Who Wants to Be a Mathematician, Round 2
Go back to the Math Jam ArchiveAoPS 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.
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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?
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.
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):
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.
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?
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.
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...
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!
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.
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.
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.
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.
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!!
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.
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.
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).
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.
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.
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
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.)
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!
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.
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.
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.
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.)
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.
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?
do you take the test on the computer?
DPatrick
2017-10-26 19:35:14
Most students do. Some take it on paper.
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.
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!
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 $2^{2017}+3^{2017}+7^{2017}$?
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?
How can we approach this?
STEM88
2017-10-26 19:36:52
find a pattern
find a pattern
ag2006
2017-10-26 19:36:52
its important to find the units digit of each and add those.
its important to find the units digit of each and add those.
avn
2017-10-26 19:36:52
find a pattern
find a pattern
goodbear
2017-10-26 19:36:52
pattern
pattern
yali306
2017-10-26 19:36:52
Find a pattern
Find a pattern
int_user
2017-10-26 19:36:59
The units digits repeat, so we can find a pattern
The units digits repeat, so we can find a pattern
mathwiz0
2017-10-26 19:36:59
find a pattern
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.
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.
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?
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
2, 4, 8, 6: cycles of 2
shootingstar8
2017-10-26 19:38:03
2 => 2,4,8,6
2 => 2,4,8,6
prajna1225
2017-10-26 19:38:03
2 4 8 6
2 4 8 6
int_user
2017-10-26 19:38:03
2, 4, 8, 6
2, 4, 8, 6
GeronimoStilton
2017-10-26 19:38:03
2,4,8,6…
2,4,8,6…
Danshim917
2017-10-26 19:38:03
2 4 8 6
2 4 8 6
SirCalcsALot
2017-10-26 19:38:03
2 -> 4 -> 8 -> 6 -> 2
2 -> 4 -> 8 -> 6 -> 2
amypham2
2017-10-26 19:38:03
2, 4, 8, 6
2, 4, 8, 6
Rushn
2017-10-26 19:38:03
2,4,8,6
2,4,8,6
Heatblast016
2017-10-26 19:38:03
2 4 8 6 2
2 4 8 6 2
DPatrick
2017-10-26 19:38:19
$2^1$ has units digit 2.
$2^1$ has units digit 2.
DPatrick
2017-10-26 19:38:37
$2^2$ has units digit 2*2 = 4.
$2^2$ has units digit 2*2 = 4.
DPatrick
2017-10-26 19:38:43
$2^3$ has units digit 2*4 = 8.
$2^3$ 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.
$2^4$ 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.
$2^5$ 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.
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:
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
$\begin{array}{r|c|c|c|c}
n \pmod{4} & 1 & 2 & 3 & 0 \\ \hline
\text{Units digit of } 2^n & 2 & 4 & 8 & 6
\end{array}$
$\begin{array}{r|c|c|c|c}
n \pmod{4} & 1 & 2 & 3 & 0 \\ \hline
\text{Units digit of } 2^n & 2 & 4 & 8 & 6
\end{array}$
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.
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?
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 $2$.
The units digit of $2^{2017$ is $2$.
DPatrick
2017-10-26 19:39:53
Exactly.
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.
$2017 \equiv 1 \pmod{4}$ (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 $3^{2017}$?
How about $3^{2017}$?
Mathforlife1
2017-10-26 19:40:17
3 9 7 1
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 …
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
3, 9, 7, 1
DPatrick
2017-10-26 19:40:28
There's a similar pattern:
There's a similar pattern:
DPatrick
2017-10-26 19:40:31
$\begin{array}{r|c|c|c|c}
n \pmod{4} & 1 & 2 & 3 & 0 \\ \hline
\text{Units digit of } 2^n & 2 & 4 & 8 & 6 \\
\text{Units digit of } 3^n & 3 & 9 & 7 & 1
\end{array}$
$\begin{array}{r|c|c|c|c}
n \pmod{4} & 1 & 2 & 3 & 0 \\ \hline
\text{Units digit of } 2^n & 2 & 4 & 8 & 6 \\
\text{Units digit of } 3^n & 3 & 9 & 7 & 1
\end{array}$
Mathforlife1
2017-10-26 19:40:40
which would mean 3
which would mean 3
shootingstar8
2017-10-26 19:40:40
3,9,7,1 => 3
3,9,7,1 => 3
amypham2
2017-10-26 19:40:40
so itd end with 3 again
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.
