Math Jams

## Who Wants to Be a Mathematician, Round 2

Go back to the Math Jam Archive

AoPS instructor David Patrick will discuss the problems on Round 2 of the 2017-2018 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 San Diego in January 2018.

#### Facilitator: Dave Patrick

ag2006 2017-10-26 19:29:53
wait... weren't you on "DO you want to be a millionaire" with the old host?
DPatrick 2017-10-26 19:29:56
I was.
DPatrick 2017-10-26 19:30:03
I once was a contestant on ABC's Who Wants to Be a Millionaire back before I started working at AoPS, way back when Regis Philbin was still the host. Here's a picture (I'm on the left, Regis is on the right):
DPatrick 2017-10-26 19:30:06
DPatrick 2017-10-26 19:30:11
Photo Credit: Maria Melin, copyright 1999 ABC Television.
ag2006 2017-10-26 19:30:36
What was the question you got out on? Or did you get the million dollars?
DPatrick 2017-10-26 19:30:42
No, I didn't win the million bucks, but I did win enough to buy a new car.
DPatrick 2017-10-26 19:30:49
And that seems like a good note on which to get started...
DPatrick 2017-10-26 19:30:53
Welcome to the 2017-18 Who Wants to Be a Mathematician Round 2 Math Jam!
DPatrick 2017-10-26 19:31:09
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 13 years, and I've written or co-written a few of our textbooks.
DPatrick 2017-10-26 19:31:20
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 2017-10-26 19:31:33
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 2017-10-26 19:31:45
This helps keep the class 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 2017-10-26 19:32:00
There are a lot of students here! Please do not take it personally if your comments do not get posted, and please do not complain about it. I expect this Math Jam to be much larger than our typical class, so please be patient with me---there are quite a few of you here tonight!!
DPatrick 2017-10-26 19:32:16
Also, we won't be going through the math quite as thoroughly as we do in our classes -- I can't teach all the material for every problem as we go.
DPatrick 2017-10-26 19:32:26
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 2017-10-26 19:32:37
Joining us tonight are the co-creators of WWTBAM, Mike Breen (mikebreen) and Bill Butterworth (TPiR).
mikebreen 2017-10-26 19:32:41
Hello, everyone.
DPatrick 2017-10-26 19:32:44
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.
TPiR 2017-10-26 19:32:55
Hi Everyone
DPatrick 2017-10-26 19:33:04
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 2017-10-26 19:33:25
As you can see, we have a lot of game show experience here tonight!
DPatrick 2017-10-26 19:33:44
We also have an assistant here to help out tonight: Erin Lipman (SqRootNegativeOne). Erin has a B.S. in math from Haverford college and is studying Statistics at the University of Chicago. Erin's favorite areas of math are knot theory and probability. Her non-mathematical hobbies include board games, knitting, and cooking.
DPatrick 2017-10-26 19:34:04
She can try to help you if you have a question or are having some other difficulty. She may open a private window with you to chat if needed.
DPatrick 2017-10-26 19:34:22
Who Wants to Be a Mathematician is a contest 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 2017-10-26 19:34:38
Tonight we'll be talking about Round 2 of the national contest, which was held earlier this month. Round 2 was only open to students who qualified with a score of at least 8 (out of 10) on Round 1, held in September. (If you missed our Round 1 discussion, the transcript is available on our website.)
DPatrick 2017-10-26 19:34:57
Round 2, like Round 1, consisted of 10 questions, with a 15-minute time limit. So the problems are pretty quick. No books, notes, calculators, or internet access was permitted during the test.
ag2006 2017-10-26 19:35:07
do you take the test on the computer?
DPatrick 2017-10-26 19:35:14
Most students do. Some take it on paper.
DPatrick 2017-10-26 19:35:34
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 also includes discussing problem-solving tactics.
DPatrick 2017-10-26 19:36:04
We'll take questions about how the contest works at the end if we have time, but for now, let's get to the problems!
DPatrick 2017-10-26 19:36:13
1. What is the ones (units) digit of the sum $2^{2017}+3^{2017}+7^{2017}$?
DPatrick 2017-10-26 19:36:33
How can we approach this?
STEM88 2017-10-26 19:36:52
find a pattern
ag2006 2017-10-26 19:36:52
its important to find the units digit of each and add those.
avn 2017-10-26 19:36:52
find a pattern
goodbear 2017-10-26 19:36:52
pattern
yali306 2017-10-26 19:36:52
Find a pattern
int_user 2017-10-26 19:36:59
The units digits repeat, so we can find a pattern
mathwiz0 2017-10-26 19:36:59
find a pattern
DPatrick 2017-10-26 19:37:16
Right. We can look for a pattern in the units digits of the powers of 2, 3, and 7.
DPatrick 2017-10-26 19:37:23
And throughout, we only care about the units digit. So that's a lot less to keep track of.
DPatrick 2017-10-26 19:37:46
What's the pattern of units digits of powers of 2?
ShreyJ 2017-10-26 19:38:03
2, 4, 8, 6: cycles of 2
shootingstar8 2017-10-26 19:38:03
2 => 2,4,8,6
prajna1225 2017-10-26 19:38:03
2 4 8 6
int_user 2017-10-26 19:38:03
2, 4, 8, 6
GeronimoStilton 2017-10-26 19:38:03
2,4,8,6…
Danshim917 2017-10-26 19:38:03
2 4 8 6
SirCalcsALot 2017-10-26 19:38:03
2 -> 4 -> 8 -> 6 -> 2
amypham2 2017-10-26 19:38:03
2, 4, 8, 6
