum of the roots of $ax^2 + bx + c = 0$ is $-\dfrac{b}{a}$.
DPatrick 2016-11-09 19:37:52
Is this equation, $b = -12$ and $a = 1$, so we get that the answer is $\boxed{12}$.
DPatrick 2016-11-09 19:38:22
We can see why this is true from the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$$
DPatrick 2016-11-09 19:38:59
When we add the two roots, the square-root part cancels (one + with one -), and what's left is $2\left(\dfrac{-b}{2a}\right)$, or just $-\dfrac{b}{a}$.
NewtonFTW 2016-11-09 19:39:07
So I'm guessing it might be useful to know the product formula as well, c/a?
EpicRuler101 2016-11-09 19:39:07
I also heard something about c/a, what would that be?
DPatrick 2016-11-09 19:39:21
Indeed, if you want the product of the two roots, that's $\dfrac{c}{a}$.
DPatrick 2016-11-09 19:39:47
Yet another approach is to complete the square:

$(x^2 - 12x + 36) = 32$

so $(x-6)^2 = 32$.
DPatrick 2016-11-09 19:39:59
When we graph this, we get a parabola that bottoms out'' at $x=6$, so the two roots average to 6, and thus their sum is 12.
DPatrick 2016-11-09 19:40:18
That was our warmup...
DPatrick 2016-11-09 19:40:22
2. What is $\tan^{-1}\left(\cos\left(\sin\left(\cos\left(\dfrac{\pi}{2}\right)\right)\right)\right)$?
DPatrick 2016-11-09 19:40:45
What's the general approach here?
mc0131 2016-11-09 19:41:18
You can just calculate each step
verum4one 2016-11-09 19:41:18
Just work from the inside out.
work from the inside out
Juliaxyzz 2016-11-09 19:41:18
work from inside out
Jillian 2016-11-09 19:41:18
Is it solve for the innermost parantheses and work your way out?
Liopleurodon 2016-11-09 19:41:18
find each value inside of the brackets
MathNerd0120 2016-11-09 19:41:18
work from inside the parentheses?
DPatrick 2016-11-09 19:41:27
Yes. We can work from the inside out.
vonnan 2016-11-09 19:41:34
cos(\pi/2)=0
DPatrick 2016-11-09 19:41:41
To start, $\cos\left(\dfrac{\pi}{2}\right)$ is 0, so our expression is $\tan^{-1}(\cos(\sin(0)))$.
verum4one 2016-11-09 19:41:53
sin(0)=0
nvpthemvp 2016-11-09 19:41:59
sin of 0 =0
DPatrick 2016-11-09 19:42:03
And then $\sin(0)$ is 0, so our expression is $\tan^{-1}(\cos(0))$.
declanhiller 2016-11-09 19:42:22
cos(0)=1
EpicDragonSlayr 2016-11-09 19:42:22
cos 0 = 1
DPatrick 2016-11-09 19:42:27
And then $\cos(0)$ is 1, so our expression is $\tan^{-1}(1)$.
verum4one 2016-11-09 19:42:52
$\tan ^{-1}(1)=\pi/4$
EpicDragonSlayr 2016-11-09 19:42:52
pi/4
mc0131 2016-11-09 19:42:52
pi/4
DPatrick 2016-11-09 19:42:56
And finally, $\tan^{-1}(1)$ is $\boxed{\dfrac{\pi}{4}}$.
DPatrick 2016-11-09 19:43:31
So this problem was mainly an exercise in being careful with your trig functions. (And obviously if you haven't seen trig functions like this yet, there's not much chance of solving this.)
DPatrick 2016-11-09 19:43:59
Next is one of the hallmarks of WWTBAM, which is that one of the questions usually involves mathematical history:
DPatrick 2016-11-09 19:44:03
3. What is the last name (family name) of the famous British mathematician who worked with Ramanujan and wrote A Mathematician's Apology?
Eugenis 2016-11-09 19:44:16
david patrick?
DPatrick 2016-11-09 19:44:31
I'm a little too young, by about a century I'm afraid.
bogstop320 2016-11-09 19:44:45
Hardy
Mathgirl03 2016-11-09 19:44:45
Hardy
Hardy
ngundotra 2016-11-09 19:44:45
Hardy
POKEMON123 2016-11-09 19:44:45
G. H. Hardy
declanhiller 2016-11-09 19:44:45
hardy
DPatrick 2016-11-09 19:44:51
Correct -- the mathematician in question is G. H. Hardy.
DPatrick 2016-11-09 19:44:57
DPatrick 2016-11-09 19:45:09
And the person mentioned by name in the question is the Indian mathematician Srinivasa Ramanujan.
DPatrick 2016-11-09 19:45:19
mikebreen 2016-11-09 19:45:30
We thought we might take advantage of the movie The Man Who Knew Infinity.
DPatrick 2016-11-09 19:45:40
Indeed, the story of Ramanujan and Hardy was recently dramatized in the movie The Man Who Knew Infinity, starring Dev Patel as Ramanujan and Jeremy Irons as Hardy.
AoPS books teach us this, which turns out to be surprisingly helpful for WWTBAM
DPatrick 2016-11-09 19:46:08
Yes, we like to sprinkle mathematical history and anecdotes into our books from time to time!
DPatrick 2016-11-09 19:46:20
As some of you know, Ramanujan was a math prodigy who grew up in India and was essentially self-taught. In 1913 at age 25, having studied Hardy's work extensively, Ramanujan wrote to Hardy and started a correspondence. Hardy immediately saw Ramanujan's mathematical talent and arrange for a position for him at the University of Cambridge in England where Hardy was a professor.
DPatrick 2016-11-09 19:46:42
Ramanujan traveled to England in 1914 and studied with Hardy for the next several years. This ultimately lead to Ramanujan's election to the Royal Society of London, an extremely prestigious award. Unfortunately, Ramanujan was in very poor health at the time, and he returned to India in 1919, only to die there just a year later at age 32.
DPatrick 2016-11-09 19:46:56
The most famous anecdote regarding Ramanujan and Hardy is told by Hardy in his biography of Ramanujan:
I remember once going to see him when he was lying ill at Putney. I had ridden in taxi cab number 1729 and remarked that the number seemed to me rather a dull one, and that I hoped it was not an unfavorable omen. "No," he replied, "it is a very interesting number; it is the smallest number expressible as the sum of two cubes in two different ways."
DPatrick 2016-11-09 19:47:36
Meanwhile, Hardy himself was also a fellow of the Royal Society of London, and a prolific mathematician in his own right, and (other than his mentorship of Ramanujan) is probably best known for his book A Mathematician's Apology, which he wrote later in life (in 1940, when he was in his 60s), which was his attempt to explain what it means to be a mathematician.
DPatrick 2016-11-09 19:47:50
Note: the photos and the stories are from the terrific math history website MacTutor History of Mathematics at www-history.mcs.st-andrews.ac.uk/, run by the University of St. Andrews in Scotland.
NewtonFTW 2016-11-09 19:47:57
Will there be history of math at the national level?
DPatrick 2016-11-09 19:48:13
There almost always is a math history question at the national finals.
DPatrick 2016-11-09 19:48:36
Anyway, on to the next problem:
DPatrick 2016-11-09 19:48:41
4. A machine makes parts, but $1\%$ of the parts are defective. If an inspector selects 100 parts at random, which of the following intervals below contains the probability that none of the 100 parts are defective?
(a) $[0,0.2)$ (b) $[0.2, 0.4)$ (c) $[0.4,0.6)$ (d) $[0.6,0.8)$ (e) $[0.8,1.0]$
Wave-Particle 2016-11-09 19:49:32
There is a $0.99$ chance it isn't defective
DPatrick 2016-11-09 19:49:51
Right, that's where we can start: given any individual part, the probability it isn't defective is 0.99.
NewtonFTW 2016-11-09 19:49:55
And that's for each part checked
bogstop320 2016-11-09 19:49:57
(99/100)^100
DPatrick 2016-11-09 19:50:11
For none of the parts to be defective, they all have to be working. Each is working with probability $0.99$, so the probability that all 100 are working is $(0.99)^{100}$.
DPatrick 2016-11-09 19:50:35
So the question becomes: which of those answer choices contains $(0.99)^{100}$. (Need I mention that you're not allowed to use a computing device?)
Math1331Math 2016-11-09 19:50:50
Rewrite as $1-.01$
DPatrick 2016-11-09 19:51:00
Good idea: why does that help?
atmchallenge 2016-11-09 19:51:13
Let's try binomial theorem
fractal161 2016-11-09 19:51:13
Binomial Theorem!!
DPatrick 2016-11-09 19:51:21
Yes, we can use the Binomial Theorem to estimate:

