Math Jams

DPatrick 2015-02-26 19:00:13
Welcome to the 2015 AMC 10B/12B Math Jam!

DPatrick 2015-02-26 19:00:19
I'm Dave Patrick, and I'll be leading our discussion tonight.

DPatrick 2015-02-26 19:00:27
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 2015-02-26 19:00:34
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.
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 2015-02-26 19:00:52
There are a lot of students here! (Over 200 so far and counting.) As I said, only (a fraction of the) well-written comments will be passed to the entire group. 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 2015-02-26 19:01:13
Also, we won't always be going through the math quite as thoroughly as we do in our classes -- I can't teach all the prerequisite 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 2015-02-26 19:01:23
We have 2 assistants tonight helping out: Jackie (3628800J) and Ray Kim (crescendo).

DPatrick 2015-02-26 19:01:28
Jackie lives in Michigan, where she won the statewide Michigan Math Prize Competition. She was also a 3-time USA(J)MO qualifier and a member of the USA team at the International Linguistics Olympiad. She plans to major in math and computer science.

DPatrick 2015-02-26 19:01:39
Ray is currently attending UNC Chapel Hill as an undergrad, looking to get a degree in math and computer science. His hometown is Asheville, NC, and while he was there he took numerous AoPS classes throughout his middle and early high school career which really boosted his understanding of math and other topics that don't get covered in the regular school curriculum.

DPatrick 2015-02-26 19:01:54
They can answer questions by whispering to you or by opening a window with you to chat 1-on-1. However, due to the incredibly large size of the session tonight, they may not be able to get to you right away (or at all). Repeating your question over and over is more likely to annoy us than to get it answered faster, so please, just ask your question once and be patient, and please understand that we may not be able to answer all the questions tonight.

DPatrick 2015-02-26 19:02:10
Please also remember that the purpose of this Math Jam is to work through the solutions to AMC problems, and not to merely present the answers. "Working through the solutions" 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, and comments that skip key steps or jump ahead in the problem, without providing explanation or motivation, won't be posted.

DPatrick 2015-02-26 19:02:25
We will work the last 5 problems from the AMC 10B, then the last 5 problems from the AMC 12B. Two pairs of these problems are the same, 10B and 12B have the same Problem 21 and 10B Problem 25 is 12B Problem 23. We'll only solve those problems once.

DPatrick 2015-02-26 19:02:36
After that, time permitting, we may do another problem or two. We'll run for about 2 hours.

DPatrick 2015-02-26 19:02:52
Let's get started!

DPatrick 2015-02-26 19:02:58
#21 was the same on both contests:

DPatrick 2015-02-26 19:03:03
21. Cozy the Cat and Dash the Dog are going up a staircase with a certain number of steps. However, instead of walking up the steps one at a time, both Cozy and Dash jump. Cozy goes two steps up with each jump (though if necessary, he will just jump the last step). Dash goes five steps up with each jump (though if necessary, he will just jump the last steps if there are fewer than 5 steps left). Suppose that Dash takes 19 fewer jumps than Cozy to reach the top of the staircase. Let $s$ denote the sum of all possible numbers of steps this staircase can have. What is the sum of the digits of $s$?
$\phantom{10B:21}$
$\text{(A) } 9 \quad \text{(B) } 11 \quad \text{(C) } 12 \quad \text{(D) } 13 \quad \text{(E) } 15$

WhaleVomit 2015-02-26 19:03:32
use ceiling functions

Deathranger999 2015-02-26 19:03:32
We can set up an equation with ceiling function.

BFYSharks 2015-02-26 19:03:32
Ceiling function

thkim1011 2015-02-26 19:03:32
let's write an equation with ceiling functions

DPatrick 2015-02-26 19:03:57
We could certainly use the ceiling function, but I think it's perhaps even clearer if we don't.

stan23456 2015-02-26 19:04:28
set up cases

DPatrick 2015-02-26 19:04:38
That's sort of what I did. Let's suppose Dash takes $n+1$ steps to climb the staircase: $n$ regular steps plus a final step of size $d$, where $1 \le d \le 5.$

DPatrick 2015-02-26 19:04:49
Then how big is the staircase?

ninjataco 2015-02-26 19:05:05
5n+d

AMN300 2015-02-26 19:05:05
5n+d

abean077 2015-02-26 19:05:05
5n + d steps

swirlykick 2015-02-26 19:05:05
5n+d

speck 2015-02-26 19:05:05
Steps = 5n+ d

MathLearner01 2015-02-26 19:05:05
5n+d

_--__--_ 2015-02-26 19:05:05
5n + d steps

geogirl08 2015-02-26 19:05:05
5n + d

DPatrick 2015-02-26 19:05:09
Each regular step is 5 stairs, so Dash's total is $5n + d.$

DPatrick 2015-02-26 19:05:15
How about Cozy the Cat?

DPatrick 2015-02-26 19:05:35
If Dash takes $n$ regular steps, how many regular steps does Cozy take?

AMN300 2015-02-26 19:05:47
n+19

Tommy2000 2015-02-26 19:05:47
n+19

mishka1980 2015-02-26 19:05:47
(n+19)

DPatrick 2015-02-26 19:05:51
Right, Cozy takes $19+n$ regular steps and a final step of size $c$, where $1 \le c \le 2.$

DPatrick 2015-02-26 19:06:05
So what's the size of the staircase using these variables?

SimonSun 2015-02-26 19:06:19
38+2n+c

swirlykick 2015-02-26 19:06:19
2n+38+c

LOTRFan123 2015-02-26 19:06:19
2n+38+c

AMN300 2015-02-26 19:06:19
2(n+19)+c

WalkerTesla 2015-02-26 19:06:19
38+2n+c

bellyflop 2015-02-26 19:06:19
2n + 38 + c

DPatrick 2015-02-26 19:06:24
Right: Cozy's total steps is $2(19+n) + c.$

DPatrick 2015-02-26 19:06:30
But Dash and Cozy climb the same staircase!

DPatrick 2015-02-26 19:06:36
So we have the equation
\[
5n + d = 2(19+n) + c.
\]

AMN300 2015-02-26 19:07:05
3n = 38+(c-d) and casework from here

geogirl08 2015-02-26 19:07:05
solve for n?

bestwillcui1 2015-02-26 19:07:05
3n=38+c-d then solve

Darn 2015-02-26 19:07:05
Rearrange the variables

DPatrick 2015-02-26 19:07:11
Exactly: let's solve for $n$:
\[
3n = 38 - d + c.
\]

DPatrick 2015-02-26 19:07:16
What do we know about $-d+c$?

lucylai 2015-02-26 19:07:57
1 mod 3

MathLearner01 2015-02-26 19:08:19
1 mod 3

mihirb 2015-02-26 19:08:19
which is 1 mod 3

Benaloo26 2015-02-26 19:08:25
greater or equal to -4, less than or equal to 1

nosyarg 2015-02-26 19:08:25
it can be from -4 to 1

Tommy2000 2015-02-26 19:08:25
Ranges from -4 to 1

DPatrick 2015-02-26 19:08:31
Yes. $d$ is between 1 and 5 (inclusive), and $c$ is either 1 or 2. So $-d+c$ is an integer between 1 and -4 (inclusive).

DPatrick 2015-02-26 19:08:50
But also, since $3n = 38-d+c$, we must have that $38-d+c$ is a multiple of 3. This means that $-d+c$ must be 1 or -2 --- those are the only numbers in our range that produce a multiple of 3.

DPatrick 2015-02-26 19:09:03
(In shorthand, $-d+c$ must be equivalent to 1 mod 3.)

Eugenis 2015-02-26 19:09:10
So we just check all the cases from there

DPatrick 2015-02-26 19:09:25
Right. We just check the two cases $-d+c=1$ and $-d+c = -2$ and list the possibilities.

DPatrick 2015-02-26 19:09:33
If $-d+c = 1$, what do we know?

TheMagician 2015-02-26 19:10:00
n=13

kunsun 2015-02-26 19:10:00
n = 13

TheMaskedMagician 2015-02-26 19:10:00

WhaleVomit 2015-02-26 19:10:00
n=13

nosaj 2015-02-26 19:10:00
n=13

amburger66 2015-02-26 19:10:00
d=1, c=2

ompatel99 2015-02-26 19:10:00
c=2,d=1

jam10307 2015-02-26 19:10:00
d=1, c=2

DPatrick 2015-02-26 19:10:04
We must have $(d,c) = (1,2)$ for this to happen.

DPatrick 2015-02-26 19:10:10
We get $3n = 39$, so $n = 13.$ This means there are $5n + d = 5(13) + 1 = 66$ stairs.

DPatrick 2015-02-26 19:10:22
(As a check, you can verify that $2(n+19)+c = 2(13+19)+2 = 2(32)+2 = 66$ is the number of steps that Cozy climbs.)

mihirb 2015-02-26 19:10:45
then we solve for n and d in the other case?