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
and 7 goes 7 9 3 1
RinoaOliver
2017-10-26 19:41:03
7, 9, 3, 1
7, 9, 3, 1
Giraffefun
2017-10-26 19:41:03
7,9,3,1
7,9,3,1
Mathforlife1
2017-10-26 19:41:03
7 9 3 1
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:
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
$\begin{array}{r|c|c|c|c}
n \pmod{4} & 1 & 2 & 3 & 0 \\ \hline
\text{Units digit of } 2^n & 2 & 4 & 8 & 6 \\
\text{Units digit of } 3^n & 3 & 9 & 7 & 1 \\
\text{Units digit of } 7^n & 7 & 9 & 3 & 1
\end{array}$
$\begin{array}{r|c|c|c|c}
n \pmod{4} & 1 & 2 & 3 & 0 \\ \hline
\text{Units digit of } 2^n & 2 & 4 & 8 & 6 \\
\text{Units digit of } 3^n & 3 & 9 & 7 & 1 \\
\text{Units digit of } 7^n & 7 & 9 & 3 & 1
\end{array}$
DPatrick
2017-10-26 19:41:21
So $7^{2017}$ has units digit 7.
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
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.
Then, the answer is the units digit of 2 + 3+ 7 which is 2.
GeronimoStilton
2017-10-26 19:41:32
add it to get 2
add it to get 2
mikebreen
2017-10-26 19:41:50
Unfortunately, some who got this wrong, but who knew what they were doing, answered "12."
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 $\boxed{2}$.
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!
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.
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?
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.
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.
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%.
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.
#9 was much harder; I'll discuss it when we get to it.
DPatrick
2017-10-26 19:44:17
On to #2:
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) $\sqrt{10}$
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?
Any we can rule out right away?
GeronimoStilton
2017-10-26 19:45:00
eliminate (d) immediately
eliminate (d) immediately
RinoaOliver
2017-10-26 19:45:00
(a) > (d)
(a) > (d)
fdas
2017-10-26 19:45:00
d is wrong. less than a
d is wrong. less than a
Emathmaster
2017-10-26 19:45:00
a>d
a>d
DPatrick
2017-10-26 19:45:05
Clearly $\sqrt[4]{101} > \sqrt[4]{100} = \sqrt{10}$. So we can eliminate (d).
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:
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) $\sqrt2 + \sqrt3$
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
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.
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.
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
\[ 5 - \sqrt3 \qquad\boxed{?}\qquad \sqrt2 + \sqrt3 \]
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
square both sides
ShreyJ
2017-10-26 19:46:53
add sqrt 3 and square it
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
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.
...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.
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:
But I'd also move the $\sqrt2$ to the left, so keep the square roots smaller:
DPatrick
2017-10-26 19:48:23
\[ 5 - \sqrt2 \qquad\boxed{?}\qquad 2\sqrt3 \]
\[ 5 - \sqrt2 \qquad\boxed{?}\qquad 2\sqrt3 \]
mathwiz0
2017-10-26 19:48:37
square both sides no
square both sides no
DPatrick
2017-10-26 19:48:41
Right: both sides are positive, so we can square without changing the inequality.
\[ 27 - 10\sqrt2 \qquad\boxed{?}\qquad 12 \]
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
\[ 15 \qquad\boxed{?}\qquad 10\sqrt2 \]
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:
\[ 225 \qquad\boxed{>}\qquad 200 \]
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 $$5 - \sqrt3 > \sqrt2 + \sqrt3.$$
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
now compare and and b
DPatrick
2017-10-26 19:49:53
2. Which of the following is largest?
(a) $\sqrt[4]{101}$ (b) $5 - \sqrt3$
2. Which of the following is largest?
(a) $\sqrt[4]{101}$ (b) $5 - \sqrt3$
GeronimoStilton
2017-10-26 19:50:12
square!
square!