\begin{align*}

(0.99)^{100} &= (1 - 0.01)^{100} \\

&= 1 - \binom{100}{1}(0.01) + \binom{100}{2}(0.01)^2 - \binom{100}{3}(0.01)^3 + \cdots - \binom{100}{99}(0.01)^{99} + \binom{100}{100}(0.01)^{100}.

\end{align*}
DPatrick 2016-11-09 19:51:37
yuvi18 2016-11-09 19:51:59
they cancel eachother out
DPatrick 2016-11-09 19:52:03
The first two terms cancel out.
Wave-Particle 2016-11-09 19:52:16
Some parts are very very small
atmchallenge 2016-11-09 19:52:16
They begin to get small pretty fast
Math1331Math 2016-11-09 19:52:16
Eventually the .01^x power gets so small it doesn't have an impact
DPatrick 2016-11-09 19:52:29
Well, I'm not so sure. We've got all those 100's in the top of $\dbinom{100}{k}$ too.
fractal161 2016-11-09 19:53:01
All negated by each $\frac{1}{100}$ term, though.
DPatrick 2016-11-09 19:53:14
Well...let's take a close look. The third term is $\dfrac{100 \cdot 99}{2 \cdot 1}(0.01)^2$.
DPatrick 2016-11-09 19:53:43
The numerator of the $\dbinom{100}{2}$ part is just about canceled out by the $(0.01)^2$.
Wave-Particle 2016-11-09 19:53:49
Like 0.495?
atmchallenge 2016-11-09 19:53:49
Which is .495
vonnan 2016-11-09 19:53:52
1/2
DPatrick 2016-11-09 19:54:04
It's exactly 0.495, but probably $\dfrac12$ is close enough.
DPatrick 2016-11-09 19:54:12
Does this sort of thing continue?
awin 2016-11-09 19:54:26
Yes
mathpractice2011 2016-11-09 19:54:26
yes
MathNerd0120 2016-11-09 19:54:26
yes
vonnan 2016-11-09 19:54:26
yes
Liopleurodon 2016-11-09 19:54:26
yes
yuvi18 2016-11-09 19:54:26
yes!
atmchallenge 2016-11-09 19:54:28
yes, then it is $1/3!$, then $1/4!$, and so on...
DPatrick 2016-11-09 19:54:34
In fact, in all of the first few terms of the form $\binom{100}{k}(0.01)^k$, we can write it as