DPatrick 2015-02-26 19:10:51
In the other case we have $-d+c = -2.$ This gives $3n = 38-2 = 36,$ so $n=12.$ What are the possibilities?

ompatel99 2015-02-26 19:11:13
(d,c)=(4,2) and (3,1) for -d+c=-2

Richardq 2015-02-26 19:11:13
(3,1) (4,2)

abean077 2015-02-26 19:11:13
d = 4 c = 2, d = 3 c = 1

_--__--_ 2015-02-26 19:11:13
c = 1 and d = 3, or c = 2 and d = 4

DPatrick 2015-02-26 19:11:17
We can have $(d,c)$ equal to (3,1) or to (4,2).

DPatrick 2015-02-26 19:11:27
If $d=3$, then $5n+d = 5(12)+3 = 63$ stairs are climbed.

DPatrick 2015-02-26 19:11:41
If $d=4$, then $5n+d = 5(12)+4 = 64$ stairs are climbed.

DPatrick 2015-02-26 19:11:48
So how do we finish?

Deathranger999 2015-02-26 19:12:05
Add them.

ninjataco 2015-02-26 19:12:05
66+64+63=193, so the answer is 13 (D)

MinsPar 2015-02-26 19:12:05
s = 66+63+64

Imagineny1 2015-02-26 19:12:05
add 63 64 66

acegikmoqsuwy2000 2015-02-26 19:12:05
so 193 total => 1+9+3=13

mathperson9 2015-02-26 19:12:05
63+66+64= 193

DPatrick 2015-02-26 19:12:09
The possible numbers of stairs are 63, 64, or 66. These sum to 193. The sum of the digits is $1+9+3 = 13.$ Answer (D).

DPatrick 2015-02-26 19:12:37
You could certainly have done this problem by trial-and-error, but the algebraic approach ensures that we catch all the solutions.

DPatrick 2015-02-26 19:12:46
On to 10B #22:

DPatrick 2015-02-26 19:12:51
22. In the figure shown below, $ABCDE$ is a regular pentagon and $AG=1$. What is $FG+JH+CD$?
$\phantom{10B:22}$
$\text{(A) } 3 \quad \text{(B) } 12-4\sqrt5 \quad \text{(C) } \dfrac{5+2\sqrt5}3 \quad \text{(D) } 1+\sqrt5 \quad \text{(E) } \dfrac{11+11\sqrt5}{10}$

DPatrick 2015-02-26 19:12:57

bluephoenix 2015-02-26 19:13:20
Similar triangles!

InLhings 2015-02-26 19:13:20
Similar Triangles?

math-rules 2015-02-26 19:13:20
Similar Triangles!!!

BFYSharks 2015-02-26 19:13:20
Everything is isoceles!

Richardq 2015-02-26 19:13:20
Similar triangles?

abean077 2015-02-26 19:13:20
Similar triangles?

DivideBy0 2015-02-26 19:13:20
similar triangles with common vertex at A

nosyarg 2015-02-26 19:13:20
similar triangles?

DPatrick 2015-02-26 19:13:43
There are a gazillion similar isosceles triangles in this picture. (That'd be a fun problem too -- counting how many there are!)

DPatrick 2015-02-26 19:13:53
Just so we have some notation to work with:

DPatrick 2015-02-26 19:13:56
Let's set $r$ to be a side of the big pentagon (so $AB = r$) and $s$ to be a side of the small pentagon (so $FG = s$).

DPatrick 2015-02-26 19:14:03
How can we relate $r$ and $s$?

numbertheory 2015-02-26 19:14:44
s = r-1

DPatrick 2015-02-26 19:15:07
There are lots and lots of ways to proceed, but that's how I started. We can look at triangle $ABH$:

DPatrick 2015-02-26 19:15:13

akim99 2015-02-26 19:15:49
Isoceles!

DPatrick 2015-02-26 19:15:51
Would you believe that triangle $ABH$ is isosceles?

DPatrick 2015-02-26 19:15:58
It looks "obvious" from the picture...but you can see it more rigorously too if you note that it's similar to triangle $CHI$, which is isosceles by symmetry.

ingridzhang97 2015-02-26 19:16:22
its a 36 72 72 triangle

zheyu1 2015-02-26 19:16:22
both 72 degree base angles

mathperson9 2015-02-26 19:16:22
You can solve for the angles

DPatrick 2015-02-26 19:16:30
So since $AH = AB$, we have $1+s = r.$

DPatrick 2015-02-26 19:16:54
And as many of you suggested earlier, there are lots of other ways to relate $r$ and $s$ too.

BFYSharks 2015-02-26 19:17:03
Then similar triangles using AFG and ADC

akim99 2015-02-26 19:17:09
AF:AD = FG

C

DPatrick 2015-02-26 19:17:23
I like this too: I found it easiest to look at similar triangles $AFG$ and $ADC$.

DPatrick 2015-02-26 19:17:29

DPatrick 2015-02-26 19:17:44
$AFG$ has base $s$ and legs of length 1.
$ADC$ has base $r$ and legs of length $2+s.$

DPatrick 2015-02-26 19:17:57
So $s = \dfrac{r}{2+s},$ and hence $r = s(2+s).$

TheMaskedMagician 2015-02-26 19:18:13

DPatrick 2015-02-26 19:18:21
Exactly: earlier we had $r = 1+s$, so now we've got the equation $1+s = s(s+2)$ just in terms of $s.$

DPatrick 2015-02-26 19:18:31
This simplifies to $s^2 + s - 1 = 0.$ So what is $s$?

ninjataco 2015-02-26 19:18:53

_--__--_ 2015-02-26 19:18:53
quadratic formula

TheMaskedMagician 2015-02-26 19:18:53

ompatel99 2015-02-26 19:18:53
(-1+sqrt(5))/2

MathStudent2002 2015-02-26 19:18:53
$s=\frac{-1+\sqrt{5}}{2}$

DPatrick 2015-02-26 19:18:57
By the quadratic formula,
\[
s = \frac{-1 \pm \sqrt{5}}{2},
\]
and since $s>0$, we must have $s = \dfrac{\sqrt{5} - 1}{2}.$

DPatrick 2015-02-26 19:19:08
How do we finish from here?

abean077 2015-02-26 19:19:32
figure out JH

Not_a_Username 2015-02-26 19:19:47
JH=1

123456789agarwal 2015-02-26 19:19:47
JH =1

ScottBusche 2015-02-26 19:19:47
Use ADC and AJH to find JH.

DPatrick 2015-02-26 19:19:51
Right. We want $FG + JH + CD$, so we can use the three similar triangles shown in red below:

DPatrick 2015-02-26 19:19:57

DPatrick 2015-02-26 19:20:10
The top triangle $AFG$ has legs of length 1 and a base of length $s$.

AMN300 2015-02-26 19:20:20
similar triangles, s*(s+1)

DPatrick 2015-02-26 19:20:28
Right, $AJH$ has legs of length $1+s$, so by similarity its base is $s(1+s)$.

vstar99 2015-02-26 19:20:57
so s+s^2

DPatrick 2015-02-26 19:21:01
Yes, and at this point, you might recall that $s^2 + s - 1 = 0$, so $s(1+s)$ is just 1. That is, $JH = 1.$

mihirb 2015-02-26 19:21:32
now for DC

DPatrick 2015-02-26 19:21:43
And we also remember that $DC = r = 1+s.$

DPatrick 2015-02-26 19:22:05
(Again, there are so many different ways in this problem that you could get these quantities. I'm just doing what seemed simplest to me.)

wolfpencil 2015-02-26 19:22:13
1+2s+1

DPatrick 2015-02-26 19:22:16
Therefore,
\[
FG + JH + CD = s + 1 + (1+s) = 2s + 2 = 2(s+1).
\]

Darn 2015-02-26 19:22:36
So add and get ;$1+\sqrt{5}\rightarrow\boxed{\textbf{D}}$

WalkerTesla 2015-02-26 19:22:36
Which is equal to $1+\sqrt{5}$ or $D$

DPatrick 2015-02-26 19:22:40
Now we compute:
\[
2(s+1) = 2\left(\frac{\sqrt5 - 1}{2} + 1\right) = (\sqrt5 - 1) + 2 = \sqrt5 + 1,
\]
answer (D).

DPatrick 2015-02-26 19:23:13
On to 10B #23:

DPatrick 2015-02-26 19:23:20
23. Let $n$ be a positive integer greater than 4 such that the decimal representation of $n!$ ends in $k$ zeros and the decimal representation of $(2n)!$ ends in $3k$ zeros. Let $s$ denote the sum of the four least possible values of $n$. What is the sum of the digits of $s$?
$\phantom{10B:23}$
$\text{(A) } 7 \quad \text{(B) } 8 \quad \text{(C) } 9 \quad \text{(D) } 10 \quad \text{(E) } 11$

DPatrick 2015-02-26 19:23:41
How do we determine the number of zeros at the end of $n!$?