DPatrick
2017-10-26 19:50:19
Let's try the same tactic. We can square both sides and preserve the inequality:
\[ \sqrt{101} \qquad\boxed{?}\qquad 28 - 10\sqrt3 \]
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.
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 $(1.75)^2 = 3.0625$.)
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
$\sqrt{101} \approx 10$
$\sqrt{101} \approx 10$
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.
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
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.
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.
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!
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?
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?
Are there any 1-digit emirps?
Cidkip
2017-10-26 19:53:09
No, thats the same prime number
No, thats the same prime number
int_user
2017-10-26 19:53:09
No because it has to be different
No because it has to be different
Danshim917
2017-10-26 19:53:09
No because theyre not differnet
No because theyre not differnet
AlcumusGuy
2017-10-26 19:53:09
No, reversing gives the same number.
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.
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.
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
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.
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.
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.
And it's clear from the definition that emirps come in pairs.
int_user
2017-10-26 19:54:08
and have distinct digits
and have distinct digits
DPatrick
2017-10-26 19:54:11
Right.
Right.
DPatrick
2017-10-26 19:54:15
So we can just look at the six different pairs of digits:
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
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?
Which of these pairs have both numbers prime?
mathwiz0
2017-10-26 19:55:13
39 and 93 are divisible by 3
39 and 93 are divisible by 3
int_user
2017-10-26 19:55:13
13x7=91 so its not prime
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
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.
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!)
(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.
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!
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.
So there are 4 pairs giving $\boxed{8}$ emirps between 1 and 100.
ag2006
2017-10-26 19:56:09
are emirps real?
are emirps real?
mikebreen
2017-10-26 19:56:29
uoy rof dooG
uoy rof dooG
DPatrick
2017-10-26 19:56:43
Moving on...
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 $|a + bi| = \sqrt{a^2+b^2}$.
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?
What's going to be a really useful fact for us to use?
Cidkip
2017-10-26 19:57:36
$|1+\sqrt{3}i| = |1-\sqrt{3}i| = 2$
$|1+\sqrt{3}i| = |1-\sqrt{3}i| = 2$
EulerMacaroni
2017-10-26 19:57:36
norm is multiplicative
norm is multiplicative
ppu
2017-10-26 19:57:41
magnitude of top divided by bottom
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 $y$.
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|
|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.
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 $\dfrac{(|1+\sqrt3\,i|)^8}{(|1-\sqrt3\,i|)^6}$.
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, $|1 \pm \sqrt{3}\,i| = 2$.
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!)
(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
2^8/2^6 = 2^2 = 4
Kreps
2017-10-26 19:59:50
so its 4
so its 4
DPatrick
2017-10-26 19:59:53
So this is just $\dfrac{2^8}{2^6} = 2^2 = \boxed{4}$.
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 $n$.
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!
Write out the sum!
DPatrick
2017-10-26 20:00:49
This is a matter of taste, but I agree with Geronimo.
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.
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:
$$\log_a(1^n) + \log_a(2^n) + \cdots + + \log_a(n^n) = n.$$
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)
log (a) + log(b) = log(ab)
DPatrick
2017-10-26 20:01:26
Aha, that rule is going to come in handy!
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, $\log_a(x) + \log_a(y) = \log_a(xy)$.
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.
So our sum of logs becomes the log of a product.
