Math Jams

## Who Wants to Be a Mathematician, Qualifying Round

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AoPS instructor David Patrick will discuss the problems on Round 1 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 2016-10-05 19:31:03
Welcome to the 2016-17 Who Wants to Be a Mathematician Round 1 Math Jam!
DPatrick 2016-10-05 19:31:15
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.
DPatrick 2016-10-05 19:31:28
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 2016-10-05 19:32:05
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 2016-10-05 19:32:18
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 2016-10-05 19:32:33
This helps keep the session organized and on track. This also means that only well-written comments will be shared into the classroom, so please take time writing responses that are complete and easy to read.
DPatrick 2016-10-05 19:32:57
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. 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 2016-10-05 19:33:22
Joining us tonight is the co-creator of WWTBAM, Mike Breen (mikebreen).
mikebreen 2016-10-05 19:33:38
DPatrick 2016-10-05 19:33:42
Mike taught at Alfred University and Tennessee Technological University before becoming AMS Public Awareness Officer in 2000. He and Bill Butterworth 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.
caro2be 2016-10-05 19:33:57
Hi Mike!
yiqun 2016-10-05 19:33:57
HI MIKE!!!
piphi 2016-10-05 19:33:57
Hi mikebreen
DPatrick 2016-10-05 19:34:11
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.
mikebreen 2016-10-05 19:34:19
Hello, everyone. Love all the interest and enthusiasm.
DPatrick 2016-10-05 19:34:28
Round 1 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 2016-10-05 19:34:51
We'll take a bit longer than 15 minutes tonight, probably more like 45-60 minutes, 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.
Python54 2016-10-05 19:35:10
What is the difficulty of the problems?
DPatrick 2016-10-05 19:35:46
You'll probably get a good idea as we go. Some are more-or-less basic algebra, but some are quite tricky. (And I've seen the Round 2 problems, and they're even trickier still.)
DPatrick 2016-10-05 19:35:56
Several of the questions have interesting sidetracks, so we'll also stop and view some of the scenery along the way.
mikebreen 2016-10-05 19:36:10
Easy to difficult. Round two much harder.
DPatrick 2016-10-05 19:36:38
Please hold your questions about the contest administration until the end, when Mike and I will try to answer them. Right now, let's get on with the math!
DPatrick 2016-10-05 19:36:49
1. What is the only positive solution to $3x^2 + 17x = 28$?
DPatrick 2016-10-05 19:37:17
Like I said, the early questions on the contest, like this one, are often more-or-less just basic algebra.
hwl0304 2016-10-05 19:37:25
Christina26 2016-10-05 19:37:25
AOPS12142015 2016-10-05 19:37:25
QF
yiqun 2016-10-05 19:37:25
DPatrick 2016-10-05 19:37:41
Certainly we can write it in a more standard form as $3x^2 + 17x - 28 = 0$ and use the quadratic formula.
pkandarpa 2016-10-05 19:37:50
Factor the question.
supermathkid 2016-10-05 19:37:50
(3x-4)*(x+7)
DPatrick 2016-10-05 19:37:54
Or, if we guess that the solutions will be rational numbers, we can try to factor it.
DPatrick 2016-10-05 19:37:59
Indeed, it factors as $(3x - 4)(x + 7) = 0$.
DPatrick 2016-10-05 19:38:08
So what are the solutions, and in particular what is the positive solution?
Patrickzcy 2016-10-05 19:38:30
4/3
Fridaychimp 2016-10-05 19:38:30
4/3
Liopleurodon 2016-10-05 19:38:30
4/3
EpicRuler101 2016-10-05 19:38:30
4/3
Intellectuality123 2016-10-05 19:38:30
4/3
rafa2be 2016-10-05 19:38:30
Positive one is $4/3$.
Shaktiala 2016-10-05 19:38:30
pramodmitikiri 2016-10-05 19:38:30
the only positive number would be 4/3, although both answers are 4/3 and -7
Python54 2016-10-05 19:38:30
$-7,4/3 \implies 4/3$ is answer
DPatrick 2016-10-05 19:38:40
Right. The solutions are the solutions to $3x - 4 = 0$ and $x + 7 = 0$.
DPatrick 2016-10-05 19:38:52
The second factor gives us $x = -7$, which is negative.
DPatrick 2016-10-05 19:39:02
But the first factor gives us $x = \boxed{\dfrac43}$, which is positive, so this is our final answer.
DPatrick 2016-10-05 19:39:21
By the way, how can we tell quickly, without much computation, that there must be a positive root?
Patrickzcy 2016-10-05 19:40:13
(b^2)-4ac
themoocow 2016-10-05 19:40:13
discrinimant
cynjinli 2016-10-05 19:40:13
b is positive but c is negative
Mega-bot 2016-10-05 19:40:19
-a/b in the equation is negative, meaning there must be a negative root and a positive one
The product of the roots is negative as the constant is negative, so one must be positive
DPatrick 2016-10-05 19:40:30
Right. Because the constant term is negative and the other two terms are positive, we can quickly see that the discriminant $17^2 - 4(3)(-28)$ is positive.
DPatrick 2016-10-05 19:40:41
That tells us that there must be two distinct real roots.
DPatrick 2016-10-05 19:40:54
And, one of Vieta's Formulas tells us that the product of the roots is $-\dfrac{28}{3}$, which is negative. So one root must be positive and one root must be negative.
DPatrick 2016-10-05 19:41:06
In general, Vieta's Formulas tell us that, given the quadratic $ax^2 + bx + c = 0$, the sum of the roots is $-\dfrac{b}{a}$ and the product of the roots is $\dfrac{c}{a}$. (See if you can prove this yourself!) Vieta's Formulas also generalize to higher-degree polynomials, but that's a topic for another day.
DPatrick 2016-10-05 19:41:35
2. What is the ones digit of $2017^{2015}$?
Christina26 2016-10-05 19:42:13
find pattern for 7's
letsgomath 2016-10-05 19:42:13
ones digit of 7^2015
popcorn1 2016-10-05 19:42:13
Ones digit of $7^2015$
guluguluga 2016-10-05 19:42:13
We can look at just the 7, because the other parts doesn't affect the ones digi
yiqun 2016-10-05 19:42:13
we only need the units digit. So, 2017 turns into 7!
DPatrick 2016-10-05 19:42:20
Great. The units digits of any product only depends on the units digits of the numbers that we're multiplying.
DPatrick 2016-10-05 19:42:29
So the units digit of $2017^{2015}$ is the same as the units digit of $7^{2015}$.
DPatrick 2016-10-05 19:42:35
That helps, but still I don't think we can compute $7^{2015}$.
try the first few powers and find a pattern
csasomachar 2016-10-05 19:42:56
patterns!
NewtonFTW 2016-10-05 19:43:13
There are 4 cycles of the units digit when a number that ends in 7 is raised to a power: 7, 9, 3, and 1.
BobaFett101 2016-10-05 19:43:13
the units digit cycles with period 4: 7, 9, 3, 1
winnertakeover 2016-10-05 19:43:13
The units digit follows this pattern 7,9,3,1
letsgomath 2016-10-05 19:43:13
pattern for powers of 7: 7,9,3,1,7,9,3,1,.....
geniusofart 2016-10-05 19:43:13
Note the cycle of 7, 9, 3, 1 of the units digits.
caro2be 2016-10-05 19:43:13
the possible units digits are 7, 9, 3, and 1
DPatrick 2016-10-05 19:43:20
Right. We can find the units digits of small powers of 7, and see if we can find a pattern.
DPatrick 2016-10-05 19:43:32
\begin{align*}