Deathranger999 2015-02-26 19:23:56
It's a function of how many fives are in each factorial.

ktbroborg 2015-02-26 19:23:56
number of 0s is number of pairs of 2s and 5s

nosyarg 2015-02-26 19:23:56
look at 5's and 2's

donot 2015-02-26 19:23:56
Multiples of 5

AMN300 2015-02-26 19:23:56
number of 5's

TheMaskedMagician 2015-02-26 19:23:56
Number of factors of $5$

willalphagamma 2015-02-26 19:23:56
The factors of 5

IDontWannaPlay 2015-02-26 19:23:56
see how many factors of 10 there are

WhaleVomit 2015-02-26 19:23:56
how many factors of 5 it has

hibiscus 2015-02-26 19:23:56
number of factors of 5

akim99 2015-02-26 19:23:56
The number of 5s.

Benaloo26 2015-02-26 19:23:56
Factors of 5

DPatrick 2015-02-26 19:24:02
We need to count the number of factors of 10 in $n!$.

DPatrick 2015-02-26 19:24:07
But 10's are made of up 2's and 5's, and $n!$ will have lots more 2's than 5's.

DPatrick 2015-02-26 19:24:14
So the number of zeros at the end of $n!$ is just the number of factors of 5 of $n!$.

anandiyer12 2015-02-26 19:24:31
start off with small numbers?

flyrain 2015-02-26 19:24:31
you can just grind it out and find the 4 numbers

fractal161 2015-02-26 19:24:31
Organized casework.

DPatrick 2015-02-26 19:24:40
Sure, one idea is to just experiment a little bit.

DPatrick 2015-02-26 19:24:47
For example: for what values of $n$ does $n!$ have 1 zero at the end?

math-rules 2015-02-26 19:25:27
5-9

nosaj 2015-02-26 19:25:27
5-9

BFYSharks 2015-02-26 19:25:27
5-9

fluffyanimal 2015-02-26 19:25:27
5,6,7,8,9

_--__--_ 2015-02-26 19:25:27
5,6,7,8,9

DrMath 2015-02-26 19:25:27
5,6,7,8,9

tdeng 2015-02-26 19:25:27
5-9

ingridzhang97 2015-02-26 19:25:27
5-9

acegikmoqsuwy2000 2015-02-26 19:25:27
5,6,7,8,9

bluephoenix 2015-02-26 19:25:27
5,6,7,8,9

geogirl08 2015-02-26 19:25:27
5, 6, 7, 8, 9

DPatrick 2015-02-26 19:25:48
Right: $n = 5,6,7,8,9$ have only 1 factor of 5.

DPatrick 2015-02-26 19:25:55
For any of these, does $(2n)!$ have 3 zeros at the end?

FrozenFire31415 2015-02-26 19:26:22
8 and 9

baseballlogic 2015-02-26 19:26:22
8, 9

abishek99 2015-02-26 19:26:22
8 and 9

rjiang16 2015-02-26 19:26:22
8,9

antler 2015-02-26 19:26:22
only 8 and 9

ninjataco 2015-02-26 19:26:22
8 and 9

DPatrick 2015-02-26 19:26:33
Note $(2n)!$ will have 3 zeros if and only if $2n = 15,16,17,18,19$.

DPatrick 2015-02-26 19:26:40
So $n=8,\,2n=16$ works, and $n=9,\,2n=18$ works.

DPatrick 2015-02-26 19:26:49
That's two values! We only need two more!

DPatrick 2015-02-26 19:26:58
Moving on...for what values of $n$ does $n!$ have 2 zeros at the end?

aadavi 2015-02-26 19:27:21
10-14

yamyamx2 2015-02-26 19:27:21
10-14

bengals 2015-02-26 19:27:21
10-14

gradysocool 2015-02-26 19:27:21
10-14

Turtwig123 2015-02-26 19:27:21
10-14

123456789agarwal 2015-02-26 19:27:21
10-14

tau172 2015-02-26 19:27:21
10 to 14

speck 2015-02-26 19:27:21
10,11,12,13,14

LOTRFan123 2015-02-26 19:27:21
10 -14

az_phx_brandon_jiang 2015-02-26 19:27:21
10-14

DPatrick 2015-02-26 19:27:25
$n=10,11,12,13,14$

DPatrick 2015-02-26 19:27:29
For which (if any) of these does $(2n)!$ have 6 zeros?

Shark77 2015-02-26 19:28:00
13, 14

kunsun 2015-02-26 19:28:00
13 and 14, because 10-14 has 2 zeroes and 25-29 has 6 zeroes

Not_a_Username 2015-02-26 19:28:00
13, 14

problemsolver12 2015-02-26 19:28:00
13, 14

Tuxianeer 2015-02-26 19:28:00
13,14,

Mathlete55 2015-02-26 19:28:00
13, 14

lucylai 2015-02-26 19:28:00
13,14

AlcumusGuy 2015-02-26 19:28:00
13 and 14

abean077 2015-02-26 19:28:00
13 and 14

DPatrick 2015-02-26 19:28:04
Note that $2n=25,26,27,28,29$ each satisfies $(2n)!$ with 6 zeros, because the factor of 25 gives an "extra" factor of 5.

DPatrick 2015-02-26 19:28:13
So $n=13,\,2n=26$ works, as does $n=14,\,2n=28$.

DPatrick 2015-02-26 19:28:21
But now we're done! We've found the 4 smallest values of $n$: 8, 9, 13, and 14.

Darn 2015-02-26 19:28:52
Thus our answer is $8+9+13+14=44\rightarrow\boxed{\textbf{(B) }8}$

wolfpencil 2015-02-26 19:28:52
add them up, then find the sum of the digits

Modest_Ked 2015-02-26 19:28:52
The answer is 8

akim99 2015-02-26 19:28:52
8+9+13+14 = 44 ==> 8, so (B).

pad 2015-02-26 19:28:52
s=8+9+13+14=44

DPatrick 2015-02-26 19:28:57
Their sum is $s=8+9+13+14 = 44,$ and the sum of the digits of $s$ is $4+4 = 8.$ Answer (B).

Turtwig123 2015-02-26 19:29:05
are these the only values?

DPatrick 2015-02-26 19:29:26
That's an interesting question. (Of course, on the AMC contest itself, you wouldn't waste time thinking about it.)

DPatrick 2015-02-26 19:29:46
It turns out these are the only 4 values that work!

DPatrick 2015-02-26 19:29:52
Let's quickly digress and see why.

DPatrick 2015-02-26 19:30:07
Given a value $n$, what's a formula for the number of factors of 5 in $n!$ (and thus the number of zeros at the end of $n!$)?

AMN300 2015-02-26 19:30:29
floor(n/5)+floor(n/25)+...

speck 2015-02-26 19:30:29
floor(n/5)+floor(n/25)+...

akim99 2015-02-26 19:30:29
Floor function of n/5 + floor of n/25 + ...

_--__--_ 2015-02-26 19:30:29
floor(n/5) + floor(n/25) + floor(n/125) + ....

DPatrick 2015-02-26 19:30:36
Right. If $n!$ ends in $k$ zeros, then
\[
k
= \left\lfloor\frac{n}{5}\right\rfloor
+ \left\lfloor\frac{n}{5^2}\right\rfloor
+ \left\lfloor\frac{n}{5^3}\right\rfloor
+ \cdots.
\]

DPatrick 2015-02-26 19:30:54
That symbol is the floor function, which (for positive numbers) means round DOWN to the nearest integer.

DPatrick 2015-02-26 19:31:02
The first term counts the number of terms in $n!$ that are multiples of 5.
The second term counts the number of terms in $n!$ that are multiples of $5^2$ (and thus contribute an extra factor of 5).
The third term counts the number of terms in $n!$ that are multiples of $5^3$ (and thus contribute an extra factor of 5 beyond the two already counted).
And so on.

DPatrick 2015-02-26 19:31:20
So if $(2n)!$ has $3k$ zeros, we must have
\[
3k
= \left\lfloor\frac{2n}{5}\right\rfloor
+ \left\lfloor\frac{2n}{5^2}\right\rfloor
+ \left\lfloor\frac{2n}{5^3}\right\rfloor
+ \cdots.
\]

DPatrick 2015-02-26 19:31:31
But how is $\left\lfloor\dfrac{n}{5^j}\right\rfloor$ related to $\left\lfloor\dfrac{2n}{5^j}\right\rfloor$ for some $j$?

DPatrick 2015-02-26 19:31:37
More generally, how are $\lfloor x \rfloor$ and $\lfloor 2x \rfloor$ related for any positive $x$?

WalkerTesla 2015-02-26 19:32:00
The second is double the first or $1$ more than double the first

hibiscus 2015-02-26 19:32:00
It depends on the fractional part of x.

flyrain 2015-02-26 19:32:04
they're at most 1 apart

DPatrick 2015-02-26 19:32:10
Right. If the fractional part of $x$ is less than 0.5, then $\lfloor 2x \rfloor = 2\lfloor x \rfloor$. (For example, $\lfloor 3.2 \rfloor = 3$ and doubling it gives $\lfloor 6.4 \rfloor = 6$.)