DPatrick
2017-10-26 20:01:51
$$\log_a(1^n \cdot 2^n \cdot \,\cdots\, \cdot n^n) = n.$$
$$\log_a(1^n \cdot 2^n \cdot \,\cdots\, \cdot n^n) = n.$$
DPatrick
2017-10-26 20:02:04
But what's the product of a bunch of $n$th powers?
But what's the product of a bunch of $n$th powers?
GeronimoStilton
2017-10-26 20:02:21
$n!^n$
$n!^n$
ShreyJ
2017-10-26 20:02:21
$n!^n$
$n!^n$
GeronimoStilton
2017-10-26 20:02:25
That's $$\log_a(n!^n) = n.$$
That's $$\log_a(n!^n) = n.$$
Kreps
2017-10-26 20:02:27
(n!)^n
(n!)^n
DPatrick
2017-10-26 20:02:43
Right, we just have a big $n$th power: $$\log_a\left((1 \cdot 2 \cdot \,\cdots\, \cdot n)^n\right) = n.$$
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 $n!$.
And that product inside the $n$th power is just $n!$.
DPatrick
2017-10-26 20:03:04
So we have \[ \log_a \left((n!)^n\right) = n \]
So we have \[ \log_a \left((n!)^n\right) = n \]
AOPS12142015
2017-10-26 20:03:18
$a^n=(n!)^n$
$a^n=(n!)^n$
int_user
2017-10-26 20:03:18
now we can bring the n out front!
now we can bring the n out front!
Emathmaster
2017-10-26 20:03:18
we can now pull the nth power out
we can now pull the nth power out
Kreps
2017-10-26 20:03:18
rewrite it as an exponential equation
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:
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 $a^n = (n!)^n$.
It's just $a^n = (n!)^n$.
awesomemaths
2017-10-26 20:03:42
a = n!
a = n!
Danshim917
2017-10-26 20:03:48
so a=n!
so a=n!
DPatrick
2017-10-26 20:03:50
So clearly $a = \boxed{n!}$.
So clearly $a = \boxed{n!}$.
mikebreen
2017-10-26 20:04:00
Bill and I were very pleased that so many got this right.
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 $\text{googol}^{\text{googol}} = \text{googolplex}^n$?
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.
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?
What's a better way to write a googol?
Abch
2017-10-26 20:05:02
Googol=10^100
Googol=10^100
kvedula2004
2017-10-26 20:05:02
10^100
10^100
goodbear
2017-10-26 20:05:02
10^100
10^100
fishy15
2017-10-26 20:05:02
$10^{100}$
$10^{100}$
maverick8
2017-10-26 20:05:02
$10^{100}$
$10^{100}$
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).
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?
And a googolplex?
ShreyJ
2017-10-26 20:05:51
10^10^100
10^10^100
ppu
2017-10-26 20:05:51
10^10^100
10^10^100
kvedula2004
2017-10-26 20:05:51
10^10^100
10^10^100
goodbear
2017-10-26 20:05:51
10^10^100
10^10^100
DPatrick
2017-10-26 20:05:54
It's $10^{\text{googol}} = 10^{10^{100}}$.
It's $10^{\text{googol}} = 10^{10^{100}}$.
DPatrick
2017-10-26 20:06:01
So our equation is $\left(10^{100}\right)^{10^{100}} = \left(10^{10^{100}}\right)^n$.
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.
This looks a lot scarier that it really is.
GeronimoStilton
2017-10-26 20:06:25
exponent properties!
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.
Right. In particular, we can now use the identity $(a^b)^c = a^{bc}$ on both sides.
DPatrick
2017-10-26 20:06:41
\[ 10^{(100 \cdot 10^{100})} = 10^{(10^{100} \cdot n)} \]
\[ 10^{(100 \cdot 10^{100})} = 10^{(10^{100} \cdot n)} \]
Kreps
2017-10-26 20:06:59
n = 100
n = 100
awesomemaths
2017-10-26 20:06:59
n = 100
n = 100
AOPS12142015
2017-10-26 20:06:59
$n=100$
$n=100$
SirCalcsALot
2017-10-26 20:06:59
Then, n clearly equals 100.