Rushn 2017-10-26 19:38:03
2,4,8,6
Heatblast016 2017-10-26 19:38:03
2 4 8 6 2
DPatrick 2017-10-26 19:38:19
$2^1$ has units digit 2.
DPatrick 2017-10-26 19:38:37
$2^2$ has units digit 2*2 = 4.
DPatrick 2017-10-26 19:38:43
$2^3$ has units digit 2*4 = 8.
DPatrick 2017-10-26 19:38:46
$2^4$ has the same units digit as 2*8 = 16, which is 6.
DPatrick 2017-10-26 19:38:50
$2^5$ has the same units digit as 2*6 = 12, which is 2.
DPatrick 2017-10-26 19:38:57
Aha, we're back to a units digit of 2! So the units digits of powers of 2 go: 2, 4, 8, 6, 2, 4, 8, 6, 2, ..., with the "2,4,8,6" block of 4 repeating.
DPatrick 2017-10-26 19:39:08
Indeed, we can make a little table of the units digits, depending on the the relationship of the exponent modulo 4:
DPatrick 2017-10-26 19:39:10
$\begin{array}{r|c|c|c|c} n \pmod{4} & 1 & 2 & 3 & 0 \\ \hline \text{Units digit of } 2^n & 2 & 4 & 8 & 6 \end{array}$
DPatrick 2017-10-26 19:39:24
If you don't know the termonology, "mod 4" just means the remainder when we divide by 4.
amypham2 2017-10-26 19:39:39
so 4 patterns... 2017/4 = 504 with 1 remaining... does it mean the ones digit is a 2?
SirCalcsALot 2017-10-26 19:39:48
The units digit of $2^{2017$ is $2$.
DPatrick 2017-10-26 19:39:53
Exactly.
DPatrick 2017-10-26 19:39:57
$2017 \equiv 1 \pmod{4}$ (that is, 2017 is 1 more than a multiple of 4), so $2^{2017}$ has units digit 2, by our table.
DPatrick 2017-10-26 19:40:07
How about $3^{2017}$?
Mathforlife1 2017-10-26 19:40:17
3 9 7 1
goodbear 2017-10-26 19:40:17
3 9 7 1 3 9 7 1 3 9 7 1 3 9 7 1 3 9 7 1 3 9 7 1 3 9 7 1 3 …
int_user 2017-10-26 19:40:17
3, 9, 7, 1
DPatrick 2017-10-26 19:40:28
There's a similar pattern:
DPatrick 2017-10-26 19:40:31
$\begin{array}{r|c|c|c|c} n \pmod{4} & 1 & 2 & 3 & 0 \\ \hline \text{Units digit of } 2^n & 2 & 4 & 8 & 6 \\ \text{Units digit of } 3^n & 3 & 9 & 7 & 1 \end{array}$
Mathforlife1 2017-10-26 19:40:40
which would mean 3
shootingstar8 2017-10-26 19:40:40
3,9,7,1 => 3
amypham2 2017-10-26 19:40:40
so itd end with 3 again
DPatrick 2017-10-26 19:40:47
Right. The units digit of $3^n$ also repeats in blocks of 4: 3,9,7,1,... So $3^{2017}$ has units digit 3, by our table.
Danshim917 2017-10-26 19:40:57
and 7 goes 7 9 3 1
RinoaOliver 2017-10-26 19:41:03
7, 9, 3, 1
Giraffefun 2017-10-26 19:41:03
7,9,3,1
Mathforlife1 2017-10-26 19:41:03
7 9 3 1
DPatrick 2017-10-26 19:41:04
Finally, we can see that the units digits of $7^n$ also repeat in a block of four:
DPatrick 2017-10-26 19:41:07
$\begin{array}{r|c|c|c|c} n \pmod{4} & 1 & 2 & 3 & 0 \\ \hline \text{Units digit of } 2^n & 2 & 4 & 8 & 6 \\ \text{Units digit of } 3^n & 3 & 9 & 7 & 1 \\ \text{Units digit of } 7^n & 7 & 9 & 3 & 1 \end{array}$
DPatrick 2017-10-26 19:41:21
So $7^{2017}$ has units digit 7.
Danshim917 2017-10-26 19:41:32
when we add 2, 3, and 7 we get a units digit of 2
SirCalcsALot 2017-10-26 19:41:32
Then, the answer is the units digit of 2 + 3+ 7 which is 2.
GeronimoStilton 2017-10-26 19:41:32
mikebreen 2017-10-26 19:41:50
Unfortunately, some who got this wrong, but who knew what they were doing, answered "12."
DPatrick 2017-10-26 19:41:51
Right. We add the units digits: $2+3+7 = 12$, and thus our answer is the units digit of the sum is the units digit of 12, or $\boxed{2}$.
DPatrick 2017-10-26 19:42:24
Yes, Mike has a good reminder to do a quick check of your final answer, and make sure you're answering what the question asked!
DPatrick 2017-10-26 19:42:52
By the way, as a few of you mentioned to me: if you know a little more advanced number theory, you might know some more sophisticated tools that you can use with this problem. I'll skip that discussion now, but we'll see some of these tools in a later problem.
GeronimoStilton 2017-10-26 19:43:09
by the way, what are the stats on these problems?
DPatrick 2017-10-26 19:43:21
We don't have the full stats because the paper submissions are still being scored.
DPatrick 2017-10-26 19:43:37
But on the ones that were completed electronically, this one was the easiest: about 90% correct.
DPatrick 2017-10-26 19:44:03
I'm not going to list the stats individually for #2-#8: they're all in the band of 55% to a little over 80%.
DPatrick 2017-10-26 19:44:11
#9 was much harder; I'll discuss it when we get to it.
DPatrick 2017-10-26 19:44:17
On to #2:
DPatrick 2017-10-26 19:44:22
2. Which of the following is largest?
(a) $\sqrt[4]{101}$ (b) $5 - \sqrt3$ (c) $\sqrt2 + \sqrt3$ (d) $\sqrt{10}$
DPatrick 2017-10-26 19:44:45
Any we can rule out right away?
GeronimoStilton 2017-10-26 19:45:00
eliminate (d) immediately
RinoaOliver 2017-10-26 19:45:00
(a) > (d)
fdas 2017-10-26 19:45:00
d is wrong. less than a
Emathmaster 2017-10-26 19:45:00
a>d
DPatrick 2017-10-26 19:45:05
Clearly $\sqrt[4]{101} > \sqrt[4]{100} = \sqrt{10}$. So we can eliminate (d).
DPatrick 2017-10-26 19:45:11
Let's update the problem:
DPatrick 2017-10-26 19:45:14
2. Which of the following is largest?
(a) $\sqrt[4]{101}$ (b) $5 - \sqrt3$ (c) $\sqrt2 + \sqrt3$
ShreyJ 2017-10-26 19:45:29
compare b and c
DPatrick 2017-10-26 19:45:55
There are a few different ways you can go from here. You can use $\sqrt2 \approx 1.4$ and $\sqrt3 \approx 1.7$ and estimate from there.
DPatrick 2017-10-26 19:46:18
But if we want to be more rigorous (since we have the luxury of a little more time tonight), we can compare (b) and (c) since they're the two that only use square roots.
DPatrick 2017-10-26 19:46:30
So which way does the inequality go in