$\frac{100 \cdot 99 \cdot \cdots \cdot (101-k)}{k!} \cdot \frac{1}{100^k}.$
DPatrick 2016-11-09 19:54:52
The $100^k$ term very nearly cancels out the numerator of this fraction.
DPatrick 2016-11-09 19:55:10
So we can approximate the expansion as

$(0.99)^{100} \approx 1 - 1 + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \cdots.$
DPatrick 2016-11-09 19:55:29
As we noticed before, the first two terms cancel out. Just using the other three terms I've written, what do we get?
Liopleurodon 2016-11-09 19:56:04
$\frac38$
EpicRuler101 2016-11-09 19:56:04
1/2 - 1/6 + 1/24
thedoge 2016-11-09 19:56:04
3/8
NewtonFTW 2016-11-09 19:56:04
3/8
EulerMacaroni 2016-11-09 19:56:04
$\frac{3}{8}$
fractal161 2016-11-09 19:56:04
~$\frac{3}{8}$?
DPatrick 2016-11-09 19:56:08
$\dfrac12 - \dfrac16 + \dfrac{1}{24} = \dfrac{9}{24} = \dfrac38 = 0.375$
DPatrick 2016-11-09 19:56:28
And the terms start to get really small after that, so the quantity won't change much.
atmchallenge 2016-11-09 19:56:35
So b!
Wave-Particle 2016-11-09 19:56:35
So it's b?
DPatrick 2016-11-09 19:56:39
This is solidly inside the interval $[0.2,0.4)$ so the answer is $\boxed{(b)}$.
mikebreen 2016-11-09 19:56:49
Third-hardest question as it turns out.
DPatrick 2016-11-09 19:57:13
Indeed, #10 is the hardest (and you'll see why when we get there), the Hardy question was 2nd-hardest, and this was 3rd-hardest with around 36% correct.
DPatrick 2016-11-09 19:57:34
Now if you've played with these sorts of expressions before, you might notice something familiar...
Kpatton 2016-11-09 19:57:51
B, the probability is about 1/e
EulerMacaroni 2016-11-09 19:57:51
$$\bigg(\frac{99}{100}\bigg)^{100}=\bigg(1-\frac{1}{100}\bigg)^{100}\approx e^{-1}$$
EulerMacaroni 2016-11-09 19:57:51
which is $\frac{1}{e}$
thedoge 2016-11-09 19:57:51
that looks like the formula for dearrangements...
DPatrick 2016-11-09 19:58:13
Indeed, this is related to one of the definitions of $e$ as $e = \displaystyle\lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n$.
DPatrick 2016-11-09 19:58:34
That is, as $n$ gets really big, the expression $(1 + 1/n)^n$ gets really close to the number $e$.
DPatrick 2016-11-09 19:58:41
This is the flip side: $\dfrac{1}{e} = \displaystyle\lim_{n \to \infty}\left(1 - \frac{1}{n}\right)^n$.
DPatrick 2016-11-09 19:59:13
And in fact $\dfrac{1}{e} \approx 0.367879\ldots$ is very close to the exact answer of $(0.99)^{100} = 0.366032\ldots$.
DPatrick 2016-11-09 20:00:15
And as thedoge also mentioned, this is also very closely related to the combinatorics problem of derangements: if $n$ people all drop their keys in a hat, and each randomly selects a key, the probability that no one draws his or her original key is about $1/e$.
DPatrick 2016-11-09 20:00:37
Let's move on:
DPatrick 2016-11-09 20:00:41
5. Find the remainder when $18^6 + 87^6$ is divided by $7$.
Wave-Particle 2016-11-09 20:01:11
Modular arithmetic
Math1331Math 2016-11-09 20:01:11
Reduce the large numbers mod7
DPatrick 2016-11-09 20:01:20
Since all we care about is the remainder when divided by 7, we can replace the numbers that we are exponentiating with their remainders upon division by 7.
DPatrick 2016-11-09 20:01:35
We have $18 = 2 \cdot 7 + 4$ and $87 = 12 \cdot 7 + 3$, so the problem becomes:
DPatrick 2016-11-09 20:01:40
5. Find the remainder when $4^6 + 3^6$ is divided by $7$.
Math1331Math 2016-11-09 20:01:58
not too hard to compute to finish
DPatrick 2016-11-09 20:02:09
Right, one option is to just compute to finish.
DPatrick 2016-11-09 20:02:22
$4^6 = 2^{12} = 4096$ and $3^6 = 729$, so $4^6 + 3^6 = 4096 + 729 = 4825$.
DPatrick 2016-11-09 20:02:28
Then $4825 = 689 \cdot 7 + 2$, so the remainder when $4825$ is divided by $7$ is $\boxed{2}$.
EpicDragonSlayr 2016-11-09 20:02:38
find power patterns
Jorvis 2016-11-09 20:02:38
4^6 = 4^3 * 4^3, and 4^3 = 1 mod 7
DPatrick 2016-11-09 20:02:47
Right, a slightly more sophisticated option is to look for a pattern among powers of 4 and powers of 3.