DPatrick 2015-02-26 19:32:17
If the fractional part of $x$ is at least 0.5, then $\lfloor 2x \rfloor = 2\lfloor x \rfloor + 1$. (For example, $\lfloor 3.8 \rfloor = 3$ and doubling it gives $\lfloor 7.6 \rfloor = 7$.)

nosaj 2015-02-26 19:32:28
either floor(2x)=2floor(x) of floor(2x)=2floor(x)+1

DPatrick 2015-02-26 19:32:37
Right, or to say it another way, $\lfloor 2x \rfloor = 2\lfloor x \rfloor + \epsilon$, where $\epsilon$ can be 0 or 1.

DPatrick 2015-02-26 19:32:48
So when we go from $n!$ to $(2n)!$, each floor term will double the number of zeros, and possibly add 1 more zero. Plus we might get one extra zero at the end if $2n$ jumps across the next power of 5 (like our example $n=13,\,2n=26$ from before.)

DrMath 2015-02-26 19:33:14
well we need to add that $1$ exactly $k$ times

DPatrick 2015-02-26 19:33:20
Right. In addition to the $2k$ zeros that we're guaranteed to have, we might have an additional $m+1$ zeros, where $m$ satisfies $5^m \le n < 5^{m+1}.$

DPatrick 2015-02-26 19:33:28
(If you know logarithms, then you know that $m = \lfloor \log_5 n \rfloor.$)

DPatrick 2015-02-26 19:33:40
But to get $3k$ zeros total, we need to have $k$ of these ``extra'' zeroes, so we need $k \le m+1.$

DPatrick 2015-02-26 19:34:17
On the other hand, we know that if $5^m \le n$, so we have
\[
k
\ge \left\lfloor\frac{5^m}{5}\right\rfloor
+ \left\lfloor\frac{5^m}{5^2}\right\rfloor
+ \left\lfloor\frac{5^m}{5^3}\right\rfloor
+ \cdots
+ \left\lfloor\frac{5^m}{5^m}\right\rfloor
= 5^{(m-1)+(m-2)+\cdots+1} = 5^{\left(\frac{m(m-1)}{2}\right)}.
\]

DPatrick 2015-02-26 19:34:56
That's a big number. It's really hard for that number to be less than $m+1$.

DPatrick 2015-02-26 19:35:08
Specifically, we must have
\[
5^\left(\frac{m(m-1)}{2}\right) \le k \le m+1.
\]

DPatrick 2015-02-26 19:35:34
Since $n>4$, we have $m>0$, and this only holds for $m=1$. (The powers-of-5 term is already bigger than $m+1$ when $m=2$, and it grows much much faster than $m+1$ as $m$ grows!)

DPatrick 2015-02-26 19:35:55
The upshot is: all this algebra proves that we only need to look at $n$ satisfying $5 \le n < 25,$ and then trial-and-error finds just the 4 solutions that we found using our original experimental method. It turns out that not only are 8,9,13,14 the four smallest values of $n$ that work, they're the only values of $n$ that work!

swirlykick 2015-02-26 19:36:06
Would if have been better to say find the sum of all n? finding the 4 values of n didnt seem very hard

DPatrick 2015-02-26 19:36:28
I think "finding them all" might have been too hard for an AMC 10. I mentioned logarithms above, for example, which is beyond the scope of the AMC 10.

DPatrick 2015-02-26 19:36:49
Anyway...on to 10B #24:

DPatrick 2015-02-26 19:36:56
24. Aaron the ant walks on the coordinate plane according to the following rules. He starts at the origin $p_0=(0,0)$ facing to the east and walks one unit, arriving at $p_1=(1,0).$ For $n=1,2,3,\ldots,$ right after arriving at the point $p_n$, if Aaron can turn $90^\circ$ left and walk one unit to an unvisited point $p_{n+1},$ he does that. Otherwise, he walks one unit straight ahead to reach $p_{n+1}$. Thus the sequence of points continues $p_2=(1,1),$ $p_3=(0,1),$ $p_4=(-1,1),$ $p_5=(-1,0),$ and so on in a counterclockwise spiral pattern. What is $p_{2015}?$
$\phantom{10B:24}$
$\text{(A) } (-22,-13) \quad \text{(B) } (-13,-22) \quad \text{(C) } (-13,22) \quad \text{(D) } (13,-22) \quad \text{(E) } (22,-13)$

PresidentJL 2015-02-26 19:37:18
Draw a diagram

mihirb 2015-02-26 19:37:18
try to find a pattern

AlcumusGuy 2015-02-26 19:37:18
Find a pattern

Not_a_Username 2015-02-26 19:37:18
Draw it out and find a pattern

ompatel99 2015-02-26 19:37:18
Picture!

bengals 2015-02-26 19:37:18
Draw a picture

Darn 2015-02-26 19:37:18
Write out the first few

15Pandabears 2015-02-26 19:37:18
pattern

mjoshi 2015-02-26 19:37:18
draw out a first few

DPatrick 2015-02-26 19:37:22
This is mainly a bookkeeping problem. We just need to keep careful track of where Aaron is after $n$ moves.

DPatrick 2015-02-26 19:37:30
Here's a picture of his path:

DPatrick 2015-02-26 19:37:35

DPatrick 2015-02-26 19:38:11
Probably a good idea is to try to quantify when he gets to the corners somehow.

_--__--_ 2015-02-26 19:38:18
Distance traveled is 1+1+2+2+3+3+4+4+...

abean077 2015-02-26 19:38:24
Notice the first two line segments are 1, then the next two are 2, and so on

DPatrick 2015-02-26 19:38:34
Yes, I found that to be a really helpful observation!

DPatrick 2015-02-26 19:38:38
He starts by taking 1 step E and 1 step N, ending at (1,1) after 2 steps and about to head W.

DPatrick 2015-02-26 19:38:44
Then he takes 2 steps W and 2 steps S, ending at (-1,-1) after 2+4 steps, and about to head E.

DPatrick 2015-02-26 19:38:51
Then he takes 3 steps E and 3 steps N, ending at (2,2) after 2+4+6 steps, and about to head W.

DPatrick 2015-02-26 19:39:01
Then he takes 4 steps W and 4 steps S, ending at (-2,-2) after 2+4+6+8 steps, and about to head E.

DPatrick 2015-02-26 19:39:06
Is there a convenient formula for the number of steps to some general point?

Deathranger999 2015-02-26 19:39:13
So triangular numbers, times 2!

RBMS14 2015-02-26 19:39:33
triangle numbers

DPatrick 2015-02-26 19:40:01
So there are a lot of choices we could make here for which point to compute.

DPatrick 2015-02-26 19:40:17
But the ones listed above all involve 2 times the triangular numbers 1+2+...+n

BFYSharks 2015-02-26 19:40:24
Bottom left corner! (-n, -n)

DPatrick 2015-02-26 19:40:42
That's what I did. We can notice that for any integer $k \ge 1,$ he's at $(-k,-k)$ after $2+4+6+\cdots+4k$ steps, and about to head E. This sum is easy to compute in terms of $k$.

DPatrick 2015-02-26 19:41:19
There are $2k$ terms in the sum, with an average value of $(2+4k)/2 = 2k+1$, so
\[
2+4+6+\cdots+4k = 2k(2k+1).
\]

blueduck1 2015-02-26 19:41:40
So P(1980) is -22,-22

mattpi 2015-02-26 19:41:40
the 1980th position is at -22, -22, so the 2015th position is 13, -22 (D)

DPatrick 2015-02-26 19:41:54
Exactly. $k=22$ gives $44(45) = 1980 < 2015.$

DPatrick 2015-02-26 19:42:00
Thus, after 1980 moves, Aaron is at (-22,-22) and is heading east.

DPatrick 2015-02-26 19:42:12
(The answer choices are also a big clue to look at $k=22$!)

AMN300 2015-02-26 19:42:35
so (13, -22) D

DPatrick 2015-02-26 19:42:37
He's got 2015 - 1980 = 35 moves to go. He won't turn again until he hits (23,-22). So he moves 35 spaces east, ending up at (-22+35,-22) = (13,-22). Answer (D).

DPatrick 2015-02-26 19:43:17
There are lots of other pattens you could have used in this problem probably just as easily.

DPatrick 2015-02-26 19:43:42
But in general I don't find these "bookkeeping" sort of problems all that interesting, so let's move on...