$5 - \sqrt3 \qquad\boxed{?}\qquad \sqrt2 + \sqrt3$
EulerMacaroni 2017-10-26 19:46:53
square both sides
ShreyJ 2017-10-26 19:46:53
add sqrt 3 and square it
DPatrick 2017-10-26 19:47:07
We can legally:
- add or subtract the same thing from both sides
- multiple or divide the same positive thing from both side
- square both sides, provided both sides are positive
DPatrick 2017-10-26 19:47:16
...meaning these all preserve the inequality, whatever it is.
DPatrick 2017-10-26 19:47:51
Adding $\sqrt3$ to both sides cleans it up a bit -- if our goal is to square, then we won't get square roots on both sides again.
DPatrick 2017-10-26 19:48:18
But I'd also move the $\sqrt2$ to the left, so keep the square roots smaller:
DPatrick 2017-10-26 19:48:23
$5 - \sqrt2 \qquad\boxed{?}\qquad 2\sqrt3$
mathwiz0 2017-10-26 19:48:37
square both sides no
DPatrick 2017-10-26 19:48:41
Right: both sides are positive, so we can square without changing the inequality.

$27 - 10\sqrt2 \qquad\boxed{?}\qquad 12$
DPatrick 2017-10-26 19:49:05
So now isolating the square root term gives

$15 \qquad\boxed{?}\qquad 10\sqrt2$
DPatrick 2017-10-26 19:49:29
We can probably eyeball it here, or we can square both sides (which are positive) again to be sure:

$225 \qquad\boxed{>}\qquad 200$
DPatrick 2017-10-26 19:49:45
So since all of our steps preserved the inequality, we can go backwards to the beginning and conclude that $$5 - \sqrt3 > \sqrt2 + \sqrt3.$$
ppu 2017-10-26 19:49:50
now compare and and b
DPatrick 2017-10-26 19:49:53
2. Which of the following is largest?
(a) $\sqrt[4]{101}$ (b) $5 - \sqrt3$
GeronimoStilton 2017-10-26 19:50:12
square!
DPatrick 2017-10-26 19:50:19
Let's try the same tactic. We can square both sides and preserve the inequality:

$\sqrt{101} \qquad\boxed{?}\qquad 28 - 10\sqrt3$
DPatrick 2017-10-26 19:50:39
At this point, if we square again the numbers will start getting big.
DPatrick 2017-10-26 19:50:48
So at this point, you might use the fact that $\sqrt3 \approx 1.7$ to the nearest tenth. (This is easy to check: $(1.7)^2 = 2.89$ and $(1.8)^2 = 3.24$, and if you want to be really careful $(1.75)^2 = 3.0625$.)
Vfire 2017-10-26 19:50:54
$\sqrt{101} \approx 10$
DPatrick 2017-10-26 19:51:07
Indeed the left side is only a little bit bigger than 10; indeed $(10.5)^2 = 110.25$ so even $\sqrt{110}$ rounds to 10, so certainly $\sqrt{101}$ rounds to 10.
prajna1225 2017-10-26 19:51:20
b is the winner
DPatrick 2017-10-26 19:51:41
The right side rounds to $28-17 = 11$ to the nearest integer. So the right side is bigger.
DPatrick 2017-10-26 19:51:45
Therefore, $\boxed{\text{(b) } 5 - \sqrt3}$ is the largest.
DPatrick 2017-10-26 19:52:13
Again, in real time, you might not have done quite so much rigor -- you might have just "eyeballed" it to the nearest tenth. But the rigor is there if you need it!
DPatrick 2017-10-26 19:52:24
3. An emirp is a prime number that is also a different prime number when its digits are reversed (so for example, 107 is an emirp but 101 is not). How many emirps are there between 1 and 100?
DPatrick 2017-10-26 19:52:51
Are there any 1-digit emirps?
Cidkip 2017-10-26 19:53:09
No, thats the same prime number
int_user 2017-10-26 19:53:09
No because it has to be different
Danshim917 2017-10-26 19:53:09
No because theyre not differnet
AlcumusGuy 2017-10-26 19:53:09
No, reversing gives the same number.
DPatrick 2017-10-26 19:53:17
No: reversing a 1-digit number gives us the same 1-digit number, and we have to make a different number when we reverse the digits.
DPatrick 2017-10-26 19:53:24
So we need only look for 2-digit emirps.
EulerMacaroni 2017-10-26 19:53:34
each digit has to be odd and not 5
AOPS12142015 2017-10-26 19:53:44
The emirp can only have digits of 1,3,7, or 9.
DPatrick 2017-10-26 19:53:56
So the only digits we can use are 1, 3, 7, or 9.
DPatrick 2017-10-26 19:54:03
And it's clear from the definition that emirps come in pairs.
int_user 2017-10-26 19:54:08
and have distinct digits
DPatrick 2017-10-26 19:54:11
Right.
DPatrick 2017-10-26 19:54:15
So we can just look at the six different pairs of digits:
DPatrick 2017-10-26 19:54:18
1 and 3: 13 and 31