\begin{align*}

4^1 &\equiv 4 \pmod{7} \\

4^2 &\equiv 16 \equiv 2 \pmod{7} \\

4^3 &\equiv 8 \equiv 1 \pmod{7} \\

4^4 &\equiv 4 \pmod{7}

\end{align*}
DPatrick 2016-11-09 20:03:09
(If you're not familiar with the notation, (mod 7) basically just means the remainder when we divide by 7.)
DPatrick 2016-11-09 20:03:25
So the pattern of remainders of powers of 4 (modulo 7) is 4,2,1,4,2,1,..., where the 1's are powers that are multiples of 3. Hence $4^6 \equiv 1 \pmod{7}$.
Jorvis 2016-11-09 20:03:30
Use similar logic for 3^6
DPatrick 2016-11-09 20:03:39
Similarly,

\begin{align*}

3^1 &\equiv 3 \pmod{7} \\

3^2 &\equiv 9 \equiv 2 \pmod{7} \\

3^3 &\equiv 6 \pmod{7} \\

3^4 &\equiv 18 \equiv 4 \pmod{7} \\

3^5 &\equiv 12 \equiv 5 \pmod{7} \\

3^6 &\equiv 15 \equiv 1 \pmod{7}

\end{align*}
DPatrick 2016-11-09 20:03:50
And we conclude that $4^6 + 3^6 \equiv 1+1 = \boxed{2} \pmod{7}$.
DPatrick 2016-11-09 20:04:01
And if you know a bit more number theory...
SkyFox163 2016-11-09 20:04:27
use fermat's little theorem
Kpatton 2016-11-09 20:04:27
Use Fermat's Little Theorem
fractal161 2016-11-09 20:04:27
Fermat's Little Theorem!
EulerMacaroni 2016-11-09 20:04:27
fermat's little theorem
primesandfractals 2016-11-09 20:04:27
use euler's theorem
Math1331Math 2016-11-09 20:04:27
also you can use a^{p-1}=1modp because a,p are relatively prime
DPatrick 2016-11-09 20:04:40
The most "slick" solution is to use Fermat's Little Theorem. This theorem states that
$a^{p-1} \equiv 1 \pmod{p},$
provided $p$ is prime and $a$ is not a multiple of $p$.
DPatrick 2016-11-09 20:04:59
(There's actually a more general version of this theorem, but I won't go into that right now.)
DPatrick 2016-11-09 20:05:07
In our example $p=7$, and indeed the exponents are $p-1 = 6$, with the bases not being multiples of 7.
DPatrick 2016-11-09 20:05:21
So each term is $1 \pmod{7}$, and their sum is $1+1 \equiv \boxed{2} \pmod{7}$.
mikebreen 2016-11-09 20:05:29
Great ideas on the solution. Were the 18 and 87 an accident?
DPatrick 2016-11-09 20:05:38
I don't know...were they an accident?
mikebreen 2016-11-09 20:05:54
A salute to Ramanujan, born in 1887
DPatrick 2016-11-09 20:06:01
Ha! Clever.
DPatrick 2016-11-09 20:06:16
Onwards:
DPatrick 2016-11-09 20:06:20
6. How many divisors (factors) of $10!$ (including $1$) are perfect squares?
Wave-Particle 2016-11-09 20:06:39
Prime factorize
thedoge 2016-11-09 20:06:39
you can use the prime factorization of 10!
themoocow 2016-11-09 20:06:39
prime factorize
Liopleurodon 2016-11-09 20:06:39
prime factorization of $10!$
DPatrick 2016-11-09 20:06:45
To answer this, we probably need the prime factorization of $10!$.
DPatrick 2016-11-09 20:06:49
\begin{align*}

10! &= 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8 \cdot 9 \cdot 10 \\

&= 2 \cdot 3 \cdot 2^2 \cdot 5 \cdot (2 \cdot 3) \cdot 7 \cdot 2^3 \cdot 3^2 \cdot (2 \cdot 5) \\

&= 2^8 \cdot 3^4 \cdot 5^2 \cdot 7

\end{align*}
DPatrick 2016-11-09 20:07:10
So using this, what divisors of $10!$ are perfect squares?
Wave-Particle 2016-11-09 20:07:17
perfect squares have even exponents (including 0)
Math1331Math 2016-11-09 20:07:17
notice perfect squares have even power of prime
DPatrick 2016-11-09 20:07:29
Right: to get a perfect square, we need each prime's exponent to be even.
DPatrick 2016-11-09 20:07:42
(So we can get the square root by taking one-half of each exponent.)
DPatrick 2016-11-09 20:08:02
Any factor of $10!$ is of the form $2^a3^b5^c7^d$, where

$0 \le a \le 8$

$0 \le b \le 4$

$0 \le c \le 2$

$0 \le d \le 1$
Wave-Particle 2016-11-09 20:08:13
For $2$, the exponent can be 0,2,4,6,8 which is 5 values. Using similar logic for 3,5,7 then multiply them
EulerMacaroni 2016-11-09 20:08:13
5 choices for 2, 3 choices for 3, 2 choices for 5, 1 choice for 7
themoocow 2016-11-09 20:08:13
for 2 we can pick $5$ possiblities, from 3 we can pick 3 possiblities, from 5 we can pick 2, and from 7 we can pick 1: 5*3*2*1=$30$
thedoge 2016-11-09 20:08:13
DPatrick 2016-11-09 20:08:36
Right, since we must have the exponents be even:

$a$ must be 0, 2, 4, 6, or 8

$b$ must be 0, 2, or 4

$c$ must be 0 or 2

$d$ must be 0
DPatrick 2016-11-09 20:08:50
This gives $5 \cdot 3 \cdot 2 = 30$ choices for exponents, and thus $\boxed{30}$ perfect square divisors.
DPatrick 2016-11-09 20:09:10
Next up:
DPatrick 2016-11-09 20:09:20
7. Spheres of equal radius are stacked inside a cylinder of the same radius so that the spheres extend from the bottom of the cylinder to the top. The volume of the space inside the cylinder that is not taken up by the spheres is equal to the volume of three of the spheres. How many spheres are inside the cylinder?
thedoge 2016-11-09 20:09:52
set up a system of equations in terms of the radius of the spheres
Liopleurodon 2016-11-09 20:09:52
make variables
DPatrick 2016-11-09 20:10:01
What variable will be probably definitely need?
Wave-Particle 2016-11-09 20:10:13
duruphi 2016-11-09 20:10:13
r
declanhiller 2016-11-09 20:10:13
r
Liopleurodon 2016-11-09 20:10:13
awin 2016-11-09 20:10:13
nvpthemvp 2016-11-09 20:10:13
r
mathpractice2011 2016-11-09 20:10:13
r
DPatrick 2016-11-09 20:10:16
Are you sure?
DPatrick 2016-11-09 20:10:28
Does the radius of the spheres matter?
caro2be 2016-11-09 20:10:40
no
awin 2016-11-09 20:10:40
No!
Liopleurodon 2016-11-09 20:10:40
not really
no it does not matter
fractal161 2016-11-09 20:10:50
Not really, because ratios are the same...
primesandfractals 2016-11-09 20:10:52
let it be 1
DPatrick 2016-11-09 20:11:05
Indeed -- the whole problem just scales if we change the radius.
DPatrick 2016-11-09 20:11:16
So let's assume the radius of the spheres are 1. (This is a useful problem solving tactic: if an unknown value in a problem doesn't matter, you are free to assign it to whatever value is convenient. Don't introduce a variable you don't need!)
thedoge 2016-11-09 20:11:23
and the number of spheres
DPatrick 2016-11-09 20:11:40
We do almost certainly need a variable for the number of spheres -- after all, that's what we're trying to find! Let's call it $n$.
NewtonFTW 2016-11-09 20:12:06
If we let r = 1, then the volume of each sphere is 4/3 * pi
DPatrick 2016-11-09 20:12:13
Indeed, the volume of a sphere of radius 1 is $\dfrac43\pi$.
Liopleurodon 2016-11-09 20:12:22
the height is $2n$
DPatrick 2016-11-09 20:12:37
Good -- this is where you could easily mess up. The height of the cylinder is $2n$ (NOT $n$!!)
DPatrick 2016-11-09 20:12:47
So what's the volume of the cylinder?
Liopleurodon 2016-11-09 20:13:17
$2n\pi$
thedoge 2016-11-09 20:13:17
$1^2*\pi*2n=2n\pi$
SkyFox163 2016-11-09 20:13:21
pi 2 n if the radius is 1
nvpthemvp 2016-11-09 20:13:21
2n times pi
2npi
DPatrick 2016-11-09 20:13:31
Yes -- the cylinder has a base of area $\pi$ and a height of $2n$ (since each sphere has diameter 2). So the cylinder's volume is $2n\pi$.
DPatrick 2016-11-09 20:13:45
How do we set up an equation now that we have all this data?
DPatrick 2016-11-09 20:14:27
"volume of cylinder not taken up by spheres" = "volume of 3 spheres"
declanhiller 2016-11-09 20:14:45
2npi-n*4/3pi=3*4/3pi?
Liopleurodon 2016-11-09 20:14:45
$2n\pi-\frac{4}{3}n\pi=4\pi$
vonnan 2016-11-09 20:14:45
2n\pi-4/3\pi*n=3*4/3\pi
gshabanets 2016-11-09 20:14:45
2n*pi - n*4/3*pi = 3*4/3*pi
DPatrick 2016-11-09 20:14:57
Exactly.
DPatrick 2016-11-09 20:15:05
The $n$ spheres together take up $\dfrac43n\pi$ of the volume.
DPatrick 2016-11-09 20:15:13
This means that the volume of the cylinder not taken up by the spheres is $2n\pi - \dfrac43n\pi = \dfrac23n\pi$.
DPatrick 2016-11-09 20:15:22
This is equal to 3 spheres' volume, which is $3 \cdot \dfrac43\pi = 4\pi$.
DPatrick 2016-11-09 20:15:32
Thus, $\dfrac23n\pi = 4\pi$.
Hanting_Z 2016-11-09 20:15:38
6
mamis511 2016-11-09 20:15:38
6
SkyFox163 2016-11-09 20:15:38
So $n=6$
Liopleurodon 2016-11-09 20:15:44
$n=6$
Wave-Particle 2016-11-09 20:15:44
so n=6
awin 2016-11-09 20:15:44
Yep, and then n = 6
duruphi 2016-11-09 20:15:44
n=6
Superman2333 2016-11-09 20:15:44
6
DPatrick 2016-11-09 20:15:47
Solving gives us $n = \boxed{6}$, so there are 6 spheres in the cylinder.
DPatrick 2016-11-09 20:16:08
Two more to go, then killer #10!
DPatrick 2016-11-09 20:16:14
8. Suppose that $a$, $b$, $c$, and $d$ are positive integers such that $a^3 = b^2$, $c^3 = d^2$, and $c-a = 9$. Find $a+b+c+d$.
DPatrick 2016-11-09 20:16:30
How can we approach this?
thedoge 2016-11-09 20:16:45
we know that a is a perfect square
atmchallenge 2016-11-09 20:16:45
We must have $a,c$ are perfect squares and $b,d$ are perfect cubes,
DPatrick 2016-11-09 20:17:06
Aha. If a perfect cube $a^3$ is also a perfect square, then $a$ itself must also be a perfect square.
DPatrick 2016-11-09 20:17:11
Think about this in terms of the prime factorization of $a^3$. Since $a^3$ is a perfect square, all of the powers of its primes must be even. When we take its cube root to get back to $a$, we divide all the exponents by 3. But dividing an even number by 3 still leaves an even number. So all of the exponents in the prime factorization of $a$ are even too, which means that $a$ is itself a perfect square.
skiboy32 2016-11-09 20:17:57
now, we find 2 positive squares that have a difference of 9
DPatrick 2016-11-09 20:18:10
Right, now the $c-a = 9$ equation becomes a difference of squares!
DPatrick 2016-11-09 20:18:17
We have $y^2 - x^2 = 9$. What can we do with this?
DPatrick 2016-11-09 20:18:45
Oops, I accidentally deleted my definitions: $a = x^2$ and $c = y^2$ for some positive integers $x$ and $y$.
Liopleurodon 2016-11-09 20:18:56
$(y-x)(y+x)=9$
fractal161 2016-11-09 20:18:56
factor $y^2-x^2=(y+x)(y-x)$
yuvi18 2016-11-09 20:18:56
(y+x)(y-x) = 9
DPatrick 2016-11-09 20:19:02
Factor! $(y-x)(y+x) = 9$.
DPatrick 2016-11-09 20:19:10
We can't have both factors be 3, because then $x=0$, and we're told all the quantities are positive.
DPatrick 2016-11-09 20:19:23
Therefore, $y-x = 1$ and $y+x = 9$ is the only possibility.
yuvi18 2016-11-09 20:19:38
y = 5 x = 4
Liopleurodon 2016-11-09 20:19:38
$y=5,x=4$
manick 2016-11-09 20:19:38
y=5 and x=4
DPatrick 2016-11-09 20:19:41
This gives $y=5$ and $x=4$.
Puzzled417 2016-11-09 20:20:02
$y^2 = 25$ and $x^2 = 16$
yuvi18 2016-11-09 20:20:02
c = 25 a = 16
DPatrick 2016-11-09 20:20:08
We have $a = x^2 = 16$ and $c = y^2 = 25$.
DPatrick 2016-11-09 20:20:26
Then $b = a^{\frac32} = 64$ and $d = c^{\frac32} = 125$.
bogstop320 2016-11-09 20:20:33
230
mamis511 2016-11-09 20:20:33
230
DPatrick 2016-11-09 20:20:36
Finally $a+b+c+d = 16 + 64 + 25 + 125 = \boxed{230}$.
DPatrick 2016-11-09 20:20:59
On to my favorite, #9:
DPatrick 2016-11-09 20:21:04
9. How many distinct real solutions are there to
$(x+1)^7 + (x+1)^6(x-1) + (x+1)^5(x-1)^2 + \cdots + (x+1)(x-1)^6 + (x-1)^7 = 0 ?$
fractal161 2016-11-09 20:21:48
Write $a=x+1$ and $b=x-1$?
DPatrick 2016-11-09 20:22:01
That is an excellent idea. Let's set $a = x+1$ and $b = x-1$ to make this equation a little easier to work with:

$a^7 + a^6b + a^5b^2 + \cdots + ab^6 + b^7 = 0.$

What can we do with this?
mathgandalfjr 2016-11-09 20:22:20
binomail theorem again??
thedoge 2016-11-09 20:22:20
looks like binomial theorem
skiboy32 2016-11-09 20:22:20
binomial theorem.
Puzzled417 2016-11-09 20:22:20
That is a binomial expansion.
DPatrick 2016-11-09 20:22:26
...except we don't have the binomial coefficients.
DPatrick 2016-11-09 20:22:36
But in fact, this is even easier to work with.
Hanting_Z 2016-11-09 20:22:45
mult by a - b?
zetaintegrator 2016-11-09 20:22:45
*(a-b)
DPatrick 2016-11-09 20:22:55
We can multiply both sides by $(a-b)$, then the whole thing telescopes!
thedoge 2016-11-09 20:23:05
we can also use (a^8-b^8)=(a-b)(a^7+a^6*b+a^5*b^2...+b^7)
DPatrick 2016-11-09 20:23:41
Exactly, if we multiply by $a-b$ (which is just $2$ if you remember what $a$ and $b$ are), we end up with the very simple $a^8 - b^8 = 0$.
Liopleurodon 2016-11-09 20:23:46
$a^8=b^8$
DPatrick 2016-11-09 20:24:00
Indeed: when do two numbers have the same 8th powers?
DPatrick 2016-11-09 20:24:07
Two real numbers, remember!
Liopleurodon 2016-11-09 20:24:18
$a=\pm b$
NewtonFTW 2016-11-09 20:24:18
when they are equal or opposites
a=b or a=-b
DPatrick 2016-11-09 20:24:49
Right, if $a$ and $b$ are real, then for $a^8 = b^8$ we need $a = \pm b$. (If they're complex number we have a lot more options, but we don't care about complex roots in this problem.)
Wave-Particle 2016-11-09 20:24:59
clearly $x-1\neq x+1$ tho
DPatrick 2016-11-09 20:25:15
Right: we can't have $a=b$, because that's $x+1 = x-1$, which is no solution for $x$.
DPatrick 2016-11-09 20:25:26
But we can have $x+1 = -(x-1)$.
thedoge 2016-11-09 20:25:42
There's only one solution--x=0
Liopleurodon 2016-11-09 20:25:42
x=0
vonnan 2016-11-09 20:25:42
x=0, one real solution
skiboy32 2016-11-09 20:25:42
so x=0 only. So the answer is $\boxed{1}$
Wave-Particle 2016-11-09 20:25:42
x=0
DPatrick 2016-11-09 20:25:54
Right: that simplifies to $x+1 = -x+1$, so $x=0$ is the solution.
DPatrick 2016-11-09 20:26:00
So there is only $\boxed{1}$ real solution: namely, $x=0$.
DPatrick 2016-11-09 20:26:33
And finally, on to #10:
DPatrick 2016-11-09 20:26:37
10. What is the smallest (minimum) value of $n$ such that $(n+1)!$ has four more digits than $n!$?
Note: this is the tie-breaking question. In the event of a tie for high scores in a region, the winner is the person with the closest to the correct answer without going over.
zetaintegrator 2016-11-09 20:26:56
the price is right :L
DPatrick 2016-11-09 20:27:17
(yes, this uses "The Price is Right" bidding rules: closest without going over wins the showcase)
TPiR 2016-11-09 20:27:25
LOL
DPatrick 2016-11-09 20:27:34
This is pretty tricky. Any ideas?
skiboy32 2016-11-09 20:27:50
well, we know that the answer is less than 10001...
DPatrick 2016-11-09 20:27:58
This is a good starting point to think about.
DPatrick 2016-11-09 20:28:04
To get from $n!$ to $(n+1)!$, we multiply by $n+1$.
DPatrick 2016-11-09 20:28:36
If $n=9999$, then multiplying by 10000 will add 4 zeros to the number. So $n=9999$ definitely satisfies the property.
DPatrick 2016-11-09 20:29:08
Similarly, if $n=999$, then multiplying by 1000 only adds 3 digits. So $n=999$ definitely does not satisfy the property.
vonnan 2016-11-09 20:29:14
1000-9999
DPatrick 2016-11-09 20:29:23
The answer is definitely in the range $1000 \le n \le 9999$.
bearytasty 2016-11-09 20:29:34
some intuition is that if your number is big enough (leading digit of 9), you can multiply by something near 1000 and it will "skip" over another digit and increase by 4 digits, not 3
DPatrick 2016-11-09 20:29:47
Yes! Multiplying by 1001 might add four digits, if the number we started with was already pretty close to needing an extra digit.