7^1 &= 7, \\

7^2 &= 49, \\

7^3 &= 243, \\

\vdots

\end{align*}
DPatrick 2016-10-05 19:43:38
But remember, we only care about the units digits!
NewtonFTW 2016-10-05 19:43:53
7, 9, 3, 1 and repeat
popcorn1 2016-10-05 19:43:53
7, 9, 3, 1
DPatrick 2016-10-05 19:43:58
$\begin{array}{c|c} \text{Power of 7} & \text{Units digit} \\ \hline 7^1 & 7 \\ 7^2 & 9 \\ 7^3 & 3 \\ 7^4 & 1 \end{array}$
DPatrick 2016-10-05 19:44:10
And then we get a repeat: the units digit of $7^5$ is the same as the units digit of $7 \cdot 7^4$, which is $7 \cdot 1 = 7$.
DPatrick 2016-10-05 19:44:21
$\begin{array}{c|c} \text{Power of 7} & \text{Units digit} \\ \hline 7^1 & 7 \\ 7^2 & 9 \\ 7^3 & 3 \\ 7^4 & 1 \\ 7^5 & 7 \end{array}$
DPatrick 2016-10-05 19:44:32
And so on. The pattern of units digits of powers of 7 will repeat in a cycle of 4:
DPatrick 2016-10-05 19:44:35
$\begin{array}{c|c} \text{Power of 7} & \text{Units digit} \\ \hline 7^1,7^5,7^9,7^{13},\ldots & 7 \\ 7^2,7^6,7^{10},7^{14},\ldots & 9 \\ 7^3,7^7,7^{11},7^{15},\ldots & 3 \\ 7^4,7^8,7^{12},7^{16},\ldots & 1 \end{array}$
DPatrick 2016-10-05 19:44:42
We need to figure out which row of the chart contains $7^{2015}$. How do we do that?
supermessi 2016-10-05 19:45:08
So we do 2015 (mod 4)
divijleisha 2016-10-05 19:45:08
so we do 2015mod4
supermathkid 2016-10-05 19:45:08
since that works then it must repeat every 4 so 2015/4 has remainder of 3 so 7^2015 must end in the same thing as 7^3 which is 3 so the answer: 3
elmaso 2016-10-05 19:45:08
i mean, 2015 is 3 mod 4, and the pattern goes: 7,9,3,1
NewtonFTW 2016-10-05 19:45:08
We can divide 2015 by 4 to find out how many cycles will be completed, and that simplifies to 503 and 3/4
Liopleurodon 2016-10-05 19:45:08
we find what 2015 mod 4 is
csasomachar 2016-10-05 19:45:08
find the reaminder when 2105 is divided by 4
seanwang2001 2016-10-05 19:45:08
2015 mod 4
DPatrick 2016-10-05 19:45:29
Right. Or, to state it more simply, you perhaps notice that the bottom row of the chart is all the powers of 7 where the exponent is a multiple of 4.
DPatrick 2016-10-05 19:45:39
Similarly:

The top row is all exponents that are 1 more than a multiple of 7.

The 2nd row is all exponents that are 2 more than a multiple of 7.

The 3rd row is all exponents that are 3 more than a multiple of 7.
divijleisha 2016-10-05 19:45:53
2012 is multiple
DPatrick 2016-10-05 19:45:55
2012 is a multiple of 4, so 2015 is 3 more than a multiple of 4.
DPatrick 2016-10-05 19:46:04
So $7^{2015}$ lies in the third row, and thus its units digit is $\boxed{3}$.
DPatrick 2016-10-05 19:46:27
If you've studying some more number theory, you might know of Euler's Theorem, which states that, if $a$ and $n$ are relatively prime, then $a^{\phi(n)} \equiv 1 \pmod{n},$ where $\phi(n)$ is the number of positive integers less than $n$ that are relatively prime to $n$.
DPatrick 2016-10-05 19:46:47
In this problem, we can use $n=10$, noting $\phi(10) = 4$ (the numbers less than 10 that are relatively prime to 10 are 1, 3, 7, and 9), and $a=2017$ to get