1 and 7: 17 and 71

1 and 9: 19 and 91

3 and 7: 37 and 73

3 and 9: 39 and 93

7 and 9: 79 and 97
DPatrick 2017-10-26 19:54:56
Which of these pairs have both numbers prime?
mathwiz0 2017-10-26 19:55:13
39 and 93 are divisible by 3
int_user 2017-10-26 19:55:13
13x7=91 so its not prime
Cidkip 2017-10-26 19:55:13
3/9 and 1/9 can be eliminated: 39 = 3*13 isn't prime, 91 = 7*13 can also be
DPatrick 2017-10-26 19:55:19
19 is prime, but 91 = 7 * 13 is not. So these are not emirps.
DPatrick 2017-10-26 19:55:34
(Beware 91. It looks like it might be prime, but it's not!)
DPatrick 2017-10-26 19:55:39
39 = 3 * 13 and 93 = 3 * 31 are not prime. So these are not emirps.
DPatrick 2017-10-26 19:55:49
The rest are all prime!
DPatrick 2017-10-26 19:56:07
So there are 4 pairs giving $\boxed{8}$ emirps between 1 and 100.
ag2006 2017-10-26 19:56:09
are emirps real?
DPatrick 2017-10-26 19:56:16
https://en.wikipedia.org/wiki/Emirp
mikebreen 2017-10-26 19:56:29
uoy rof dooG
DPatrick 2017-10-26 19:56:43
Moving on...
DPatrick 2017-10-26 19:56:47
4. Evaluate $\left|\dfrac{(1+\sqrt3\,i)^8}{(1-\sqrt3\,i)^6}\right|$ where $i = \sqrt{-1}$ and $|a + bi| = \sqrt{a^2+b^2}$.
DPatrick 2017-10-26 19:57:04
What's going to be a really useful fact for us to use?
Cidkip 2017-10-26 19:57:36
$|1+\sqrt{3}i| = |1-\sqrt{3}i| = 2$
EulerMacaroni 2017-10-26 19:57:36
norm is multiplicative
ppu 2017-10-26 19:57:41
magnitude of top divided by bottom
DPatrick 2017-10-26 19:58:01
Right, In simpler language, what EulerMacaroni said is that in complex numbers, the absolute value "commutes" with multiplication. That is, $|xy| = |x||y|$ for any complex numbers $x$ and $y$.
kvedula2004 2017-10-26 19:58:19
|a/b|=|a|/|b|
DPatrick 2017-10-26 19:58:26
This is easy to prove algebraically if you haven't seen this before, but I'll leave it as an exercise for you to do if you like. It's even easier to prove if you write complex numbers in polar form as $re^{i\theta}$ for some $r \ge 0$ and angle $\theta$, but that's a topic for another day.
DPatrick 2017-10-26 19:59:01
So using this "commutative" property of absolute value (which is more technically called "norm" or "magnitude"), we can rewrite our quantity as $\dfrac{(|1+\sqrt3\,i|)^8}{(|1-\sqrt3\,i|)^6}$.
DPatrick 2017-10-26 19:59:21
And nicely, as was mentioned previously, $|1 \pm \sqrt{3}\,i| = 2$.
DPatrick 2017-10-26 19:59:44
(If you like to think geometrically, it's the hypotenuse of a right triangle with legs $1$ and $\sqrt3$. That's a 30-60-90 triangle!)
RinoaOliver 2017-10-26 19:59:50
2^8/2^6 = 2^2 = 4
Kreps 2017-10-26 19:59:50
so its 4
DPatrick 2017-10-26 19:59:53
So this is just $\dfrac{2^8}{2^6} = 2^2 = \boxed{4}$.
DPatrick 2017-10-26 20:00:16
5. Given that $\displaystyle\sum_{k=1}^n \log_a(k^n) = n$, find $a$ in terms of $n$.
GeronimoStilton 2017-10-26 20:00:37
Write out the sum!
DPatrick 2017-10-26 20:00:49
This is a matter of taste, but I agree with Geronimo.
DPatrick 2017-10-26 20:00:53
Although the $\Sigma$ notation is compact, it sometimes obscures what's really going on.
DPatrick 2017-10-26 20:01:03
So let's get rid of the Sigma notation. What we have is a sum of logs:

$$\log_a(1^n) + \log_a(2^n) + \cdots + + \log_a(n^n) = n.$$
ppu 2017-10-26 20:01:16
log (a) + log(b) = log(ab)
DPatrick 2017-10-26 20:01:26
Aha, that rule is going to come in handy!
DPatrick 2017-10-26 20:01:35
To add logarithms, we multiply the terms that we're taking the log of. That is, $\log_a(x) + \log_a(y) = \log_a(xy)$.
DPatrick 2017-10-26 20:01:45
So our sum of logs becomes the log of a product.
DPatrick 2017-10-26 20:01:51
$$\log_a(1^n \cdot 2^n \cdot \,\cdots\, \cdot n^n) = n.$$
DPatrick 2017-10-26 20:02:04
But what's the product of a bunch of $n$th powers?
GeronimoStilton 2017-10-26 20:02:21
$n!^n$
ShreyJ 2017-10-26 20:02:21
$n!^n$
GeronimoStilton 2017-10-26 20:02:25
That's $$\log_a(n!^n) = n.$$
Kreps 2017-10-26 20:02:27
(n!)^n
DPatrick 2017-10-26 20:02:43
Right, we just have a big $n$th power: $$\log_a\left((1 \cdot 2 \cdot \,\cdots\, \cdot n)^n\right) = n.$$
DPatrick 2017-10-26 20:02:57
And that product inside the $n$th power is just $n!$.
DPatrick 2017-10-26 20:03:04
So we have $\log_a \left((n!)^n\right) = n$
AOPS12142015 2017-10-26 20:03:18
$a^n=(n!)^n$
int_user 2017-10-26 20:03:18
now we can bring the n out front!
Emathmaster 2017-10-26 20:03:18
we can now pull the nth power out
Kreps 2017-10-26 20:03:18
rewrite it as an exponential equation
DPatrick 2017-10-26 20:03:32
Either way to finish. At this point I just wrote this equation without the log:
DPatrick 2017-10-26 20:03:35
It's just $a^n = (n!)^n$.
awesomemaths 2017-10-26 20:03:42
a = n!
Danshim917 2017-10-26 20:03:48
so a=n!
DPatrick 2017-10-26 20:03:50
So clearly $a = \boxed{n!}$.
mikebreen 2017-10-26 20:04:00
Bill and I were very pleased that so many got this right.
DPatrick 2017-10-26 20:04:15
6. A googol is 1 followed by 100 zeroes. A googolplex is 1 followed by a googol zeroes. For what exponent $n$ is $\text{googol}^{\text{googol}} = \text{googolplex}^n$?
DPatrick 2017-10-26 20:04:36
Clearly we need to replace the words with actual math.
DPatrick 2017-10-26 20:04:44
What's a better way to write a googol?
Abch 2017-10-26 20:05:02
Googol=10^100
kvedula2004 2017-10-26 20:05:02
10^100
goodbear 2017-10-26 20:05:02
10^100
fishy15 2017-10-26 20:05:02
$10^{100}$
maverick8 2017-10-26 20:05:02
$10^{100}$
DPatrick 2017-10-26 20:05:22
It's $10^{100}$. That's a 1 followed by 100 zeros (the same way $10^3 = 1000$ is a 1 followed by 3 zeros).
DPatrick 2017-10-26 20:05:30
And a googolplex?
ShreyJ 2017-10-26 20:05:51
10^10^100
ppu 2017-10-26 20:05:51
10^10^100
kvedula2004 2017-10-26 20:05:51
10^10^100
goodbear 2017-10-26 20:05:51
10^10^100
DPatrick 2017-10-26 20:05:54
It's $10^{\text{googol}} = 10^{10^{100}}$.
DPatrick 2017-10-26 20:06:01
So our equation is $\left(10^{100}\right)^{10^{100}} = \left(10^{10^{100}}\right)^n$.
DPatrick 2017-10-26 20:06:11
This looks a lot scarier that it really is.
GeronimoStilton 2017-10-26 20:06:25
exponent properties!
DPatrick 2017-10-26 20:06:34
Right. In particular, we can now use the identity $(a^b)^c = a^{bc}$ on both sides.
DPatrick 2017-10-26 20:06:41
$10^{(100 \cdot 10^{100})} = 10^{(10^{100} \cdot n)}$
Kreps 2017-10-26 20:06:59
n = 100
awesomemaths 2017-10-26 20:06:59
n = 100
AOPS12142015 2017-10-26 20:06:59
$n=100$
SirCalcsALot 2017-10-26 20:06:59
Then, n clearly equals 100.
DPatrick 2017-10-26 20:07:04
The exponents on either side must match, so $100 \cdot 10^{100} = 10^{100} \cdot n$.
DPatrick 2017-10-26 20:07:09
And this gives us $n = \boxed{100}$.
DPatrick 2017-10-26 20:07:44
Note that 5 and 6 were pretty similar, in the sense that if you can fluently manipulate exponents and logs, they're pretty straightforward. Don't be scared by exponents and logs!
DPatrick 2017-10-26 20:07:54
7. What is the sum of the squares of the roots (zeroes) of $x^4 - 8x^3 + 16x^2 - 11x + 5$?
DPatrick 2017-10-26 20:08:13
Let's call the roots $p,q,r,s$ so we can talk about them. So, we're trying to find $p^2+q^2+r^2+s^2$.
GeronimoStilton 2017-10-26 20:08:36
vieta!
ShreyJ 2017-10-26 20:08:36
Vieta's
awesomemaths 2017-10-26 20:08:36
vietas
DPatrick 2017-10-26 20:08:50
Vieta is going to be our dear friend here.
DPatrick 2017-10-26 20:09:10
"Vieta's formulas" are how we relate the roots of a polynomial to its coefficients. I'll briefly explain in case you haven't seen it before.
DPatrick 2017-10-26 20:09:28
We can write the polynomial as factored in terms of its roots:

$x^4 - 8x^3 + 16x^2 - 11x + 5 = (x-p)(x-q)(x-r)(x-s).$
DPatrick 2017-10-26 20:09:46
If we multiply that right side out and compare coefficients to the left side, we get Vieta's formulas.
DPatrick 2017-10-26 20:10:04
For example, the coefficient on the right side of $x^3$, if you expand it out, is $-(p+q+r+s)$. On the left side it's $-8$.
DPatrick 2017-10-26 20:10:08
So $p+q+r+s = 8$.
DPatrick 2017-10-26 20:10:24
What about the $x^2$ coefficients? What formula does that give us?
RinoaOliver 2017-10-26 20:10:55
pq + pr + ps + qr + qs + rs
DPatrick 2017-10-26 20:11:09
Yes: on the right side, we get the sum of all the products of pairs of roots.
Kreps 2017-10-26 20:11:11
= 16
DPatrick 2017-10-26 20:11:18
And on the left side, we have the 16.
DPatrick 2017-10-26 20:11:22
Comparing those tells us that $pq + pr + ps + qr + qs + rs = 16$.
DPatrick 2017-10-26 20:11:36
How does these help? Remember, we want $p^2 + q^2 + r^2 + s^2$.
(p+q+r+s)^2-2(pq+pr+...)
RinoaOliver 2017-10-26 20:11:49
(p + q + r + s)^2
Kreps 2017-10-26 20:11:54
we can substitute these into the expansion of (p+q+r+s)^2
Emathmaster 2017-10-26 20:11:54
(p+q+r+s)^2-2(...)
DPatrick 2017-10-26 20:12:02
Exactly. We can use the fact that

$(p+q+r+s)^2 = p^2 + q^2 + r^2 + s^2 + 2(pq+pr+ps+qr+qs+rs)$
DPatrick 2017-10-26 20:12:12
But we know most of those terms, and what we don't know is what we want!
DPatrick 2017-10-26 20:12:22
Specifically,