DPatrick 2015-02-26 19:43:50
Next is 10B #25, which was also 12B #23:

DPatrick 2015-02-26 19:43:54
25. A rectangular box measures $a\times b\times c$, where $a,$ $b,$ and $c$ are integers and $1\leq a\leq b\leq c.$ The volume and the surface area of the box are numerically equal. How many ordered triples $(a,b,c)$ are possible?
$\phantom{10B:25}$
$\text{(A) } 4 \quad \text{(B) } 10 \quad \text{(C) } 12 \quad \text{(D) } 21 \quad \text{(E) } 26$

acegikmoqsuwy2000 2015-02-26 19:44:23
$abc=2(ab+bc+ac)$

Darn 2015-02-26 19:44:23
Write this as $abc=2(ab+bc+ac)$

Not_a_Username 2015-02-26 19:44:23
Set up an equation: $abc=2ab+2ac+2bc$

Deathranger999 2015-02-26 19:44:23
abc = 2(ab + bc + ca)

ompatel99 2015-02-26 19:44:23
Volume=abc=SA=2(ab+bc+ac)

AlcumusGuy 2015-02-26 19:44:23
Write an equation: abc = 2(ab + bc + ca)

_--__--_ 2015-02-26 19:44:23
abc = 2(ab + ac + bc)

DrMath 2015-02-26 19:44:23
$abc=2(ab+bc+ca)$

bluephoenix 2015-02-26 19:44:23
abc = 2ab+2bc+2ac

mathtastic 2015-02-26 19:44:23
$abc=2ab+2bc+2ca$

droid347 2015-02-26 19:44:23
We know $abc=2(ab+bc+ca)$ by formulas for volume and surface area.

DPatrick 2015-02-26 19:44:27
The volume of the box is $abc.$

DPatrick 2015-02-26 19:44:31
The box has two sides with area $ab$, two sides with area $ac$, and two sides with area $bc.$

DPatrick 2015-02-26 19:44:38
Thus we're trying to solve
\[
abc = 2(ab+ac+bc).
\]

DPatrick 2015-02-26 19:44:45
How can we work with this equation?

AMN300 2015-02-26 19:44:57
Divide both sides by 2abc to simplify a little

Not_a_Username 2015-02-26 19:44:57
Divide both sides by $2abc$

nosaj 2015-02-26 19:44:57
We must have $abc=2ab+2ac+2bc$. Dividing both sides of the equation by $2abc$ yields the (simple) equation $\frac 1a + \frac 1b + \frac 1c=\frac 12$.

bestwillcui1 2015-02-26 19:44:57
divide through by 2abc

TheMaskedMagician 2015-02-26 19:44:57

numbersandnumbers 2015-02-26 19:44:57
divide by 2abc

DPatrick 2015-02-26 19:45:03
One idea is to make the terms simpler, by dividing through by $2abc.$ This gives
\[
\frac12 = \frac1a + \frac1b + \frac1c.
\]

DPatrick 2015-02-26 19:45:07
What does this tell us?

ingridzhang97 2015-02-26 19:45:32
a,b,c>2

rjiang16 2015-02-26 19:45:32
we know a, b, c all greater than 2

vinayak-kumar 2015-02-26 19:45:32
A>2

DPatrick 2015-02-26 19:45:36
This tells us right away that $a,b,c \ge 3$. Otherwise, if any of the denominators are 1 or 2, then the right side is too big.

jigglypuff 2015-02-26 19:45:58
a is between 3 and 6

Darn 2015-02-26 19:45:58
$2\leq a \leq 6$

AlcumusGuy 2015-02-26 19:45:58
a <= 6, as otherwise we cannot sum to 1/2

amburger66 2015-02-26 19:45:58
a,b,c are less than or equal to 6

DrMath 2015-02-26 19:45:58
well the greatest value among $1/a, 1/b, 1/c$ is $\ge 1/6$

AllenWang314 2015-02-26 19:46:03
$a,b,c\le 6$

DPatrick 2015-02-26 19:46:10
Also remember that $a \le b \le c$, so that $\frac1a \ge \frac1b \ge \frac1c.$

DPatrick 2015-02-26 19:46:13
So this also tells us that $a \le 6$. Otherwise (if $a>6$) all the right-side terms are less than $\frac16$, so the right side is too small.

DPatrick 2015-02-26 19:46:33
And what happens if $a=6$?

bluehall90 2015-02-26 19:47:00
b=6 and c=6

acegikmoqsuwy2000 2015-02-26 19:47:00
$b=c=6$

bfea 2015-02-26 19:47:00
b and c are 6

stealbob 2015-02-26 19:47:00
b adn c are 6 as well

ninjataco 2015-02-26 19:47:00
a=b=c=6

MathLearner01 2015-02-26 19:47:00
a=b=c=6

Benaloo26 2015-02-26 19:47:00
b and c = 6

abean077 2015-02-26 19:47:00
b = 6 and c = 6

Imagineny1 2015-02-26 19:47:00
b=c=6

DPatrick 2015-02-26 19:47:06
It also tells us that the only solution with $a=6$ is $a=b=c=6$: all three terms on the right side are $\frac16$. That's 1 solution.

DPatrick 2015-02-26 19:47:22
So we have three cases left: $a=3$, $a=4$, or $a=5$. I don't know of a slick way to solve this except to do the cases one by one.

Darn 2015-02-26 19:47:44
Casework!

DPatrick 2015-02-26 19:47:52
If $a=3$, we get
\[
\frac{1}{6} = \frac{1}{b} + \frac{1}{c}
\]
with the restriction that $3 \le b \le c.$

bluehall90 2015-02-26 19:48:25
And b =< 12

fluffyanimal 2015-02-26 19:48:25
6<b<=12

abean077 2015-02-26 19:48:25
b = c = 12 is one way

DPatrick 2015-02-26 19:48:28
We could use the same idea: use bounding to show that we must have $6 \le b \le 12$, and see which of these work.

DPatrick 2015-02-26 19:48:36
But there's a more clever way. Let's go back to the original form of the equation and substitute in $a=3$:
\[
3bc = 2(3b+3c+bc).
\]

pnpiano 2015-02-26 19:49:04
we can rearrange and then use simon's favorite factoring trick

pnpiano 2015-02-26 19:49:04
SFFT right?

BFYSharks 2015-02-26 19:49:04
SFFT

NumberNinja 2015-02-26 19:49:04
SFFT

az_phx_brandon_jiang 2015-02-26 19:49:04
expand and factor

DPatrick 2015-02-26 19:49:09
This simplifies to $bc - 6b - 6c = 0.$

DPatrick 2015-02-26 19:49:15
And now we use Simon's Favorite Factoring Trick!

DPatrick 2015-02-26 19:49:24
If you haven't seen SFFT before, the idea is to add a constant to both sides so that one side can factor. What constant should we add?

SHARKYBOY 2015-02-26 19:49:45
36

geogirl08 2015-02-26 19:49:45
36

15Pandabears 2015-02-26 19:49:45
36

eswa2000 2015-02-26 19:49:45
36

dli00105 2015-02-26 19:49:45
36

mathboxboro 2015-02-26 19:49:45
36

DPatrick 2015-02-26 19:49:50
I'd add 36, to give $bc - 6b - 6c + 36 = 36.$ Now it factors as
\[
(b-6)(c-6) = 36.
\]
What are the integer solutions to this?

donot 2015-02-26 19:50:36
we factor bash

jam10307 2015-02-26 19:50:36
factors of 36

jigglypuff 2015-02-26 19:50:36
1*36,2*18,3*12,4*9,6*6 and add 6 to all of them

DPatrick 2015-02-26 19:50:41
Right. 36 can be factored 5 ways: $6 \cdot 6$, $4 \cdot 9$, $3 \cdot 12$, $2 \cdot 18$, $1 \cdot 36.$

DPatrick 2015-02-26 19:50:58
Since $b \le c$, all of these must assign $b-6$ to the smaller factor, and $c-6$ to the larger factor.

DPatrick 2015-02-26 19:51:06
Adding 6 to each of these gives a solution to $(b,c)$ for $a=3$: (12,12), (10,15), (9,18), (8,24), (7,42).

DPatrick 2015-02-26 19:51:13
Note that all of these have $3 \le b \le c$, so they're all valid solutions.

mathcountsprep 2015-02-26 19:51:24
so 5 ways

Not_a_Username 2015-02-26 19:51:24
5 triples.

SimonSun 2015-02-26 19:51:24
5 soltions

DPatrick 2015-02-26 19:51:27
So that's 5 more solutions. We're up to 6 solutions total.

ryansong2002 2015-02-26 19:51:39
now on the a=4

rjiang16 2015-02-26 19:51:39
a=4 case

ninjataco 2015-02-26 19:51:39
now onto a=4!

DPatrick 2015-02-26 19:51:41
Now on to $a=4$.

DPatrick 2015-02-26 19:51:47
We again plug this into the original
\[
4bc = 2(4b+4c+bc).
\]

DPatrick 2015-02-26 19:51:56
This simplifies to $2bc - 8b - 8c = 0$.
Divide by 2 to get $bc - 4b - 4c = 0$.

problemsolver12 2015-02-26 19:52:11
SFFT again

pad 2015-02-26 19:52:11
sfft!