$2017^4 \equiv 1 \pmod{10}.$
DPatrick 2016-10-05 19:47:13
From there, we get $2017^{2012} = (2017^4)^{503} \equiv 1^{503} \equiv 1 \pmod{10},$ so the units digit of $2017^{2015}$ is the same as the units digit of $2017^3$, which is the same as the units digit of $7^3 = 243$, so the answer is 3.
DPatrick 2016-10-05 19:47:35
It's basically the same work we just did, but with fancier language and notation.
DPatrick 2016-10-05 19:47:48
Onwards:
DPatrick 2016-10-05 19:47:52
3. $(15+i)(15-i) =$
csasomachar 2016-10-05 19:48:13
difference of squares!
arowaaron 2016-10-05 19:48:13
difference of squares
pramodmitikiri 2016-10-05 19:48:13
difference of sqaure
cinamon 2016-10-05 19:48:13
it's the difference of squares
DPatrick 2016-10-05 19:48:23
Yes, it's very helpful if you know the difference-of-squares formula $(a+b)(a-b) = a^2 - b^2$. This formula comes up over and over and over again in math contests.
DPatrick 2016-10-05 19:48:30
So, $(15+i)(15-i) = 15^2 - i^2$.
DPatrick 2016-10-05 19:48:47
What the heck is that $i$ thingy?
pkandarpa 2016-10-05 19:49:08
imaginary number
EpicRuler101 2016-10-05 19:49:08
i = sqrt(-1) so i^2 = -1
Intellectuality123 2016-10-05 19:49:08
$\sqrt{-1}$
Liopleurodon 2016-10-05 19:49:08
$\sqrt{-1}$
fruitarian 2016-10-05 19:49:08
the square root of negative 1
mathmaniacjulia 2016-10-05 19:49:08
square root of -1
Apurple 2016-10-05 19:49:08
square root of -1
EpicRuler101 2016-10-05 19:49:08
i is the root of -1
pkandarpa 2016-10-05 19:49:08
i=sqrt-1
rafa2be 2016-10-05 19:49:08
$\sqrt{-1}$
DPatrick 2016-10-05 19:49:14
$i$ is a special number, called an imaginary number, with the property that $i^2 = -1$. (As you probably know, no real number has a negative square, so we have to invent a new number to fill this role.)
Patrickzcy 2016-10-05 19:49:44
(15+i)(15-i)=(15^2)-(i^2)=225-(-1)=226
mathgandalfjr 2016-10-05 19:49:44
so the answer to this problem is 225 -(-1) which is 226
DPatrick 2016-10-05 19:49:47
So now we can finish: $15^2 = 225$ and $i^2 = -1$, so this is just $225 - (-1) = \boxed{226}$.
DPatrick 2016-10-05 19:50:12
That was pretty straightforward, as long as you're familiar with what $i$ means.
DPatrick 2016-10-05 19:50:23
4. A cone of radius $r$ and height $h$ has a volume equal to that of a right circular cylinder having the same height. What is the radius of the right circular cylinder?
letsgomath 2016-10-05 19:50:52
make an equation
DPatrick 2016-10-05 19:51:19
Right. It may seem obvious, but when the word "equal" is in a word problem, that's a pretty good clue that we'll have two quantities that we can equate using an equation.
EpicRuler101 2016-10-05 19:51:25
volume of cone = volume of cylinder
elmaso 2016-10-05 19:51:25
equal both equations, since it's the same volume
DPatrick 2016-10-05 19:51:43
Let's call the radius of the cylinder $x$, since that's what we're trying to find.
DPatrick 2016-10-05 19:51:49
What's the volume of the cylinder?
acegikmoqsuwy2000 2016-10-05 19:52:25
$\pi x^2h$
cynjinli 2016-10-05 19:52:25
x^2*h*pi
Mathisfun04 2016-10-05 19:52:25
x squared * pi * h
letsgomath 2016-10-05 19:52:25
x^2*h*pi
Liopleurodon 2016-10-05 19:52:25
$x^2h\pi$
cinamon 2016-10-05 19:52:25
pi x^2 * h
Christina26 2016-10-05 19:52:25
so pix^2h
DPatrick 2016-10-05 19:52:44
It's $\pi x^2 h$. (The volume of a cylinder is the area of the base times the height.)
DPatrick 2016-10-05 19:52:49
And what's the volume of the cone?
acegikmoqsuwy2000 2016-10-05 19:53:27
$\dfrac 13\pi r^2h$
AnaT129 2016-10-05 19:53:27
1/3hr^2
1/3 * r^2 * pi * h
NewtonFTW 2016-10-05 19:53:27
1/3pir^2h
DPatrick 2016-10-05 19:53:42
It's $\frac13 \pi r^2 h$. (The volume of a cone is 1/3 times the area of the base times the height.)
DPatrick 2016-10-05 19:53:49
The volumes are equal, so we must have $\pi x^2 h = \frac13 \pi r^2 h$.
divide by pi and h
NewtonFTW 2016-10-05 19:54:04
We can cancel out pi and h
Mathisfun04 2016-10-05 19:54:07
you would cancel out from here
DPatrick 2016-10-05 19:54:11
We can divide by $\pi h$ and we get $x^2 = \frac13 r^2$.
winnertakeover 2016-10-05 19:54:21
Take the root
DPatrick 2016-10-05 19:54:27
So, taking the square root, we see that $x = \boxed{\frac{1}{\sqrt3}r}$.
acegikmoqsuwy2000 2016-10-05 19:54:44
so $x=\dfrac{r\sqrt 3} {3}$
acegikmoqsuwy2000 2016-10-05 19:54:44
if we rationalized, does the answer count as correct?
DPatrick 2016-10-05 19:54:48
Absolutely.
NewtonFTW 2016-10-05 19:55:10
In the contest, will we have to rationalize the denominator
EpicRuler101 2016-10-05 19:55:10
we dont need to rationalize the denominator?
Christina26 2016-10-05 19:55:10
if you dont rationalize, is the answer wrong?
DPatrick 2016-10-05 19:55:25
It should also be correct. (Though I'm not 100% sure of the rules for answer format on the contest.)
mikebreen 2016-10-05 19:55:31
DPatrick 2016-10-05 19:55:42
There you go -- leaving it the way I did is correct too.
DPatrick 2016-10-05 19:56:02
Let's continue:
DPatrick 2016-10-05 19:56:06
5. A palindromic number is one whose digits read the same backward and forward, for example 484 or 909. Which of the following prime numbers is a factor of every four-digit palindromic number?
a. 3 b. 7 c. 11 d. 13 e. There is no such prime number
popcorn1 2016-10-05 19:56:40
Let the number be $abba$
we have to write out the base 10 form of a palindromic number
DPatrick 2016-10-05 19:57:02
Right. Our number has to look like ABBA for some digits A and B. (I like using capital letters for digits, to remind me that they're not just random variables.)
DPatrick 2016-10-05 19:57:13
Plus maybe I like 70s-era Swedish pop bands.
elmaso 2016-10-05 19:57:38
divisibility properties
popcorn1 2016-10-05 19:57:38
Divisibility rule for 3, 7, 11, 13
Mathisfun04 2016-10-05 19:57:38
from here, we could generalize and use the divisibility tricks
DPatrick 2016-10-05 19:58:00
Right. There are easy tests for divisibility by 3 or 11.
themoocow 2016-10-05 19:58:29
Obviously a-b+b-a=0 divisible by 11
csasomachar 2016-10-05 19:58:29
it should be 11, by divisibilityy rules
It should be divisible by 11 because the rule is A-B+B-A = 0, which is divisible by 11
Liopleurodon 2016-10-05 19:58:29
11 works!
DPatrick 2016-10-05 19:58:42
Yes! To test for divisibility by 11, we take the "alternating sum" of the digits: the first digit - the second digit + the third digit - the fourth digit. (And if it were a longer number, we'd keep going, alternately adding and subtracting digits.)
DPatrick 2016-10-05 19:58:49
If the alternating sum is 11, then the number is a multiple of 11.
DPatrick 2016-10-05 19:59:00
So for ABBA, the alternating sum is A - B + B - A = 0.
DPatrick 2016-10-05 19:59:11
This is a multiple of 11!
DPatrick 2016-10-05 19:59:30
So every number of the form ABBA is a multiple of 11. The answer is $\boxed{\text{c. 11}}$.
elmaso 2016-10-05 19:59:45
for ABBA to be divisible by 3: 2(A+B) must be divisible by 3
Z_Math404 2016-10-05 19:59:45
2(a+b) isn't necessarily divisible by 3
Yanlord 2016-10-05 19:59:45
3 doesn't work e.g. 4774
DPatrick 2016-10-05 19:59:56
Right, it's pretty clear that 3 doesn't work.
DPatrick 2016-10-05 20:00:07
The sum of the digits is 2A + 2B. There's no reason this needs to be a multiple of 3. (It's often not. For example, 2002 is not a multiple of 3.)
DPatrick 2016-10-05 20:00:54
There's no easy test for when a number is divisible by 7 or 13, but there are lots of examples of ABBA's that aren't divisible by 7 or 13. (2112 is such an example I believe.)
winnertakeover 2016-10-05 20:01:10
$1000a+100b+10b+a=1001a+110b=11(91a+10b)$
DPatrick 2016-10-05 20:01:22
That's a nice algebraic proof that it works!
DPatrick 2016-10-05 20:01:58
And this gives insight into why the divisibility test for 11 works: it's because the powers of 10 alternate between being 1 more and 1 less than a multiple of 11:
$10^0 = 1$ is one more than a multiple of 11 (namely, 0).
$10^1 = 10$ is one less than a multiple of 11 (namely, 11).
$10^2 = 100$ is one more than a multiple of 11 (namely, 99).
$10^3 = 1000$ is one less than a multiple of 11 (namely, 1001).
And so on.
DPatrick 2016-10-05 20:02:33
OK, on to what was the second-hardest problem on the contest:
DPatrick 2016-10-05 20:02:37
6. How many solutions are there to the equation $\cos 2x - \sin x = 1$, for $0 \le x < 2\pi$ ($x$ in radians)?
DeathLlama9 2016-10-05 20:03:13
Double angle!
Christina26 2016-10-05 20:03:13
cos and sin identities
hliu70 2016-10-05 20:03:13
Catoc17 2016-10-05 20:03:13
cos(2x) = 1-2*sin^2(x)
DPatrick 2016-10-05 20:03:33
Sounds promising. A goal is to relate the cosine and sine terms somehow.
DPatrick 2016-10-05 20:03:52
We have the identity $\cos 2x = \cos^2 x - \sin^2 x$. (This is the only way that I remember it.)
DPatrick 2016-10-05 20:04:08
But we also know that $\cos^2 x = 1 - \sin^2 x$. So $\cos 2x = 1 - 2 \sin^2 x$. How does that help?
ptes77 2016-10-05 20:04:35
We can solve a quadratic in sinx
winnertakeover 2016-10-05 20:04:35
we can solve for $\sin(x)$!
its easier to solve a trigonometric equation with the same functions
cynjinli 2016-10-05 20:04:35
set sin x as a new variable and solve quadratic
BobaFett101 2016-10-05 20:04:35
rewrite as a quadratic in sin x
pkandarpa 2016-10-05 20:04:42
get rid of the cosines and only leave the sines
divijleisha 2016-10-05 20:04:42
we could substitute for cos2x
acegikmoqsuwy2000 2016-10-05 20:04:42
substitute in, we get a quadratic in $\sin x$
DPatrick 2016-10-05 20:04:53
Good idea! It might be useful to make a new variable $u = \sin x$.
DPatrick 2016-10-05 20:05:04
Then, we just found out that $\cos 2x = 1 - 2u^2$, so our original equation becomes