$8^2 = p^2 + q^2 + r^2 + s^2 + 2(16).$
Emathmaster 2017-10-26 20:12:36
so ans is 32
int_user 2017-10-26 20:12:36
32
Danshim917 2017-10-26 20:12:36
so 32
goodbear 2017-10-26 20:12:36
32
DPatrick 2017-10-26 20:12:40
So $p^2 + q^2 + r^2 + s^2 = 8^2 - 2(16) = 64 - 32 = \boxed{32}$.
DPatrick 2017-10-26 20:13:27
This is a pretty typical "intermediate algebra" sort of problem that you're likely to see on the AMC 10/12 or AIME or WWTBAM or similar contests. It's really useful to know when and how to use Vieta's formulas!
DPatrick 2017-10-26 20:13:37
8. From each of the three corners of a 5-12-13 right triangle remove a sector of a circle of radius 2 (centered at the corresponding corner, as shown). What is the area of the resulting figure?
DPatrick 2017-10-26 20:13:40
DPatrick 2017-10-26 20:14:14
It took me longer to recreate the picture than it did for me to solve it. There's a key observation to make that makes the solution almost immediate.
EulerMacaroni 2017-10-26 20:14:29
angles add to 180 so its half of the circle
Danshim917 2017-10-26 20:14:29
AOPS12142015 2017-10-26 20:14:29
Use the fact that the sum of the angles of a triangle adds up to ${180}^{\circ}$
beastieric123 2017-10-26 20:14:29
angles of a triangle sum to 180
ppu 2017-10-26 20:14:29
Notice that the sum of the angles of a triangle is 180 degrees, so the sectors make up a semicircle of radius 2
CornSaltButter 2017-10-26 20:14:29
The sector's angles have to add up to 180 degrees
fishy15 2017-10-26 20:14:29
we can put all the chopped off corners next to one another
DPatrick 2017-10-26 20:14:51
Right. The angles of the triangle add up to $180^\circ$, so the blue regions add up to exactly one-half of a circle of radius 2.
DPatrick 2017-10-26 20:15:10
So our answer is just the area of the triangle, minus the area of a semicircle of radius 2.
Emathmaster 2017-10-26 20:15:24
30-2pi
ppu 2017-10-26 20:15:24
$30-2\pi$ is our answer!
EasyAs_Pi 2017-10-26 20:15:24
:facepalm: $30-2\pi$....
rainbow10 2017-10-26 20:15:24
30-2pi
Mathforlife1 2017-10-26 20:15:24
30-2pi
DPatrick 2017-10-26 20:15:34
The triangle area is $\frac12 \cdot 12 \cdot 5 = 30$.
DPatrick 2017-10-26 20:15:40
The total blue area is $\frac12 \cdot \pi(2)^2 = 2\pi$.
DPatrick 2017-10-26 20:15:44
Hence, the area remaining is $\boxed{30-2\pi}$.
mikebreen 2017-10-26 20:15:56
Smart group.
DPatrick 2017-10-26 20:16:04
That was an easy problem, the calm before the storm.
DPatrick 2017-10-26 20:16:10
The last 2 are really much harder.
DPatrick 2017-10-26 20:16:29
Only about a quarter of the students got #9, which is:
DPatrick 2017-10-26 20:16:33
9. What is the remainder when $1000^{2018}$ is divided by $2018$?
DPatrick 2017-10-26 20:17:04
What might help to know?
Ancy 2017-10-26 20:17:07
1009 is prime
DPatrick 2017-10-26 20:17:16
Yes, that's useful!
DPatrick 2017-10-26 20:17:30
We have the prime factorization $2018 = 2 \cdot 1009$.
DPatrick 2017-10-26 20:17:46
The other (obvious, perhaps) prime factorization that's probably useful is $1000 = 10^3 = 2^3 \cdot 5^3$.
Cidkip 2017-10-26 20:17:56
Computing things mod a composite is hard. Computing things mod prime is easier, so compute it mod 2 and mod 1009 instead of mod 2018, then use CRT
DPatrick 2017-10-26 20:18:14
Exactly. The Chinese Remainder Theorem tells us that (in theory) if we can compute this separately mod 2 and mod 1009, then we should be able to combine those results to get our answer mod 2018.
DPatrick 2017-10-26 20:18:28
First, the easy half: what is $1000^{2018} \pmod{2}$?
DPatrick 2017-10-26 20:19:00
I won't bother posting all your correct responses. It's an even number, so it's 0 mod 2.
DPatrick 2017-10-26 20:19:09
And how do we approach $1000^{2018} \pmod{1009}$?
EasyAs_Pi 2017-10-26 20:19:29
fermat's little theorem
SHARKYKESA 2017-10-26 20:19:29
Euler totient function
Emathmaster 2017-10-26 20:19:29
FLT
plshalp 2017-10-26 20:19:29
fLT
ppu 2017-10-26 20:19:36
since 1000 and 1009 are relatively prime, we can use Euler's Totient Theorem
CornSaltButter 2017-10-26 20:19:36
use fermat's little theorem, 1000^1008 is 1 mod 1009
DPatrick 2017-10-26 20:20:07
Right. Here the tool to grab is Fermat's Little Theorem (or we could use its generalization, Euler's Theorem, but it's not needed).
DPatrick 2017-10-26 20:20:14
Fermat's Little Theorem tells us that if $p$ is prime, then $a^{p-1} \equiv 1 \pmod{p}$, provided $a$ is not a multiple of $p$.
DPatrick 2017-10-26 20:20:33
(This is a bit difficult to prove unless you've had a fair amount of number theory experience, so I won't prove this here.)
Superwiz 2017-10-26 20:20:43
Note 1000^1008 is 1 mod 1009
AlcumusGuy 2017-10-26 20:20:47
1000^1008 is 1 modulo 1009
DPatrick 2017-10-26 20:20:56
Right. In our particular example, it tells us that $1000^{1008} \equiv 1 \pmod{1009}$.
DPatrick 2017-10-26 20:22:07
We then have $1000^{1009} \equiv 1000 \pmod{1009}$.
DPatrick 2017-10-26 20:22:15
But 1000 and -9 are the same in mod 1009.
DPatrick 2017-10-26 20:22:24
So $1000^{1009} \equiv -9 \pmod{1009}$.
Superwiz 2017-10-26 20:22:34
Square it to get 81
DPatrick 2017-10-26 20:22:56
And then $1000^{2018} = \left(1000^{1009}\right)^2 \equiv (-9)^2 \equiv 81 \pmod{1009}$.
DPatrick 2017-10-26 20:23:25
So we've learned that:

\begin{align*}

1000^{2018} &\equiv 0 \pmod{2} \\

1000^{2018} &\equiv 81 \pmod{1009}

\end{align*}

What does that tell us mod 2018?
Emathmaster 2017-10-26 20:23:46
and we can solve the resulting system, or eyeball it 1009+81=1090
goodbear 2017-10-26 20:23:46
1090
AOPS12142015 2017-10-26 20:23:46
CRT so $0 \pmod 2$ and $81 \pmod {1009}$ is $\boxed{1090} \pmod {2018}$
stan23456 2017-10-26 20:23:46
1090 mod 2018, and unique
DPatrick 2017-10-26 20:23:54
We need an even number that's 81 more than a multiple of 1009.
DPatrick 2017-10-26 20:23:58
That's clearly $1009 + 81 = \boxed{1090}$, which is our final answer!
DPatrick 2017-10-26 20:24:24
I'm impressed that so many people (about 25%) got this -- it's not an easy problem.
mikebreen 2017-10-26 20:24:35
I agree
DPatrick 2017-10-26 20:24:37
And finally, #10.
DPatrick 2017-10-26 20:24:41
10. What is the smallest value of $n$ such that the number of primes less than or equal to $n$ is $25{,}000$?
DPatrick 2017-10-26 20:24:46
We are told this is the tiebreaker question. To break ties, the person closest to the correct answer will win.
AOPS12142015 2017-10-26 20:24:52
May I ask how many people got the exact answer to #10?
DPatrick 2017-10-26 20:24:55
I'm told zero.
DPatrick 2017-10-26 20:25:12
What's another way of phrasing this question?
Abch 2017-10-26 20:25:28
What is the 25000th prime
Tryharder 2017-10-26 20:25:28
what is the 25000 prme
CornSaltButter 2017-10-26 20:25:30
what is the 25000th prime
DPatrick 2017-10-26 20:25:35
Right. We're asked to determine the 25,000th prime.
DPatrick 2017-10-26 20:25:58
Of course there's no way we'll actually be able to determine it exactly without a computer, so the task is to come up with a reasonable estimate.
DPatrick 2017-10-26 20:26:22
We can start with math trivia that you might know: how many of the first 100 positive integers are prime?
awesomemaths 2017-10-26 20:26:34
25
lifeisgood03 2017-10-26 20:26:34
25
plshalp 2017-10-26 20:26:34
25
EasyAs_Pi 2017-10-26 20:26:34
25
AOPS12142015 2017-10-26 20:26:34
$25$
rainbow10 2017-10-26 20:26:34
25
DPatrick 2017-10-26 20:26:39
Right, there are 25 primes less than 100:

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97

(Including our 8 emirps from problem #3!)
DPatrick 2017-10-26 20:26:51
So you might guess that about 1 out of every 4 numbers is prime.
DPatrick 2017-10-26 20:26:57
That would lead you to guess that the 25,000th prime is about 100,000.
DPatrick 2017-10-26 20:27:05
Do you think that guess is too high or too low?
EulerMacaroni 2017-10-26 20:27:21
too low
int_user 2017-10-26 20:27:21
too low
BCZ01 2017-10-26 20:27:21
too low
RinoaOliver 2017-10-26 20:27:21
no, starts to get less common later too low
Ancy 2017-10-26 20:27:21
way too low, we don't consider new primes that form.
RinoaOliver 2017-10-26 20:27:21
too low
mikebreen 2017-10-26 20:27:31
Smart group.
DPatrick 2017-10-26 20:27:33
It's quite a bit too low, because primes get rarer as numbers get larger.
DPatrick 2017-10-26 20:27:44
One way to think about this is that it's "harder" for a random large number to be prime. It has to avoid being a multiple of every prime number below it!
DPatrick 2017-10-26 20:28:01
So the question is: can we say exactly how much rarer primes get as our numbers get larger and larger?
GeronimoStilton 2017-10-26 20:28:03
It's 168 out of 1,000.
DPatrick 2017-10-26 20:28:24
Indeed, that's down to 16.8% percent (from our 25% of numbers less than 100).
EasyAs_Pi 2017-10-26 20:28:29
1229/10000
DPatrick 2017-10-26 20:28:38
And there are 1,229 primes less than 10,000. Down to 12.29%.
DPatrick 2017-10-26 20:28:55
It turns out that a number around $n^2$ is about half as likely to be prime as a number around $n$.
DPatrick 2017-10-26 20:29:26
We've seen that experimentally: 25% of numbers less than 100 are prime, and 12.29% of numbers less than 10,000 are prime.
DPatrick 2017-10-26 20:29:41
Historically, this relationship was noticed by Gauss in the late 1790s, but he didn't know how to prove it. The proof came much later.
DPatrick 2017-10-26 20:29:57
What kind of function does this look like to you? Specifically, for what sort of function can we square the input and get half the output?
EulerMacaroni 2017-10-26 20:30:20
logarithm
EulerMacaroni 2017-10-26 20:30:20
$\frac{1}{\log{n}}$
GeronimoStilton 2017-10-26 20:30:20
log!
Ancy 2017-10-26 20:30:20
logarithms
DPatrick 2017-10-26 20:30:43
Right! We know that $\log(n^2) = 2\log(n)$. If we square the input, we double the output.
DPatrick 2017-10-26 20:30:47
So the growth rate of primes is given by the reciprocal of a log function, since $\dfrac{1}{\log(n^2)} = \dfrac12 \cdot \dfrac{1}{\log(n)}$ is the behavior that we observe.
EulerMacaroni 2017-10-26 20:31:04
prime number theorem
ppu 2017-10-26 20:31:04
I think the number of primes below n is around n/natural log n
DPatrick 2017-10-26 20:31:20
And in fact, what is true is called the Prime Number Theorem, which states that the fraction of numbers less than $N$ that are prime is approximately $\dfrac{1}{\ln(N)}$, where $\ln$ is the natural logarithm (with base $e$).
DPatrick 2017-10-26 20:31:36
The word "approximately" in the theorem actually has a precise mathematical meaning, but it requires a bit more time and much more advanced mathematics to explain. (You can research it yourself if you're interested.)
DPatrick 2017-10-26 20:31:59
There's a book called "The Music of the Primes" that's particular good on this subject.
DPatrick 2017-10-26 20:32:36
What this means is: a pretty good guess for the $m$th prime is the number $N$ such that $\dfrac{N}{\ln(N)} = m$.
DPatrick 2017-10-26 20:32:50
So can we estimate a value of $N$ for which $\dfrac{N}{\ln(N)}$ is about 25,000? Without a calculator!
DPatrick 2017-10-26 20:33:45
It's a bit of an art form estimating from here. Here's what I did:
DPatrick 2017-10-26 20:34:28
I'll start us out with $e \approx \dfrac83$. (Actually $e = 2.718281828\ldots$ is irrational.)
DPatrick 2017-10-26 20:34:44
Squaring gives $e^2 \approx \dfrac{64}{9}$ which is a wee bit more than 7.
DPatrick 2017-10-26 20:34:50
So $e^4 \approx 50$ is probably pretty close, which gives $\ln(50) \approx 4$.
DPatrick 2017-10-26 20:35:32
Can you go from $\ln(50) \approx 4$ to an approximate solution for $\dfrac{N}{\ln(N)} \approx 25000$?
Danshim917 2017-10-26 20:35:39
DPatrick 2017-10-26 20:35:46
That's useful. Squaring gives $\ln(2500) \approx 8$, and cubing gives $\ln(125{,}000) \approx 12$.
GeronimoStilton 2017-10-26 20:36:17
since we know it's going to be more than 100,000, we can say that it's roughly $25,000$ times $10$, or $250,000$.
DPatrick 2017-10-26 20:36:29
Let's go with $N = 250000$ for a second.
DPatrick 2017-10-26 20:37:03
If we try to estimate $\dfrac{250{,}000}{\ln(250{,}000)}$...
GeronimoStilton 2017-10-26 20:37:09
We also randomly know that $\ln(2) \approx 0.7$.
DPatrick 2017-10-26 20:37:15
That is pretty random.
DPatrick 2017-10-26 20:37:31
I'd probably go with $\ln(2) \approx 0.5$ because 12.5 is easier to divide by.
DPatrick 2017-10-26 20:37:37
This suggests that 250,000 is about the 20,000th prime, since $\dfrac{250{,}000}{\ln(250{,}000)} \approx \dfrac{250{,}000}{12.5} = 20{,}000$.
DPatrick 2017-10-26 20:38:11
And now I might eyeball: if we need 250,000 integers to get the first 20,000 primes, then we probably need another 80,000 primes (or so) to get 25% more primes to get us to 25,000. (I totally made that number up, by the way. We're all just estimating here.)
DPatrick 2017-10-26 20:38:21
So my guess was 330,000.
DPatrick 2017-10-26 20:38:52
Of course, there's no way that can be exactly right, because it's not prime! But getting it exact is hopeless anyway.
RinoaOliver 2017-10-26 20:38:59
awesomemaths 2017-10-26 20:38:59
ppu 2017-10-26 20:38:59
awesomemaths 2017-10-26 20:39:02
drumroll
DPatrick 2017-10-26 20:39:08
Turns out my guess is not too bad, but not great either.
DPatrick 2017-10-26 20:39:13
The actual answer is $\boxed{287{,}117}$.
mikebreen 2017-10-26 20:39:33
250,000 is around what we were expecting for good answers, but we actually got answers to #10 within 40 or 50 of 287,117.
DPatrick 2017-10-26 20:39:41
Part of my error is the inherent fact that the $\dfrac{N}{\ln(N)}$ formula is just an approximation. Indeed, 287,117 is the 25,000th prime, but the formula predicts $\dfrac{287117}{\ln(287117)} = 22845.7279\ldots$. So the formula already is about 10% too low.
DPatrick 2017-10-26 20:40:03
Another approach is to use the fact that the $n$th prime is "approximately" $n \ln(n)$. (Again, the word "approximately" has a much more precise definition that I'm omitting.)
DPatrick 2017-10-26 20:40:14
That means that our answer is approximately $25000 \ln(25000)$. So we need to approximate $\ln(25000)$.Using $\ln(2500) \approx 8$ and $\ln(10) \approx 2$ gives $\ln(25000) \approx 8+2 = 10$.