Then, n clearly equals 100.
DPatrick
2017-10-26 20:07:04
The exponents on either side must match, so $100 \cdot 10^{100} = 10^{100} \cdot n$.
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 $n = \boxed{100}$.
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!
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 $x^4 - 8x^3 + 16x^2 - 11x + 5$?
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 $p^2+q^2+r^2+s^2$.
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!
vieta!
ShreyJ
2017-10-26 20:08:36
Vieta's
Vieta's
awesomemaths
2017-10-26 20:08:36
vietas
vietas
DPatrick
2017-10-26 20:08:50
Vieta is going to be our dear friend here.
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.
"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:
\[ x^4 - 8x^3 + 16x^2 - 11x + 5 = (x-p)(x-q)(x-r)(x-s).\]
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.
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 $-8$.
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 $p+q+r+s = 8$.
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?
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
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.
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
= 16
DPatrick
2017-10-26 20:11:18
And on the left side, we have the 16.
And on the left side, we have the 16.
DPatrick
2017-10-26 20:11:22
Comparing those tells us that $pq + pr + ps + qr + qs + rs = 16$.
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 $p^2 + q^2 + r^2 + s^2$.
How does these help? Remember, we want $p^2 + q^2 + r^2 + s^2$.
pad
2017-10-26 20:11:49
(p+q+r+s)^2-2(pq+pr+...)
(p+q+r+s)^2-2(pq+pr+...)
RinoaOliver
2017-10-26 20:11:49
(p + q + r + s)^2
(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
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(...)
(p+q+r+s)^2-2(...)
DPatrick
2017-10-26 20:12:02
Exactly. We can use the fact that
\[ (p+q+r+s)^2 = p^2 + q^2 + r^2 + s^2 + 2(pq+pr+ps+qr+qs+rs)\]
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!
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,
\[ 8^2 = p^2 + q^2 + r^2 + s^2 + 2(16).\]
Specifically,
\[ 8^2 = p^2 + q^2 + r^2 + s^2 + 2(16).\]
Emathmaster
2017-10-26 20:12:36
so ans is 32
so ans is 32
int_user
2017-10-26 20:12:36
32
32
Danshim917
2017-10-26 20:12:36
so 32
so 32
goodbear
2017-10-26 20:12:36
32
32
DPatrick
2017-10-26 20:12:40
So $p^2 + q^2 + r^2 + s^2 = 8^2 - 2(16) = 64 - 32 = \boxed{32}$.
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!
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?
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.
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
angles add to 180 so its half of the circle
Danshim917
2017-10-26 20:14:29
the angles add to 180
the angles add to 180
AOPS12142015
2017-10-26 20:14:29
Use the fact that the sum of the angles of a triangle adds up to ${180}^{\circ}$
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
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
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
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
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.
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.
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
30-2pi
ppu
2017-10-26 20:15:24
$30-2\pi$ is our answer!
$30-2\pi$ is our answer!
EasyAs_Pi
2017-10-26 20:15:24
:facepalm: $30-2\pi$....
:facepalm: $30-2\pi$....
rainbow10
2017-10-26 20:15:24
30-2pi
30-2pi
Mathforlife1
2017-10-26 20:15:24
30-2pi
30-2pi
DPatrick
2017-10-26 20:15:34
The triangle area is $\frac12 \cdot 12 \cdot 5 = 30$.
The triangle area is $\frac12 \cdot 12 \cdot 5 = 30$.
DPatrick
2017-10-26 20:15:40
The total blue area is $\frac12 \cdot \pi(2)^2 = 2\pi$.
The total blue area is $\frac12 \cdot \pi(2)^2 = 2\pi$.
DPatrick
2017-10-26 20:15:44
Hence, the area remaining is $\boxed{30-2\pi}$.