$1 - 2u^2 - u = 1.$
acegikmoqsuwy2000 2016-10-05 20:05:27
rewrite as $u(2u+1)=0$
NewtonFTW 2016-10-05 20:05:30
-2u^2-u = 0
supermathkid 2016-10-05 20:05:30
2u^2 + u= 0
DPatrick 2016-10-05 20:05:34
This simplifies to $2u^2 + u = 0$, or $u(2u+1) = 0$.
therefore u = 0 or 2u + 1 = 0
Liopleurodon 2016-10-05 20:05:52
u=0,u=-1/2
DPatrick 2016-10-05 20:05:55
Right. So either $u=0$ or $u = -\frac12$.
DPatrick 2016-10-05 20:06:08
And so back to the original question: how many solutions does this give us with $0 \le x < 2\pi$?
yamyamx2 2016-10-05 20:06:17
u=sinx
DPatrick 2016-10-05 20:06:22
Right. We need $\sin x = 0$ or $\sin x = -\frac12$.
ImpossibleCube 2016-10-05 20:06:41
Each gives $2$ solutions, so $\boxed{4}$
acegikmoqsuwy2000 2016-10-05 20:06:41
4 solutions, but just to be safe we should check they all work
DPatrick 2016-10-05 20:06:55
That's right. There are two $x$ for for each: $\sin x = 0$ for $x=0$ or $x=\pi$, and $\sin x = -\frac12$ for $x = \frac{7\pi}{6}$ and $x = \frac{11\pi}{6}$.
DPatrick 2016-10-05 20:07:17
The actual values of $x$ don't matter, of course -- all that matters is that there are 2 values of $u$ that work, and then 2 values of $x$ for each value of $u$. But I also agree that it's a good check of our work.
DPatrick 2016-10-05 20:07:25
So in total there are $\boxed{4}$ solutions.
DPatrick 2016-10-05 20:07:36
Did anybody solve this problem in a different way? (I did!)
you can also draw a picture
DPatrick 2016-10-05 20:08:01
Yes!
DPatrick 2016-10-05 20:08:06
I wrote the equation as $\cos 2x = \sin x + 1$, and sketched the graphs:
DPatrick 2016-10-05 20:08:10
DPatrick 2016-10-05 20:08:30
Even though I drew this graph with software for tonight's Math Jam, you don't need it to be so accurate to see how many times they intersect along $0 \le x < 2\pi$.
DPatrick 2016-10-05 20:08:38
Note that the graph of $\cos 2x$ (in red) is just the graph of $\cos x$, but compressed horizontally by a factor of 2.
DPatrick 2016-10-05 20:08:50
And the graph of $\sin x + 1$ is the graph of $\sin x$ shifted up 1 unit.
DPatrick 2016-10-05 20:09:07
We can clearly see 4 intersection points: one at $x=0$, one at $x=\pi$, and then two with $\pi < x < 2\pi$.
DPatrick 2016-10-05 20:09:29
7. Which of the following is closest to $1 + \dfrac{1}{1+\frac{1}{1+\frac{1}{1+1}}}$?
a. $\dfrac{1+\sqrt3}{2}$ b. $\sqrt2$ c. $\dfrac{1+\sqrt5}{2}$ d. $\sqrt{\pi}$ e. $\dfrac{2\pi}{3}$
caro2be 2016-10-05 20:10:03
Start from the bottom and work up!
DPatrick 2016-10-05 20:10:17
I suppose we could try to simplify it.
supermathkid 2016-10-05 20:10:33
8/5 = the big fraction
EpicRuler101 2016-10-05 20:10:33
8/5
BobaFett101 2016-10-05 20:10:33
8/5 is the final result
NewtonFTW 2016-10-05 20:10:33
8/5
DPatrick 2016-10-05 20:10:35
To start, $\dfrac{1}{1+1}$ is just $\dfrac12$:

$1 + \frac{1}{1+\frac{1}{1+\frac12}}.$
DPatrick 2016-10-05 20:10:46
Then $1 + \frac12 = \frac32$, so $\frac{1}{1+\frac12} = \frac23$:

$1 + \frac{1}{1+\frac23}.$
DPatrick 2016-10-05 20:10:59
Finally, $1+ \frac23 = \frac53$, so $\frac{1}{1+\frac23} = \frac35$, and the whole thing is equal to $\frac85$.
DPatrick 2016-10-05 20:11:08
Which of those answer choices is closest to $\frac85$ (or $1.6$ if you prefer -- decimals are often easier to work with when approximating)?
hliu70 2016-10-05 20:11:58
C=1.61
ImpossibleCube 2016-10-05 20:11:58
c, it's the golden ratio, which is about 1.618
mathguy5041 2016-10-05 20:11:58
The choices are 1.36, 1.41, 1.62, 1.78, >2 so obviously c
DPatrick 2016-10-05 20:12:15
Right, we can eyeball these quantities, even without a calculator (which isn't permitted).
DPatrick 2016-10-05 20:12:27
$\sqrt3$ is less than 2, so (a) is less than the average of 1 and 2, which is 1.5.
DPatrick 2016-10-05 20:12:40
Similarly (b) is a bit less than 1.5.
DPatrick 2016-10-05 20:12:53
$\sqrt5$ is a little larger than 2, so (c) is a little larger than the average of 1 and 2, which is 1.5. That's promising.
DPatrick 2016-10-05 20:13:00
Indeed, to be a little more precise, $\sqrt5$ is between 2.2 and 2.3, so (c) is between 1.6 and 1.65. That's certainly closer than (a) and (b), which are each less than 1.5.
DPatrick 2016-10-05 20:13:28
(And many of you know more exactly what (c) is, because it's a famous number. I'll say more about that in just a minute.)
DPatrick 2016-10-05 20:13:41
In (d), $\sqrt{\pi} > \sqrt3 > 1.7$, so that's further than (c).
NewtonFTW 2016-10-05 20:13:49
But 2pi/3 is certainly too large
DPatrick 2016-10-05 20:14:02
Right, (e) is easiest: it's more than 2 (since $\pi > 3$), so it's a lot farther than (c).
DPatrick 2016-10-05 20:14:12
So the answer is $\boxed{\text{c}}$.
DPatrick 2016-10-05 20:14:38
That was the ugly way to plow through the answer choices. But as many of you noted, there's a more "big-picture" way to think about this problem.
yamyamx2 2016-10-05 20:14:57
I just assumed infinitly going on
winnertakeover 2016-10-05 20:14:57
You could instead evaluate it as an infinite series, and use that...
hinna 2016-10-05 20:14:57
What happens when we continue that fraction infinitely?
DPatrick 2016-10-05 20:15:11
Suppose that, instead of being a finite fraction, the fraction continued on with this pattern repeated infinitely often. Let's name this new number $z$:

$z = 1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\ddots}}}}.$
DPatrick 2016-10-05 20:15:25
This should be very close to the given number, because the stuff we've added is buried under several layers of denominator. (Each "layer" makes the overall denominator a little bit bigger, but the deeper we go, the less and less effect it has on the overall quantity.)
bobjoe123 2016-10-05 20:15:50
$z = 1 + \frac{1}{z}$
csasomachar 2016-10-05 20:15:50
z=1+(1/z)
ImpossibleCube 2016-10-05 20:15:50
$z = 1 + \frac{1}{z}$
AnaT129 2016-10-05 20:15:50
since it is infinite, you can substitute z in, getting 1+(1/z)
winnertakeover 2016-10-05 20:15:50
$z=1+\frac{1}{z}$
mathgandalfjr 2016-10-05 20:15:50
z=1+(1/z)
DPatrick 2016-10-05 20:15:57
Aha! Notice that $z$ is "self-similar" -- everything other than the first two 1's (the initial "1+" and the 1 in the numerator) is just another copy of $z$!
DPatrick 2016-10-05 20:16:09
That is, $z = 1+\dfrac{1}{z}$.
DPatrick 2016-10-05 20:16:16
This is an equation that we can solve!
popcorn1 2016-10-05 20:16:44
z^2 = z+1
csasomachar 2016-10-05 20:16:44
z= (1+sqrt5)/2
supermathkid 2016-10-05 20:16:44
z^2 - z -1 =0
Shaktiala 2016-10-05 20:16:44
Z^2 = z + 1
DPatrick 2016-10-05 20:16:50
Multiplying by $z$ and simplifying gives us $z^2 - z - 1 = 0$.
DPatrick 2016-10-05 20:16:59
This is a quadratic, so there are two solutions: $z = \dfrac{1 \pm \sqrt{5}}{2}$.
DPatrick 2016-10-05 20:17:09
But we know $z$ is positive, so we exclude the negative solution and we conclude $z = \dfrac{1+\sqrt5}{2}$.
Liopleurodon 2016-10-05 20:17:15
That's the golden ratio! C.
DPatrick 2016-10-05 20:17:41
That's answer choice (c)! While this certainly doesn't prove that (c) is the closest, because our $z$ is not exactly the quantity we started with, it's pretty strong evidence.
DPatrick 2016-10-05 20:17:57
As you may know, this number $\dfrac{1+\sqrt5}{2}$ is a pretty famous number in mathematics.