DPatrick 2016-11-09 20:29:52
For example, $9999 \cdot 1001 = 10008999$, gaining 4 digits in the process.
DPatrick 2016-11-09 20:30:13
But how can we quantify this in a way that lets us make a reasonable guess?
thedoge 2016-11-09 20:30:22
we could use logarithms
DPatrick 2016-11-09 20:30:39
There's a thought. Using logs, what's the formula for the number of digits of a number $x$?
DPatrick 2016-11-09 20:30:58
For example, if $2 \le \log_{10} x < 3$, what can we conclude?
thedoge 2016-11-09 20:31:10
x has 3 digits
Wave-Particle 2016-11-09 20:31:10
3 digits
Liopleurodon 2016-11-09 20:31:19
it has 3 digits
3 digits
manick 2016-11-09 20:31:19
3 digits
DPatrick 2016-11-09 20:31:22
Right. Removing the logs (by taking 10 to the power of each term) gives $100 \le x < 1000$, which exactly means that $x$ has 3 digits.
DPatrick 2016-11-09 20:31:52
So the formula for the number of digits of $x$ is $$\lfloor \log_{10} x \rfloor + 1,$$ where $\lfloor \; \rfloor$ means to round down to the nearest integer.
NewtonFTW 2016-11-09 20:32:06
floor function
DPatrick 2016-11-09 20:32:15
Yes, that's usually called the floor function.
DPatrick 2016-11-09 20:32:21
Great. How are $\log_{10} n!$ and $\log_{10} (n+1)!$ related?
thedoge 2016-11-09 20:32:54
log((n+1)!)=log(n!)+log(n+1)
Jorvis 2016-11-09 20:32:59
log n! + log (n+1) = log (n+1)!
DPatrick 2016-11-09 20:33:03
Since $(n+1)! = (n+1) \cdot n!$, we have $\log_{10} (n+1)! = \log_{10} n! + \log_{10} (n+1)$.
DPatrick 2016-11-09 20:33:23
So if we think of $\log_{10} x$ as one less than the digit count (where we through away "fractional" digits), we're adding $\log_{10} (n+1)$ digits every time we go from $n!$ to $(n+1)!$.
DPatrick 2016-11-09 20:33:44
Let's recap: $\log_{10} 1000 = 3$, which is consistent with "multiplying by 1000 adds 3 digits always".
DPatrick 2016-11-09 20:33:55
$\log_{10} 1001 = 3 + \text{ a tiny bit extra}$, so multiplying by 1000 will almost always add just 3 digits, unless the "tiny bit extra" rolls the fractional part of $\log_{10}(1000!)$ over to a new integer, in which case we get a 4th digit.
DPatrick 2016-11-09 20:34:05
Then continuing:

\begin{align*}

\log_{10} 1002 &= 3 + \text{ a slightly larger but still tiny bit extra} \\

\log_{10} 1003 &= 3 + \text{ an even larger but still pretty tiny bit extra }

\end{align*}
DPatrick 2016-11-09 20:34:18
The question is: at what point does adding up all those "extras" result in an extra digit?
DPatrick 2016-11-09 20:34:36
To me, the biggest problem in this computation is that we have no good way of telling what the fractional part of $\log_{10}(1000!)$ is to begin with.
DPatrick 2016-11-09 20:35:06
Certainly if we want to be absolutely sure that we don't go over (under these Price is Right rules), we could guess 1000. (This is the equivalent of the "ONE DOLLAR!" bid on TPiR.)
DPatrick 2016-11-09 20:35:38
But if we want to try something a bit more clever, here is maybe what I'd do.
DPatrick 2016-11-09 20:36:12
$\log_{10}(1001)$ is about $3.0005$. (It turns out that if $\epsilon$ is very small, then $10^\epsilon \approx 1 + 2\epsilon$ is a pretty good approximation.)
DPatrick 2016-11-09 20:36:30
Actually, if you know a little calculus, you may know that $10^\epsilon \approx 1 + \ln(10)\epsilon$ is an even better approximation, and $\ln(10) \approx 2.3$.
DPatrick 2016-11-09 20:36:57
But let's stick with $\log_{10}(1001) \approx 3.0005$. And more generally, $\log_{10}(1000+k) = 3 + .0005k$.
DPatrick 2016-11-09 20:37:17
Admittedly, you kind of have to know this from experience in working with these sorts of calculations.
DPatrick 2016-11-09 20:37:34
So the fractional amount of digits we get by multiplying by 1001, 1002, etc. up through $1000+(k+1)$ is approximately