Hence, the area remaining is $\boxed{30-2\pi}$.
mikebreen
2017-10-26 20:15:56
Smart group.
Smart group.
DPatrick
2017-10-26 20:16:04
That was an easy problem, the calm before the storm.
That was an easy problem, the calm before the storm.
DPatrick
2017-10-26 20:16:10
The last 2 are really much harder.
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:
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 $2018$?
9. What is the remainder when $1000^{2018}$ is divided by $2018$?
DPatrick
2017-10-26 20:17:04
What might help to know?
What might help to know?
Ancy
2017-10-26 20:17:07
1009 is prime
1009 is prime
DPatrick
2017-10-26 20:17:16
Yes, that's useful!
Yes, that's useful!
DPatrick
2017-10-26 20:17:30
We have the prime factorization $2018 = 2 \cdot 1009$.
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 $1000 = 10^3 = 2^3 \cdot 5^3$.
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
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.
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 $1000^{2018} \pmod{2}$?
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.
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 $1000^{2018} \pmod{1009}$?
And how do we approach $1000^{2018} \pmod{1009}$?
EasyAs_Pi
2017-10-26 20:19:29
fermat's little theorem
fermat's little theorem
SHARKYKESA
2017-10-26 20:19:29
Euler totient function
Euler totient function
Emathmaster
2017-10-26 20:19:29
FLT
FLT
plshalp
2017-10-26 20:19:29
fLT
fLT
ppu
2017-10-26 20:19:36
since 1000 and 1009 are relatively prime, we can use Euler's Totient Theorem
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
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).
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 $p$.
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.)
(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
Note 1000^1008 is 1 mod 1009
AlcumusGuy
2017-10-26 20:20:47
1000^1008 is 1 modulo 1009
1000^1008 is 1 modulo 1009
DPatrick
2017-10-26 20:20:56
Right. In our particular example, it tells us that $1000^{1008} \equiv 1 \pmod{1009}$.
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 $1000^{1009} \equiv 1000 \pmod{1009}$.
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.
But 1000 and -9 are the same in mod 1009.
DPatrick
2017-10-26 20:22:24
So $1000^{1009} \equiv -9 \pmod{1009}$.
So $1000^{1009} \equiv -9 \pmod{1009}$.
Superwiz
2017-10-26 20:22:34
Square it to get 81
Square it to get 81
DPatrick
2017-10-26 20:22:56
And then $1000^{2018} = \left(1000^{1009}\right)^2 \equiv (-9)^2 \equiv 81 \pmod{1009}$.
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?
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
and we can solve the resulting system, or eyeball it 1009+81=1090
goodbear
2017-10-26 20:23:46
1090
1090
AOPS12142015
2017-10-26 20:23:46
CRT so $0 \pmod 2$ and $81 \pmod {1009}$ is $\boxed{1090} \pmod {2018}$
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
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.
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!
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.
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
I agree
DPatrick
2017-10-26 20:24:37
And finally, #10.
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 $25{,}000$?
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.
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?
May I ask how many people got the exact answer to #10?
DPatrick
2017-10-26 20:24:55
I'm told zero.
I'm told zero.
DPatrick
2017-10-26 20:25:12
What's another way of phrasing this question?
What's another way of phrasing this question?
Abch
2017-10-26 20:25:28
What is the 25000th prime
What is the 25000th prime
Tryharder
2017-10-26 20:25:28
what is the 25000 prme
what is the 25000 prme
CornSaltButter
2017-10-26 20:25:30
what is the 25000th prime
what is the 25000th prime
DPatrick
2017-10-26 20:25:35
Right. We're asked to determine the 25,000th prime.
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.
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?
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
25
lifeisgood03
2017-10-26 20:26:34
25
25
plshalp
2017-10-26 20:26:34
25
25
EasyAs_Pi
2017-10-26 20:26:34
25
25
AOPS12142015
2017-10-26 20:26:34
$25$
$25$
rainbow10
2017-10-26 20:26:34
25
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!)