So that makes our guess 250,000.
PandaPower 2017-10-26 20:40:28
how can you be exact though?
DPatrick 2017-10-26 20:40:35
You can't. Primes are unexact things.
DPatrick 2017-10-26 20:40:53
The way you can actually get the 25,000th prime is to count the first 25,000 primes.
AOPS12142015 2017-10-26 20:41:04
Is it true that if you go over the exact number, you lose tiebreakers?
DPatrick 2017-10-26 20:41:08
No.
DPatrick 2017-10-26 20:41:16
That's it for Round 2!
DPatrick 2017-10-26 20:41:26
The national finals of WWTBAM will be held on Saturday, January 13 at 1 PM Pacific (4 PM Eastern) as part of the 2018 Joint Mathematics Meetings in San Diego. (A 30-minute drive from my house!) If you're in southern California, you're welcome to attend the finals for free! They are part of "Mathematicon" happening all day Saturday:
http://jointmathematicsmeetings.org/meetings/national/jmm2018/2197_mathcon
DPatrick 2017-10-26 20:41:42
If you won't be in San Diego that day, no worries: the finals will be live-streamed on the AMS website.
http://www.ams.org/wwtbam
DPatrick 2017-10-26 20:41:54
There will be 12 finalists chosen to participate in the national finals. 10 of the 12 will be the highest scorer (broken by the tie-breaker question #10 if necessary) on Round 2 in each of the 10 regions shown on the map below:
DPatrick 2017-10-26 20:41:57
DPatrick 2017-10-26 20:42:08
The 11th finalist will be the highest scorer in the United Kingdom, and the 12th finalist will be the highest scorer in the San Diego metro area.
DPatrick 2017-10-26 20:42:28
I'm told it will take another week or so to finish and double-check the scoring, and to notify the finalists.
Ancy 2017-10-26 20:42:35
san diego getting its own region ;-;
DPatrick 2017-10-26 20:42:44
The home city of the finals always gets a local contestant.
mikebreen 2017-10-26 20:42:49
Great to see everyone working out the answers. Thanks, Dave, and to the AoPS Jammers.
DPatrick 2017-10-26 20:42:51
The finalists will compete live on-stage with Mike as the host. I'll be in the front row of the audience cheering them on!
DPatrick 2017-10-26 20:43:21
Again, if you want to learn more about the contest and how to participate next year, visit http://ams.org/wwtbam.
TPiR 2017-10-26 20:43:32
Thanks, everyone. Great job!
DPatrick 2017-10-26 20:43:44
Thanks for coming tonight to our Math Jam. Good night!