It's called the golden ratio and is often denoted by the Greek letter $\varphi$. Indeed, this ratio was known to the ancient Greeks. A rectangle whose sides are in the golden ratio was widely considered to be aesthetically pleasing:
DPatrick 2016-10-05 20:18:07
DPatrick 2016-10-05 20:18:23
The geometric property here is that if we cut off a square from one side of this rectangle, what remains is a smaller rectangle whose sides also satisfy the golden ratio:
DPatrick 2016-10-05 20:18:29
DPatrick 2016-10-05 20:18:48
And this pattern can be repeated indefinitely, to form a pleasing spiral:
DPatrick 2016-10-05 20:18:52
DPatrick 2016-10-05 20:19:11
We have a bunch of white squares, and the small red rectangle is still in the golden ratio, so we could continue the process forever.
EpicRuler101 2016-10-05 20:19:19
Fibonnaci Sequence!
BCharwhis50 2016-10-05 20:19:19
this relates to the fibonacci sequence
DPatrick 2016-10-05 20:19:35
Indeed, the golden ratio is very closely related to the Fibonacci numbers!
DPatrick 2016-10-05 20:20:10
That is a really interesting topic to explore deeply -- I encourage you to investigate further on your own!
DPatrick 2016-10-05 20:20:16
Let's move on:
DPatrick 2016-10-05 20:20:20
8. A right triangle has legs $a$ and $b$, hypotenuse $c$ and perimeter $2d$. Find $\sqrt{d(d-a)(d-b)(d-c)}$.
supermathkid 2016-10-05 20:20:49
heron's formula
ballislife21 2016-10-05 20:20:49
This looks like herons formula
this is herons formula
hliu70 2016-10-05 20:20:49
heron's formula!
mathgandalfjr 2016-10-05 20:20:49
isn't that the Herons formula or something
Catoc17 2016-10-05 20:20:49
Heron's Formula
heron
ImpossibleCube 2016-10-05 20:20:49
It's heron's formula
letsgomath 2016-10-05 20:20:49
heron's formula!
NewtonFTW 2016-10-05 20:20:49
That is Heron's Formula
math129 2016-10-05 20:20:49
Looks like heron's
hwl0304 2016-10-05 20:20:49
heron's
sas4 2016-10-05 20:20:49
Heron's Formula!
DPatrick 2016-10-05 20:20:58
Yes. This is actually a famous formula involving triangles, called Heron's Formula.
DPatrick 2016-10-05 20:21:19
Heron's Formula states that this quantity equals the area of the triangle. (You often see Heron's Formula with an $s$, for semiperimeter.)
divijleisha 2016-10-05 20:21:33
area of a triangle
NewtonFTW 2016-10-05 20:21:33
Therefore, the formula equals to the area of the triangle
DPatrick 2016-10-05 20:21:39
Right: our answer is just the area of the triangle.
NewtonFTW 2016-10-05 20:21:58
ab/2
bobjoe123 2016-10-05 20:21:58
ab/2
winnertakeover 2016-10-05 20:21:58
it's ab/2
cinamon 2016-10-05 20:21:58
so a*b/2
Intellectuality123 2016-10-05 20:21:58
it's essentially asking for the area of the triangle which is also equal to ab/2
acegikmoqsuwy2000 2016-10-05 20:21:58
$\dfrac{ab}{2}$
Catoc17 2016-10-05 20:21:58
which is 1/2 * a * b
hinna 2016-10-05 20:21:58
Or $ab/2$
supermathkid 2016-10-05 20:21:58
a * b /2
Shaktiala 2016-10-05 20:21:58
The area is (a*b)/2
DPatrick 2016-10-05 20:22:02
So, since this is a right triangle with legs $a$ and $b$, the area (and thus our answer) is $\boxed{\dfrac12ab}$.
DPatrick 2016-10-05 20:22:27
Heron's Formula is not that difficult to prove: it's mostly just using the Pythagorean Theorem and then plowing through a bunch of algebra. (I say "not that difficult" in the present day -- it's remarkable that the ancient Greeks knew how to prove this over 2000 years ago.)
DPatrick 2016-10-05 20:22:44
Next up is what turned out to be the hardest problem on the contest:
DPatrick 2016-10-05 20:22:50
9. A perfect number is a number greater than 1 that is equal to the sum of its proper factors/divisors (including 1, but not including itself). Example: $6 = 1 + 2 + 3$. How many perfect numbers are less than $10{,}000$?
DPatrick 2016-10-05 20:23:09
6 is the smallest perfect number: as in the given example, its proper divisors are 1, 2, 3, and $1+2+3=6$.
ballislife21 2016-10-05 20:23:14
We could find a pattern
DPatrick 2016-10-05 20:23:28
Indeed. Does anyone know the next perfect number?
popcorn1 2016-10-05 20:23:42
The next is 28
j2005 2016-10-05 20:23:42
As is 28
letsgomath 2016-10-05 20:23:42
28
ImpossibleCube 2016-10-05 20:23:42
28
mathgandalfjr 2016-10-05 20:23:42
28
acegikmoqsuwy2000 2016-10-05 20:23:42
28
hwl0304 2016-10-05 20:23:42
28
Catoc17 2016-10-05 20:23:42
28
popcorn1 2016-10-05 20:23:42
$28$
j2005 2016-10-05 20:23:42
28
ptes77 2016-10-05 20:23:42
28
28
DPatrick 2016-10-05 20:23:48
The next one is 28. Its prime factorization is $2^2 \cdot 7$, so its proper divisors are 1, 2, 4, 7, and 14, and indeed $1+2+4+7+14 = 28$.
hwl0304 2016-10-05 20:24:10
there is no reasonable pattern, it's just you know it or you don't
DPatrick 2016-10-05 20:24:21
Actually, there is a pattern to the perfect numbers!
DPatrick 2016-10-05 20:24:38
It's not easy, though.
DPatrick 2016-10-05 20:25:04
We have $6 = 2 \cdot 3$ and $28 = 2^2 \cdot 7$. Do you see any pattern in those prime factorizations?
hinna 2016-10-05 20:25:20
Both have 2
sathvi_k 2016-10-05 20:25:20
powers of 2
scienceFTW 2016-10-05 20:25:33
the 2s are increasing by a power of 1
hinna 2016-10-05 20:25:33
Also 2^2-1=3
mathgandalfjr 2016-10-05 20:25:43
yes, 3 or 7 plus 1 is a power of 2
DPatrick 2016-10-05 20:25:58
Indeed. It's a power of 2, times a prime that is one less than the next power of 2.
ptes77 2016-10-05 20:26:09
Catoc17 2016-10-05 20:26:18
(2^(n-1))(2^n -1)
ninjataco 2016-10-05 20:26:25
of the form $2^k (2^{k+1} - 1)$
DPatrick 2016-10-05 20:26:38
Right: the examples we've found are of the form $2^n(2^{n+1}-1)$ for some $n$. Can we say any more?
EpicRuler101 2016-10-05 20:26:51
so the next would be 2^3 x 15
DPatrick 2016-10-05 20:27:07
Right, we can try $2^3 \cdot 15 = 120$.
winnertakeover 2016-10-05 20:27:10
$2^{n+1}-1$ must be prime.
DPatrick 2016-10-05 20:27:23
Right: this doesn't work because 15 isn't prime. This has too many factors.
popcorn1 2016-10-05 20:27:42
n = 4, then 496
divijleisha 2016-10-05 20:27:42
and then would be 2^4*31
EpicDragonSlayr 2016-10-05 20:27:42
then 2^4*31
ImpossibleCube 2016-10-05 20:27:42
$2^4 *31$ works
agbdmrbirdyface 2016-10-05 20:27:42
the next is actually 496
DPatrick 2016-10-05 20:27:52
Next is 496. Its prime factorization is $2^4 \cdot 31$, so its proper divisors are 1, 2, 4, 8, 16, 31, 62, 124, and 248. Indeed $1+2+4+8+16+31+62+124+248 = 496$.
DPatrick 2016-10-05 20:28:35
Before we go on...why does this work? Why is $2^{n-1}(2^n-1)$ perfect if $2^n-1$ is prime?
DPatrick 2016-10-05 20:29:24
I'm going to write it as $2^{p-1}(2^p-1)$ where $p$ is prime and $2^p-1$ is prime, because $2^p-1$ can only be prime if $p$ is prime (it's a neat exercise to try to prove this yourself).
Derive_Foiler 2016-10-05 20:29:33
Derive_Foiler 2016-10-05 20:29:33
try adding all factors to get the same number again
DPatrick 2016-10-05 20:29:46
Right, we can just list the factors for this number!
DPatrick 2016-10-05 20:30:08
If we set $q = 2^p -1$, then the proper factors are just $1, 2, 4, \ldots 2^{p-1}$ and $q, 2q, 4q, \ldots 2^{p-2}q$.
NewtonFTW 2016-10-05 20:30:23
1 + 2 + 4...+ 2^n-1 = 2^n-1
DPatrick 2016-10-05 20:30:44
Right. Moreover, we've got two geometric series worth of factors:
\begin{align*}
1 + 2 + \cdots + 2^{p-1} &= 2^p - 1, \\
q + 2q + \cdots + 2^{p-2}q &= q(2^{p-1}-1).
\end{align*}
popcorn1 2016-10-05 20:30:58
DPatrick 2016-10-05 20:31:06
That first sum is just $q$ itself, so the divisors sum to $q + q(2^{p-1}-1) = 2^{p-1}q$, which is the number we started with!
DPatrick 2016-10-05 20:31:19
So we've proved a theorem:

If $2^p-1$ is prime, then $2^{p-1}(2^p-1)$ is a perfect number.
Derive_Foiler 2016-10-05 20:31:31
and we're done! but how do we know that these are the $only$ perfect numbers?
DPatrick 2016-10-05 20:32:06
Well, we don't based just on what we've done above -- that's a good point. But it is known that these are the only possibilities for even perfect numbers.
DPatrick 2016-10-05 20:32:12
In fact, the theorem goes the other way too: all even perfect numbers are of this form. This is called the Euclid-Euler Theorem. (Euclid proved the direction that we just proved, over 2000 years ago! Euler proved the converse in the 18th century.)
DPatrick 2016-10-05 20:32:40
So let's see if we can generate any other perfect numbers under 10000 using this theorem. Recall we've already got:

\begin{align*}

6 &= 2 \cdot 3 = 2^1(2^2-1) \\

28 &= 2^2 \cdot 7 = 2^2(2^3-1) \\

496 &= 2^4 \cdot 31 = 2^4(2^5-1)