Zynaria 2015-02-26 19:52:11
16

bluephoenix 2015-02-26 19:52:11
add 16

fractal161 2015-02-26 19:52:11
SFFT with 16

AMN300 2015-02-26 19:52:11
add 16 to both sides

mathymath 2015-02-26 19:52:11
sfft again

pisgood 2015-02-26 19:52:11
Add by 16

hibiscus 2015-02-26 19:52:11
(b-4)(c-4)=16

ingridzhang97 2015-02-26 19:52:11
(b-4)(c-4)=16

DPatrick 2015-02-26 19:52:29
Good -- we SFFT again! Add 16 to both sides:
\[
bc - 4b - 4c + 16 = 16.
\]
This factors to $(b-4)(c-4) = 16.$

DPatrick 2015-02-26 19:52:39
What are the positive integer solutions to this?

_--__--_ 2015-02-26 19:52:57
(5,20),(6,12),(8,8)

bestwillcui1 2015-02-26 19:52:57
so just 3

flyrain 2015-02-26 19:52:57
and some 3 or so ways from this

rjiang16 2015-02-26 19:52:57
(4,5,20)(4,6,12)(4,8,8)

Darn 2015-02-26 19:52:57
Notice that we can factor $16$ in 3 distinct ways

hibiscus 2015-02-26 19:52:57
(1, 16) (2, 8) (4, 4)

Jyzhang12 2015-02-26 19:52:57
factor 16 and add 4

jigglypuff 2015-02-26 19:52:57
1*16,2*8,4*4 and add 4 to all of them

DPatrick 2015-02-26 19:53:03
16 can be factored 3 ways: $4 \cdot 4$, $2 \cdot 8$, $1 \cdot 16.$

DPatrick 2015-02-26 19:53:08
Adding 4 to each of these gives a solution to $(b,c)$ for $a=4$: (8,8), (6,12), (5,20).

mathperson9 2015-02-26 19:53:27
All are valid answers

rjiang16 2015-02-26 19:53:30
so we have 9 ways total now

DPatrick 2015-02-26 19:53:32
All of these have $4 \le b \le c,$ so they're all legal. That's 3 more solutions. We're up to 9 solutions total.

song2sons 2015-02-26 19:53:42
now a=5

mihirb 2015-02-26 19:53:42
now for a = 5

DPatrick 2015-02-26 19:53:47
Finally we have $a=5$:
\[
5bc = 2(5b+5c+bc).
\]

DPatrick 2015-02-26 19:53:52
This simplifies to $3bc - 10b - 10c = 0.$

crosby87 2015-02-26 19:54:02
SFFT

problemsolver12 2015-02-26 19:54:08
SFFT once more

rjiang16 2015-02-26 19:54:08
harder to SFFT...

DPatrick 2015-02-26 19:54:14
We'd like to use SFFT again, but there doesn't seem to be a nice integer we can add to make it factor.

nosaj 2015-02-26 19:54:28
Multiply by 3

jigglypuff 2015-02-26 19:54:28
multiply by 3 then sfft

nosaj 2015-02-26 19:54:28
9bc-30b-30c=0

jigglypuff 2015-02-26 19:54:28
multiply by 3 then simons

DPatrick 2015-02-26 19:54:35
Right. The problem is that $3bc$ term. Since everything is symmetric in $b$ and $c$, we'd like to make that $3bc$ term factor symmetrically too. But the only way to factor it using integer coefficients is $3b \cdot c$ or $b \cdot 3c$, and that's not symmetric.

DPatrick 2015-02-26 19:54:44
But we can multiply through by 3:
\[
9bc - 30b - 30c = 0.
\]

DPatrick 2015-02-26 19:54:57
Now $9bc = 3b \cdot 3c$ is a symmetric factorization. Can we apply SFFT?

pnpiano 2015-02-26 19:55:21
100

Deathranger999 2015-02-26 19:55:21
Add 100.

pnpiano 2015-02-26 19:55:21
add 100

droid347 2015-02-26 19:55:21
yes, add 100

abean077 2015-02-26 19:55:21
(3b - 10)(3c - 10)

jameswangisb 2015-02-26 19:55:21
(3b-10)(3c-10)=100

DPatrick 2015-02-26 19:55:26
Yes! Add 100:
\[
9bc - 30b - 30c + 100 = 100.
\]
And now we can factor: $(3b-10)(3c-10) = 100.$

DPatrick 2015-02-26 19:55:36
What are the integer solutions to this?

TheMagician 2015-02-26 19:56:06
b=5 and c=10

Darn 2015-02-26 19:56:06
There is only 1 way that satisfies our bounds: $(5,10)$

rjiang16 2015-02-26 19:56:06
(5,10) and that's it?

ZekromReshiram 2015-02-26 19:56:06
5 ways to factor 100

RegretDragunity 2015-02-26 19:56:06
1*100 5*20 2*50 4*25 10*10

MathLearner01 2015-02-26 19:56:06
(5,10)

Not_a_Username 2015-02-26 19:56:06
$(b, c)=(5, 10)$

DPatrick 2015-02-26 19:56:11
100 can be factored 5 ways: $10 \cdot 10$, $5 \cdot 20$, $4 \cdot 25$, $2 \cdot 50$, and $1 \cdot 100$.

DPatrick 2015-02-26 19:56:35
But we need pairs of factors that are each 10 less than a multiple of 3. So only some of these work.

DPatrick 2015-02-26 19:56:42
$5 \cdot 20$ works, giving $b = 5$ and $c = 10$.

DPatrick 2015-02-26 19:56:56
$2 \cdot 50$ also works, giving $b = 4$ and $c = 20$.

nosaj 2015-02-26 19:57:10
4, 20 doesn't work because 4<5

meteor88 2015-02-26 19:57:13
but b is less than a

abean077 2015-02-26 19:57:13
But 4 is too low, it has to be at least a

DPatrick 2015-02-26 19:57:16
...but wait, that second one doesn't actually work, since we need $5 \le b \le c.$

DPatrick 2015-02-26 19:57:30
So there's only 1 more solution in the $a=5$ case.

RBMS14 2015-02-26 19:57:39
The answer is 10

crosby87 2015-02-26 19:57:39
10 solutions therefore $B$

problemsolver12 2015-02-26 19:57:39
10 total then?

nosaj 2015-02-26 19:57:39
Therefore answer is (B) 10.

mathcountsprep 2015-02-26 19:57:39
9+1=10 (B)

ryansong2002 2015-02-26 19:57:39
so only 10?

DPatrick 2015-02-26 19:57:42
That brings our total up to 10 overall across all the cases. Answer (B).

DPatrick 2015-02-26 19:58:07
OK, I'm going to take a 4 minute break. We'll resume at :02 past the hour!

DPatrick 2015-02-26 20:02:02
Hi, we're back!

DPatrick 2015-02-26 20:02:15
We did 12B #21 (it was also 10B #21), so on to 12B #22:

DPatrick 2015-02-26 20:02:20
22. Six chairs are evenly spaced around a circular table. One person is seated in each chair. Each person gets up and sits down in a chair that is not the same chair and is not adjacent to the chair he or she originally occupied, so that again one person is seated in each chair. In how many ways can this be done?
$\phantom{12B:22}$
$\text{(A) } 14 \quad \text{(B) } 16 \quad \text{(C) } 18 \quad \text{(D) } 20 \quad \text{(E) } 24$

trumpeter 2015-02-26 20:02:44
casework!

rjiang16 2015-02-26 20:02:44
You can draw a tree map and count all 20 cases that work xD

nosyarg 2015-02-26 20:02:44
cases?

abean077 2015-02-26 20:02:54
This looks like a caswework problem

Satyaprakash2009rta 2015-02-26 20:02:54
cases

DPatrick 2015-02-26 20:02:56
You can solve this directly by careful casework, as I did when I first attempted the problem, but it's a little messy.

DPatrick 2015-02-26 20:03:07
Afterwards, my colleague Jeremy Copeland showed me a clever trick to make the problem a lot simpler.

DPatrick 2015-02-26 20:03:21
You can recast the problem as a 2-step process:
First, everyone gets up and sits in the chair immediately opposite them.
Second, everyone either stays put, or moves to the chair to their immediate left or right.

DPatrick 2015-02-26 20:03:42
Is it clear that this is exactly the same result as the original problem?

DPatrick 2015-02-26 20:03:55
And since there are no choices in the first step, we just have to count the number of ways to do the second step.

DPatrick 2015-02-26 20:04:10
So we can recast the problem as problem 22J:

DPatrick 2015-02-26 20:04:14
22J. Six chairs are evenly spaced around a circular table. One person is seated in each chair. Some people (perhaps none) get up and sit down in an adjacent chair, so that again one person is seated in each chair. In how many ways can this be done?
$\phantom{12B:22}$
$\text{(A) } 14 \quad \text{(B) } 16 \quad \text{(C) } 18 \quad \text{(D) } 20 \quad \text{(E) } 24$

MSTang 2015-02-26 20:04:30
J for Jeremy?

DPatrick 2015-02-26 20:04:33
Of course!