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.
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.
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?
Do you think that guess is too high or too low?
EulerMacaroni
2017-10-26 20:27:21
too low
too low
int_user
2017-10-26 20:27:21
too low
too low
BCZ01
2017-10-26 20:27:21
too low
too low
RinoaOliver
2017-10-26 20:27:21
no, starts to get less common later too low
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.
way too low, we don't consider new primes that form.
RinoaOliver
2017-10-26 20:27:21
too low
too low
mikebreen
2017-10-26 20:27:31
Smart group.
Smart group.
DPatrick
2017-10-26 20:27:33
It's quite a bit too low, because primes get rarer as numbers get larger.
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!
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?
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.
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).
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
1229/10000
DPatrick
2017-10-26 20:28:38
And there are 1,229 primes less than 10,000. Down to 12.29%.
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 $n$.
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.
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.
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?
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
logarithm
EulerMacaroni
2017-10-26 20:30:20
$\frac{1}{\log{n}}$
$\frac{1}{\log{n}}$
GeronimoStilton
2017-10-26 20:30:20
log!
log!
Ancy
2017-10-26 20:30:20
logarithms
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.
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.
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
prime number theorem
ppu
2017-10-26 20:31:04
I think the number of primes below n is around n/natural log n
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 $e$).
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.)
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.
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 $\dfrac{N}{\ln(N)} = m$.
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!
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:
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.)
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.
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 $\ln(50) \approx 4$.
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 $\dfrac{N}{\ln(N)} \approx 25000$?
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
so ln2500 is about 8
so ln2500 is about 8
DPatrick
2017-10-26 20:35:46
That's useful. Squaring gives $\ln(2500) \approx 8$, and cubing gives $\ln(125{,}000) \approx 12$.
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 $250,000$.
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.
Let's go with $N = 250000$ for a second.
DPatrick
2017-10-26 20:37:03
If we try to estimate $\dfrac{250{,}000}{\ln(250{,}000)}$...
If we try to estimate $\dfrac{250{,}000}{\ln(250{,}000)}$...
GeronimoStilton
2017-10-26 20:37:09
We also randomly know that $\ln(2) \approx 0.7$.
We also randomly know that $\ln(2) \approx 0.7$.
DPatrick
2017-10-26 20:37:15
That is pretty random.
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.
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 $\dfrac{250{,}000}{\ln(250{,}000)} \approx \dfrac{250{,}000}{12.5} = 20{,}000$.
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.)
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.
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.
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
what was the answer?
what was the answer?
awesomemaths
2017-10-26 20:38:59
and the answer is...
and the answer is...
ppu
2017-10-26 20:38:59
what is the real answer?
what is the real answer?
awesomemaths
2017-10-26 20:39:02
drumroll
drumroll
DPatrick
2017-10-26 20:39:08
Turns out my guess is not too bad, but not great either.
Turns out my guess is not too bad, but not great either.
DPatrick
2017-10-26 20:39:13
The actual answer is $\boxed{287{,}117}$.
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.
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.
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.)
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.
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?
how can you be exact though?
DPatrick
2017-10-26 20:40:35
You can't. Primes are unexact things.
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.
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?
Is it true that if you go over the exact number, you lose tiebreakers?
DPatrick
2017-10-26 20:41:08
No.
No.
DPatrick
2017-10-26 20:41:16
That's it for Round 2!
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
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
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:
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.
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.
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 ;-;
san diego getting its own region ;-;
DPatrick
2017-10-26 20:42:44
The home city of the finals always gets a local contestant.
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.
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!
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.
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!
Thanks, everyone. Great job!
DPatrick
2017-10-26 20:43:44
Thanks for coming tonight to our Math Jam. Good night!
Thanks for coming tonight to our Math Jam. Good night!
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