\end{align*}
DPatrick 2016-10-05 20:32:50
What's next to check?
supermathkid 2016-10-05 20:32:58
8128
mathgandalfjr 2016-10-05 20:32:58
and 8128
mathguy5041 2016-10-05 20:32:58
$127*64=8128$
NewtonFTW 2016-10-05 20:33:01
2^6 * 127
Mathisfun04 2016-10-05 20:33:03
8128?
DPatrick 2016-10-05 20:33:08
We'll try $p=7$, giving us the number $2^6(2^7-1)$.
DPatrick 2016-10-05 20:33:13
This is $64 \cdot 127 = 8128$. (You don't actually need to compute it -- it's enough for our purposes to see that it's less than 10000).
DPatrick 2016-10-05 20:33:23
(We do need to check that 127 is prime, and indeed it is.)
csasomachar 2016-10-05 20:33:34
these are the only 4
divijleisha 2016-10-05 20:33:34
DPatrick 2016-10-05 20:33:41
The next possibility on our list would be $2^{10}(2^{11}-1) = 1024 \cdot 2047$, but that's already way too big. And it's not perfect, since $2047 = 23 \cdot 89$ isn't prime.
DPatrick 2016-10-05 20:33:51
So, the only perfect numbers less than 10000 are 6, 28, 496, and 8128. There are $\boxed{4}$ of them.
Hero_of_Hyrule 2016-10-05 20:34:01
Are there any odd perfect numbers?
DPatrick 2016-10-05 20:34:17
This is one of the great unsolved problems in mathematics. Nobody knows if there any odd perfect numbers. Our theorem only works for even perfect numbers. No one has found an odd perfect number yet, but no one has been able to prove that they don't exist.
DPatrick 2016-10-05 20:34:45
To add one more point: you may have heard of Mersenne primes: these are primes of the form $2^p - 1$ where $p$ is itself prime. So the Euclid-Euler Theorem tells us that all even perfect numbers are Mersenne primes times a power of 2.
hliu70 2016-10-05 20:35:05
do we know there aren't any under 10,000?
DPatrick 2016-10-05 20:35:27
Yes -- we know in fact that there are no odd perfect numbers less than $10^{1500}$.
DPatrick 2016-10-05 20:35:39
But there might be a bigger one out there somewhere!
DPatrick 2016-10-05 20:36:00
It is also unknown if there are infinitely many perfect numbers.
DPatrick 2016-10-05 20:36:05
Only 49 of them are known so far.
DPatrick 2016-10-05 20:36:15
The largest known Mersenne prime is $2^{74207281}-1$, which means that the largest known perfect number is $2^{74207280}(2^{74207281}-1)$. This number has over 44 million digits!
acegikmoqsuwy2000 2016-10-05 20:36:29
GIMPS!
DPatrick 2016-10-05 20:36:39
Yes: the Great Internet Mersenne Prime Search
shouldnt there be infinite mersenne primes if there are infinite primes?
DPatrick 2016-10-05 20:36:53
That's exactly right: these two questions are equivalent.
DPatrick 2016-10-05 20:37:01
Anyway, it's getting late, so on to #10:
DPatrick 2016-10-05 20:37:05
10. Which of the following is largest?
a. $2016^{2016}$ b. $2016!$ c. $20^{(16^{20})}$ d. $16^{(20^{20})}$ e. $20^{(20^{16})}$
NewtonFTW 2016-10-05 20:37:32
Eliminate b easily
cinamon 2016-10-05 20:37:32
I can see already that a>b
scienceFTW 2016-10-05 20:37:32
We can immediately rule out B as A is larger
somepersonoverhere 2016-10-05 20:37:32
rule out b, and look at logarithms of the answer choices
DPatrick 2016-10-05 20:37:45
Let start with the easiest comparison. We can easily compare (a) and (b).
DPatrick 2016-10-05 20:37:52
Both are the product of 2016 numbers. But the numbers in (a)'s product are all 2016's, whereas almost all of the numbers in (b)'s product are less than 2016.
DPatrick 2016-10-05 20:38:00
So (a) is greater than (b). Thus (b) cannot be the answer. Let's delete it from our choices:
DPatrick 2016-10-05 20:38:04
10. Which of the following is largest?
a. $2016^{2016}$ c. $20^{(16^{20})}$ d. $16^{(20^{20})}$ e. $20^{(20^{16})}$
DPatrick 2016-10-05 20:38:18
(a) looks a lot different than the others, so let's see if we can rule it in or out.
DPatrick 2016-10-05 20:38:30
About how big is (a)? Can we estimate its number of digits?
Derive_Foiler 2016-10-05 20:39:08
2016 ~ 2*1000, so 3000 digits? 600?
Derive_Foiler 2016-10-05 20:39:08
6000?
ptes77 2016-10-05 20:39:08
6666
to_chicken 2016-10-05 20:39:08
6400 ish
jasonhu4 2016-10-05 20:39:08
less than 8000
DPatrick 2016-10-05 20:39:13
Pretty good estimates.
DPatrick 2016-10-05 20:39:35
An often-useful estimating tool is $2^{10} \approx 10^3$. So we get that $$2016^{10} \approx (2 \cdot 10)^{10} \approx 10^{33}.$$
DPatrick 2016-10-05 20:40:14
So, $2016^{2016} \approx (10^{33})^{201.6} \approx 10^{6653}$. Therefore, (a) has approximately 6654 digits.
DPatrick 2016-10-05 20:40:30
(This is a pretty good approximation. I put it into my computer and it actually has 6662 digits.)
DPatrick 2016-10-05 20:40:43
How does this compare to the number of digits for any of (c) through (e)?
acegikmoqsuwy2000 2016-10-05 20:40:54
6654 is way less than the n umber of digits of c,d,e
ptes77 2016-10-05 20:40:54
A whole lot smaller
somepersonoverhere 2016-10-05 20:40:54
It pales in comparison
DPatrick 2016-10-05 20:41:05
Not even close. (c),(d),(e) all have a base larger than 10, so the number of digits is larger than the value of the exponent.
DPatrick 2016-10-05 20:41:18
But those exponents are huge -- they're all at least 20-digit numbers themselves, so each of (c),(d),(e) have a number of digits that itself is at least a 20-digit number. (Way more than the 4-digit number of digits given by 6654.)
DPatrick 2016-10-05 20:41:34
So the answer must be one of (c), (d), or (e).
DPatrick 2016-10-05 20:41:38
10. Which of the following is largest?
c. $20^{(16^{20})}$ d. $16^{(20^{20})}$ e. $20^{(20^{16})}$
DPatrick 2016-10-05 20:42:22
Before we try to compute anything, do you have intuition? These all use the same three numbers (a 16 and two 20's). Do we want bigger numbers in the base or in the exponent?
ninjataco 2016-10-05 20:42:46
exponent
Liopleurodon 2016-10-05 20:42:46
exponent
popcorn1 2016-10-05 20:42:46
exponent
Z_Math404 2016-10-05 20:42:46
exponent
mathgandalfjr 2016-10-05 20:42:46
the bigger exponent will do so I think d is the answer
supermathkid 2016-10-05 20:42:46
exponent
divijleisha 2016-10-05 20:42:46
exponent
Intellectuality123 2016-10-05 20:42:46
exponent
UrInvalid 2016-10-05 20:42:46
exponent
DPatrick 2016-10-05 20:42:56
The intuitive idea is that, if the numbers are big enough, we're generally better off making the exponent as big as possible.