$\dfrac{1+2+\cdots+(k+1)}{2000} = \dfrac{(k+1)(k+2)}{4000}.$
DPatrick 2016-11-09 20:37:52
And since $\sqrt{4000} \approx 63$, this suggests $n \le 62$ will make the extra fractional digits add up to about 1.
DPatrick 2016-11-09 20:38:22
So now I've got a really crude estimate that the answer can't possibly be too much bigger than about 1062.
DPatrick 2016-11-09 20:38:32
But we also have to keep in mind that $\log_{10}(1000!)$ probably started with some fractional digits to begin with.
DPatrick 2016-11-09 20:38:52
So we might hedge our bets and guess 1031 or even 1020. And we want to be absolutely sure that we don't go over, we could still always guess 1000.
DPatrick 2016-11-09 20:39:02
Anybody want to guess at the actual exact answer?
zetaintegrator 2016-11-09 20:39:14
1042
1-1 is 3 2016-11-09 20:39:21
1042?
1042 xd
DPatrick 2016-11-09 20:39:28
Indeed, it's 1042.
zetaintegrator 2016-11-09 20:39:36
42 is the number of the gods
DPatrick 2016-11-09 20:39:42
Yep, it's a good guess.
DPatrick 2016-11-09 20:39:48
That's it for Round 2!
DPatrick 2016-11-09 20:39:55
I do have some stats on the contest that Mike Breen kindly provided.
mikebreen 2016-11-09 20:39:59
Successful guesses were about what you said: 1020 and 1030, but we did get 1042.
DPatrick 2016-11-09 20:40:07
377 students qualified for Round 2. The median score was 6 (out of 10) and the mean was 5.9.
DPatrick 2016-11-09 20:40:31
As I mentioned earlier, Problem 3 (the history question about Ramanujan and Hardy) was the hardest of the first nine, with about 33% getting it right. Problem 4 (the machine making defective parts) was the next hardest with about 36% getting it right.
DPatrick 2016-11-09 20:40:46
All of the remaining problems (except #10) had success rates of over 50%.
Wave-Particle 2016-11-09 20:40:56
how many people got 10
DPatrick 2016-11-09 20:41:11
I believe just a single person had all of the first nine right and also was correct on #10.
mikebreen 2016-11-09 20:41:24
Students do much better than I would have in high school. Yes, one person got a 10.
DPatrick 2016-11-09 20:41:27
The top scorer on Round 2 in each of 9 regions, plus the top scorer in the Atlanta area, will advance to the National semifinals, to be held live at the 2017 Joint Mathematics Meetings in Atlanta in January.
DPatrick 2016-11-09 20:41:31
zetaintegrator 2016-11-09 20:41:47
how many 9's moo..
DPatrick 2016-11-09 20:42:07
Mike sent me that data, but I lost where I put it. Perhaps he knows.
mikebreen 2016-11-09 20:42:25
About 9% of the 377 scored 9.
bogstop320 2016-11-09 20:42:34
When will the winners be announced?
DPatrick 2016-11-09 20:42:56
I believe they're in the processing of contacting winners to be sure they can travel to Atlanta in January.
mikebreen 2016-11-09 20:43:13
We're waiting for the contestants to confirm that they can participate. We hope at the end of this week or beginning of next.
DPatrick 2016-11-09 20:43:42
And Mike, Bill, and I will all be in Atlanta in person at the finals as well. (Mike had better be -- he's the host!)
DPatrick 2016-11-09 20:43:53
Visit http://www.ams.org/programs/students/wwtbam/wwtbam to learn more about the game and to visit the archive of past years' contests.
mikebreen 2016-11-09 20:44:21
Yes, we'll be there. Watch us live at http://www.livestream.com/wwtbam2017.
DPatrick 2016-11-09 20:44:28
That's it for our Math Jam tonight -- thanks for attending!
DPatrick 2016-11-09 20:44:46
If you missed any part of our discussion, a transcript will be up on the AoPS website soon.
mikebreen 2016-11-09 20:44:51
Thanks, everyone! Great work by you and Dave.
TPiR 2016-11-09 20:44:51
Thanks, Dave. Great job, as always.
mathsm 2016-11-09 20:48:05
How do you access the transcript?
DPatrick 2016-11-09 20:48:26
Once it's up (probably in about 15-30 minutes), go to "Math Jams" under the Online School menu of the website.
DPatrick 2016-11-09 20:48:32
Then click on the "Previous Math Jams" button.
zetaintegrator 2016-11-09 20:49:21
can you release who was the perfect scorer? xD
DPatrick 2016-11-09 20:49:37
I suspect they will soon. Maybe not though.
DPatrick 2016-11-09 20:49:53
(I don't know who it is.)
mikebreen 2016-11-09 20:50:35
Not now. He is one of the 10, though. We may post it on the page profiling the 10 contestants and talk about it at the game in Atlanta.
DPatrick 2016-11-09 20:52:35
It's getting late in the east and I suspect many of you were up late last night. So let's end it here. Have a great evening!