DPatrick 2015-02-26 20:04:42
What are the possibilities? Are there any "obvious" possibilities that we can get out of the way right up front?

AMN300 2015-02-26 20:04:58
no one moves

willalphagamma 2015-02-26 20:04:58
Everyone stays put

Frinkahedron 2015-02-26 20:04:58
everyone stays put

Imagineny1 2015-02-26 20:04:58
all stay seated

az_phx_brandon_jiang 2015-02-26 20:04:58
No one moves again

DPatrick 2015-02-26 20:05:02
Everyone could just stay put. That's 1 possible outcome.

Tuxianeer 2015-02-26 20:05:15
everyone moves left/right

Darn 2015-02-26 20:05:15
Rotate 1 to right or left

abean077 2015-02-26 20:05:15
Everyone shifts one space (counter)clockwise

ninjataco 2015-02-26 20:05:15
everyone moves 1 right or 1 left

noobynoob 2015-02-26 20:05:15
everyone sits to the right or left

chenmeister22 2015-02-26 20:05:15
everyone moves to the left or right

ZekromReshiram 2015-02-26 20:05:15
Everyone moves to the left or right

DPatrick 2015-02-26 20:05:21
Or, everyone could move one seat to their left. That's another possible outcome.
Or, everyone could move one seat to their right. That's a third possible outcome.

DPatrick 2015-02-26 20:05:30
Otherwise, what has to happen?

mathwrath 2015-02-26 20:05:49
Pairs of switching

ompatel99 2015-02-26 20:05:49
Or one group of two adjacent could switch places

UrInvalid 2015-02-26 20:05:49
two people swap

Imagineny1 2015-02-26 20:05:49
switching seats

willalphagamma 2015-02-26 20:05:49
Some two people have to swap

DPatrick 2015-02-26 20:06:30
Right. The only cases left to count are where 1, 2, or 3 pairs of neighboring people swap seats. (And the rest stay put.)

SimonSun 2015-02-26 20:06:50
1 swap is 6 ways

DPatrick 2015-02-26 20:07:06
Right: if just one pair of people swap, there are 6 ways: just pick the pair of neighboring people to swap. (Alternatively, just pick a single person to move left; his neighbor to the left must then move right.)

DPatrick 2015-02-26 20:07:13
In how many ways can 2 pairs of people swap?

MSTang 2015-02-26 20:07:59
6 + 3 = 9

UrInvalid 2015-02-26 20:07:59
6*3/2=9 ways

DPatrick 2015-02-26 20:08:20
Right. There are 6 ways to pick a first pair (an example choice is in red below), and then 3 ways to pick a second pair (the blue choices below):

DPatrick 2015-02-26 20:08:25

DPatrick 2015-02-26 20:08:35
But this double-counts, since we could have chosen the pairs in either order. So there are (6*3)/2 = 9 possibilities in this case.

DPatrick 2015-02-26 20:09:03
Or, you could count the way to pick the two people who stay put: they are either neighbors (6 ways) or opposite each other (3 ways), for 6+3 = 9 total ways.

DPatrick 2015-02-26 20:09:12
Finally, in how many ways can 3 pairs of people swap seats?

AMN300 2015-02-26 20:09:29
2 ways for 3 pairs to swap also

SimonSun 2015-02-26 20:09:29
3 swap is 2 choices

ingridzhang97 2015-02-26 20:09:29
2 ways

sparkles257 2015-02-26 20:09:29
2 ways to do 3 people

ScottBusche 2015-02-26 20:09:29
2.

Hydroxide 2015-02-26 20:09:29
2

TheStrangeCharm 2015-02-26 20:09:29
2

az_phx_brandon_jiang 2015-02-26 20:09:29
2 ways

RegretDragunity 2015-02-26 20:09:29
2

DPatrick 2015-02-26 20:09:33
Just 2: either all the red pairs swap, or all the blue pairs swap:

DPatrick 2015-02-26 20:09:38

SimonSun 2015-02-26 20:10:01
so it is (D) 20

DPatrick 2015-02-26 20:10:04
Adding the different cases, we get 3 + 6 + 9 + 2 = 20 possibilities. Answer (D).

DPatrick 2015-02-26 20:10:58
Again, you certainly could have counted the possibilities from the original problem directly, as I did when I first solved the problem, but it's a lot messier than this nice solution Jeremy came up with!

DPatrick 2015-02-26 20:11:17
We already did 12B #23 as 10B #25, so on to 12B #24:

DPatrick 2015-02-26 20:11:20
24. Four circles, no two of which are congruent, have centers at $A$, $B$, $C$, and $D$, and points $P$ and $Q$ lie on all four circles. The radius of circle $A$ is $\dfrac58$ times the radius of circle $B,$ and the radius of circle $C$ is $\dfrac58$ times the radius of circle $D.$ Furthermore, $AB=CD=39$ and $PQ=48.$ Let $R$ be the midpoint of $\overline{PQ}.$ What is $AR+BR+CR+DR?$
$\phantom{12B:24}$
$\text{(A) } 180 \quad \text{(B) } 184 \quad \text{(C) } 188 \quad \text{(D) } 192 \quad \text{(E) } 196$

SimonSun 2015-02-26 20:11:47
draw it

nosyarg 2015-02-26 20:11:47
lets draw a picture

DPatrick 2015-02-26 20:11:50
It's hard to draw a picture of this...

DPatrick 2015-02-26 20:11:54
We don't really know where $A,$ $B,$ $C,$ and $D$ are.

DPatrick 2015-02-26 20:12:03
What do we know about them?

trumpeter 2015-02-26 20:12:22
the four are collinear

noobynoob 2015-02-26 20:12:22
collinear

ScottBusche 2015-02-26 20:12:22
They're collinear.

swirlykick 2015-02-26 20:12:27
ABCD are colinear

DPatrick 2015-02-26 20:12:37
More specifically, what line to they all lie on?

giftedbee 2015-02-26 20:13:01
The perpendicular bisector of PQ

UrInvalid 2015-02-26 20:13:01
the perpendicular bisector of PQ

celery2 2015-02-26 20:13:01
perpendicular bisector of PQ

DPatrick 2015-02-26 20:13:07
Right. If a circle passes through $P$ and $Q$, then its center has to lie on the perpendicular bisector of $\overline{PQ}.$ So $A$, $B$, $C$, $D$ all lie on this line (which also passes through $R$).

DPatrick 2015-02-26 20:13:29
From here there are some different approaches we can take. One that works nicely is to imagine this all on the coordinate plane.

DPatrick 2015-02-26 20:13:40
We probably want R to be the origin, and let's set it up so that $\overline{PQ}$ lies along the $y$-axis. Something like this:

DPatrick 2015-02-26 20:13:45

DPatrick 2015-02-26 20:14:03
Warning: this picture might not be accurate! Note that $A,$ $B,$ $C,$ and $D$ all lie on the $x$-axis, but $A$ and $B$ might not be on the same side like I drew it above.

DPatrick 2015-02-26 20:14:13
But it's a place to start.

DPatrick 2015-02-26 20:14:34
Also we notice that $A,B$ and $C,D$ have identical conditions but are themselves completely separate from each other. So let's try to solve for $A,B$ first, and hopefully we'll actually get two solutions: one will be $A,B$ and the other will be $C,D.$

DPatrick 2015-02-26 20:15:06
So if $R$ is the origin, then what are $P$ are $Q$?

UrInvalid 2015-02-26 20:15:34
(0,24)and(0,-24)

acegikmoqsuwy2000 2015-02-26 20:15:34
(0,24) and (0,-24)

Imagineny1 2015-02-26 20:15:34
(0,24) (0, -24)

ingridzhang97 2015-02-26 20:15:34
(0,24) and (0,-24

pnpiano 2015-02-26 20:15:34
(0,24) and (0,-24)

eswa2000 2015-02-26 20:15:34
(0,24), (0,-24)

AMN300 2015-02-26 20:15:34
(0, 24) and (0, -24)

az_phx_brandon_jiang 2015-02-26 20:15:39
(0,24) and (0,-24)

DPatrick 2015-02-26 20:15:51
Right. $P$ is $(0,24)$ and $Q$ is $(0,-24)$.

DPatrick 2015-02-26 20:16:00
We can also set $A$ to be $(x,0)$ for some $x$. Then what's $B$?

Imagineny1 2015-02-26 20:16:24
(x+39,0)

Amrutamaya 2015-02-26 20:16:24
x+39,0

AMN300 2015-02-26 20:16:24
(x+39, 0) or (x-39, 0)

mathperson9 2015-02-26 20:16:24
(x+39,0)

hnkevin42 2015-02-26 20:16:24
(x + 39, 0)

ingridzhang97 2015-02-26 20:16:24
(x+39,0)

DPatrick 2015-02-26 20:16:27
Then since $AB = 39$ we have $B = (x+39,0).$ (Note that we can assume that $B$ is to the right of $A$ by symmetry; if not, just reflect it across the $y$-axis.)