DPatrick 2016-10-05 20:43:01
To look at some simple examples, $3^5 = 243$ is larger than $5^3 = 125$, and $2^{10} = 1024$ is larger than $10^2 = 100$.
winnertakeover 2016-10-05 20:43:11
d) could be the largest.
Apurple 2016-10-05 20:43:11
DPatrick 2016-10-05 20:43:32
This intuition should lead us to answer choice (d) -- it puts the smaller number 16 in the base, and the bigger numbers in both exponents.
jasonhu4 2016-10-05 20:43:45
this is hardly rigorous
DPatrick 2016-10-05 20:43:52
I agree. Let's see if we can approximate it better.
DPatrick 2016-10-05 20:44:09
A lot of you earlier suggested using logarithms. That's a good idea.
DPatrick 2016-10-05 20:44:17
We have the helpful rule that $\log(a^b) = b\log(a)$.
DPatrick 2016-10-05 20:44:23
So we can replace our question with:
DPatrick 2016-10-05 20:44:27
10. Which of the following is largest?
c. $16^{20} \log 20$ d. $20^{20} \log 16$ e. $20^{16} \log 20$
DPatrick 2016-10-05 20:44:51
Any why not use that rule again?
DPatrick 2016-10-05 20:46:28
$\log(16^{20} \log 20) = \log(16^{20}) + \log(\log 20) = 20 \log 16 + \log(\log 20)$
DPatrick 2016-10-05 20:47:18
10. Which of the following is largest?
c. $20 \log(16) + \log(\log 20)$ d. $20 \log(20) + \log(\log 16)$ e. $16 \log(20) + \log(\log 20)$
acegikmoqsuwy2000 2016-10-05 20:47:31
now $\log(\log(x))$ is very close to 1 for such small numbers like 16 and 20
DPatrick 2016-10-05 20:47:42
Right. If, say, we use 10 as the base of our logs, roughly what are $\log_{10}(16)$ and $\log_{10}(20)$?
yamyamx2 2016-10-05 20:48:03
1.2
jasonhu4 2016-10-05 20:48:03
1.2 or so
DPatrick 2016-10-05 20:48:08
Estimating using logs is a bit of an art form. Here's how you can estimate $\log_{10}(20)$: write it as $\log_{10}(20) = \log_{10}(10 \cdot 2) = 1 + \log_{10}2 = 1 + \frac{1}{\log_2{10}}.$ Then $2^3 = 8$ and $2^4 = 16$, so $\log_2{10}$ is about $3\frac13$, so $\frac{1}{\log_2{10}}$ is about $0.3$, and hence $\log_{10}(20) \approx 1.3$ should be a pretty good estimate.
DPatrick 2016-10-05 20:48:19
A similar exercise gives the approximation $\log_{10}(16) \approx 1.2$.
DPatrick 2016-10-05 20:48:59
And then $\log(\log 16)$ and $\log(\log 20)$ are each about $\log(1.2)$ and $\log(1.3)$, so these are really close to 0.
DPatrick 2016-10-05 20:49:33
Derive_Foiler 2016-10-05 20:49:40
So, we want the 20log 20, and we get d
Liopleurodon 2016-10-05 20:49:40
so the answer is d, since the multiplies the largest numbers
DPatrick 2016-10-05 20:49:48
Right. $\boxed{(\text{d})}$ is the largest quantity.
DPatrick 2016-10-05 20:50:03
That's it for Round 1!
DPatrick 2016-10-05 20:50:15
If you officially participated in the contest, and you got at least 8 of these 10 problems correct, then you'll move on to Round 2 later this month.
DPatrick 2016-10-05 20:50:26
Official notification of this should go out to your teacher early next week.
DPatrick 2016-10-05 20:50:34
We will have a Round 2 Math Jam on Wednesday, November 2 at 7:30 PM Eastern / 4:30 PM Pacific to discuss the Round 2 problems and solutions.
DPatrick 2016-10-05 20:50:45
Then, 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 2016-10-05 20:50:54
If you're curious, here are the 9 regions:
DPatrick 2016-10-05 20:50:57
DPatrick 2016-10-05 20:51:11
Perhaps I'll see you in Atlanta in January!
DPatrick 2016-10-05 20:51:19
DPatrick 2016-10-05 20:51:37
That's all for tonight's Math Jam -- have a good evening!
mikebreen 2016-10-05 20:51:43
Good luck in Round Two!
Is your round 1 score taken into account when deciding what top scorer (in case of ties) will advance to semifinals?
DPatrick 2016-10-05 20:52:27
I don't believe so. The tiebreaker criteria are on the AMS's website that I linked to above.
mikebreen 2016-10-05 20:52:52
Right. Round One doesn't count for final selection.
DPatrick 2016-10-05 20:53:42
For those who missed the intro at the beginning, Mike Breen is the AMS Public Awareness Officer, and the co-creator and host of Who Wants to Be a Mathematician. So his answers are definitive.
Shaktiala 2016-10-05 20:53:57
Are there any sample questions to prepare?
DPatrick 2016-10-05 20:54:06
There is an extensive archive of past contests on the AMS's website.
mikebreen 2016-10-05 20:54:36
bobjoe123 2016-10-05 20:54:41
WHEN IS IT NEXT YEAR
DPatrick 2016-10-05 20:54:59
The national contest qualifying round always runs in late September through early October.
DPatrick 2016-10-05 20:55:18
There are also various regional contests throughout the year. Check the AMS website for a schedule (as they become announced).
DPatrick 2016-10-05 20:55:43
I see that there's one in late October in Fort Worth, Texas.
seanwang2001 2016-10-05 20:55:51
will there be a transcript for this
DPatrick 2016-10-05 20:56:09
Yes, the full transcript will be posted as soon as we're done answering questions (which we'll do for a few more minutes).
Hero_of_Hyrule 2016-10-05 20:56:14
Will we review the semifinal problems later here?
DPatrick 2016-10-05 20:56:34
Yes, there's a Math Jam scheduled for November 2 to go over the Round 2 problems.
DPatrick 2016-10-05 20:56:44
And the Finals from Atlanta in January will be webcast live.
mikebreen 2016-10-05 20:56:56
RI on Pi Day, MI in March (?), DC in late April (National Math Festival)
DPatrick 2016-10-05 20:57:08
These are some upcoming regional contests coming this winter/spring.
cinamon 2016-10-05 20:57:14
Did Mr. Breen's hand actually bleed on the wheel of fortune?
DPatrick 2016-10-05 20:57:23
I don't think he made it up.
mikebreen 2016-10-05 20:57:46
Yes. I didn't realize it until I sat back down in the audience and saw the stage hands cleaning up
DPatrick 2016-10-05 20:58:04
Mike tells me that the national Final will be on Saturday, January 7 at 1 PM Eastern.
DPatrick 2016-10-05 20:59:03
I think we'll end it there for tonight. If you have further questions, please check out the website at http://www.ams.org/programs/students/wwtbam/wwtbam.
DPatrick 2016-10-05 20:59:14
Thanks for coming, and hopefully I'll see you again in 4 weeks for Round 2!
mikebreen 2016-10-05 20:59:17
Thanks, everyone.
bobjoe123 2016-10-05 20:59:57
THANK YOU!!!!!!!!!!!!!!!!!
Liopleurodon 2016-10-05 21:01:57
thank you for the Math Jam!
yiqun 2016-10-05 21:01:57
thank you!! @DPatrick and Naysh and mikebreen!!!!