DPatrick 2015-02-26 20:16:38
So we've got this:

DPatrick 2015-02-26 20:16:42

DPatrick 2015-02-26 20:16:47
Now what?

Satyaprakash2009rta 2015-02-26 20:17:02
ratio of radius

nosaj 2015-02-26 20:17:02
solve for the radius

AMN300 2015-02-26 20:17:14
We need to use the radius length condition

RahulKrishnan 2015-02-26 20:17:14
solve for radius

DPatrick 2015-02-26 20:17:17
We're given that $AP = \frac58 \cdot BP,$ or $8(AP) = 5(BP).$

DivideBy0 2015-02-26 20:17:26
use pythag on radius

DPatrick 2015-02-26 20:17:37
Right, we can compute $AP$ and $BP$ using our coordinates!

DPatrick 2015-02-26 20:17:47
Squaring first, though, seems like a sensible thing to do: $64(AP)^2 = 25(BP)^2.$

DPatrick 2015-02-26 20:18:01
(No nasty square roots to deal with now.)

DPatrick 2015-02-26 20:18:08
Now we just plug in the formulas for the distances:
\[
64(x^2 + 24^2) = 25((x+39)^2 + 24^2).
\]

EulerMacaroni 2015-02-26 20:18:33
that's an ugly quadratic

SimonSun 2015-02-26 20:18:33
too big

Darn 2015-02-26 20:18:33
Rearrange

DPatrick 2015-02-26 20:18:42
This simplifies (if you can call it that) to $39x^2 - 50(39)x + (39(24^2)-25(39^2)) = 0.$

ingridzhang97 2015-02-26 20:19:00
divide by 39

EulerMacaroni 2015-02-26 20:19:00
Factor out a 39

mathtastic 2015-02-26 20:19:00
DIVIDE BY 39!!

dli00105 2015-02-26 20:19:00
divide by 39

numbersandnumbers 2015-02-26 20:19:00
divide by 39

DPatrick 2015-02-26 20:19:04
How convenient! We can divide by 39 to get just $x^2 - 50x - 399 = 0.$

DPatrick 2015-02-26 20:19:12
399 = 3 * 7 * 19, so can you see how it factors?

bharatputra 2015-02-26 20:19:40
x=57 or -7 but only 57 works

Tuxianeer 2015-02-26 20:19:40
57,-7

ninjataco 2015-02-26 20:19:40
(x-57)(x+7)

sssssssssssssss 2015-02-26 20:19:40
x-57, x+7

celery2 2015-02-26 20:19:40
(x-57)(x+7)=0

DPatrick 2015-02-26 20:19:44
It factors as $(x-57)(x+7) = 0.$

DPatrick 2015-02-26 20:19:49
And, as hoped, we get two solutions: $x=57$ and $x=-7.$

DPatrick 2015-02-26 20:20:10
Of course, my picture doesn't match the $x = -7$ case, but that doesn't matter, because all the algebra works just fine!

DPatrick 2015-02-26 20:20:26
So one pair of circles are centered at (57,0) and (96,0), and the other pair of circles are centered at (-7,0) and (32,0). (Up to reflection, of course -- the second pair might be at (7,0) and (-32,0), for instance.)

DPatrick 2015-02-26 20:20:42
And what's the final answer?

celery2 2015-02-26 20:20:58
192

giftedbee 2015-02-26 20:20:58
D

mattpi 2015-02-26 20:20:58
D

abean077 2015-02-26 20:20:58
Now just add 57+96+7+32

DPatrick 2015-02-26 20:21:04
We just sum the absolute values of the x-coordinates: 57+96+7+32 = 192. Answer (D).

DPatrick 2015-02-26 20:21:33
They were kind, I thought, not to put 178 as an answer choice, since 178 = 57 + 96 - 7 + 32 might be a common mistake.

DPatrick 2015-02-26 20:21:53
I thought this was a cute problem.

DPatrick 2015-02-26 20:22:03
And on to 12B #25:

DPatrick 2015-02-26 20:22:09
25. A bee starts flying from point $P_0.$ She flies 1 inch due east to point $P_1.$ For $j\geq1,$ once the bee reaches point $P_j,$ she turns $30^\circ$ counterclockwise and then flies $j+1$ inches straight to point $P_{j+1}.$ When the bee reaches $P_{2015}$ she is exactly $a\sqrt b+c\sqrt d$ inches away from $P_0,$ where $a,$ $b,$ $c,$ and $d$ are positive integers and $b$ and $d$ are not divisible by the square of any prime. What is $a+b+c+d?$
$\phantom{12B:25}$
$\text{(A) } 2016 \quad \text{(B) } 2024 \quad \text{(C) } 2032 \quad \text{(D) } 2040 \quad \text{(E) } 2048$

DPatrick 2015-02-26 20:22:30
(I wonder why Aaron the ant got a name from the 10A problem, but this bee is anonymous. I'm going to call her Betty. And she can call me Al.)

EulerMacaroni 2015-02-26 20:22:52
Complex numbers here

AMN300 2015-02-26 20:22:52
complex plane!

AMN300 2015-02-26 20:22:52
Since we are rotating, we could use the complex plane

hnkevin42 2015-02-26 20:22:52
These rotations are making me want to use the complex numbers.

DPatrick 2015-02-26 20:23:18
Good idea! Trying to track this sort of rotational movement is a great candidate for using the complex plane.

DPatrick 2015-02-26 20:23:38
Let's set $w = \dfrac{\sqrt3}{2} + \dfrac12 i = e^{(\pi i)/6}$ to be the complex number with $|w| = 1$ that is along the $30^\circ$ ray. Note that $w$ is a primitive 12th root of unity, so that $w^{12} = 1.$

DPatrick 2015-02-26 20:23:57
The key fact is that rotating by $30^\circ$ is the same as multiplying by $w$.

DPatrick 2015-02-26 20:24:38
(I should warn you -- if you haven't yet seen this sort of math before, you're probably going to be mostly lost in this problem. See what you can pick up, and I'm going to do a couple more AMC 10 problems after this one.)

DPatrick 2015-02-26 20:25:05
So on move $k$, what complex number are we adding to Betty's position?

MSTang 2015-02-26 20:25:34
$kw^{k-1}$

trumpeter 2015-02-26 20:25:34
$kw^{k-1}$

mathtastic 2015-02-26 20:25:39
$kw^{k-1}$

DPatrick 2015-02-26 20:25:48
Good. On move $k$, we are moving along the direction $kw^{k-1}$, and "moving" means adding $kw^{k-1}$ to our current position.

DPatrick 2015-02-26 20:26:07
(The first move is 1, the second move is $2w$, the third move is $3w^2$, and so on.)

mathtastic 2015-02-26 20:26:13
the answer is just $|1+2w+3w^2+\cdots +2015w^{2014}|$

DPatrick 2015-02-26 20:26:19
Therefore, after 2015 moves, we're at the point
\[
1 + 2w + 3w^2 + 4w^3 + \cdots + 2015w^{2014}.
\]
Let's call this number $s$. We want to compute $|s|$.

UrInvalid 2015-02-26 20:26:33
arithmeto-geometric sequence

ingridzhang97 2015-02-26 20:26:33
arithmetic-geometric sequence

DPatrick 2015-02-26 20:26:53
Indeed, it looks a bit like a geometric series, and a bit like an arithmetic sequence...what's the standard technique we would use for this?

dhruv 2015-02-26 20:27:19
multiply by ratio

az_phx_brandon_jiang 2015-02-26 20:27:19
multiply by w and subtract

ninjataco 2015-02-26 20:27:19
multiply s by w, and then subtract w from sw

eswa2000 2015-02-26 20:27:19
multiply by w

DivideBy0 2015-02-26 20:27:19
multiply by w and subtract

DPatrick 2015-02-26 20:27:25
We multiply by the common ratio. So let's do that:
\[
ws = w + 2w^2 + 3w^3 + \cdots + 2015w^{2015}.
\]

DPatrick 2015-02-26 20:27:33
And then subtracting makes it (almost) into an actual geometric sequence:
\begin{align*}
s &= 1 + 2w + 3w^2 + 4w^3 + \cdots + 2015w^{2014} \\
ws &= \phantom{1 + }\;\;\, w + 2w^2 + 3w^3 + \cdots + 2014w^{2014} + 2015w^{2015} \\ \hline
s - ws &= 1 + \phantom{2}w + \phantom{3}w^2 + \phantom{4}w^3 + \cdots + \phantom{2015}w^{2014} - 2015w^{2015}.
\end{align*}
We've got that extra term at the end that's annoying, but the rest is nice and geometric.

DPatrick 2015-02-26 20:27:49
How can we make the geometric series part even nicer?

mathtastic 2015-02-26 20:28:10
Well now let's simplify $1+w+w^2+\cdots+w^{2014}$

bharatputra 2015-02-26 20:28:16
w^2015-1/w-1

ninjataco 2015-02-26 20:28:16
write it as (w^2015 - 1)/(w-1)

AMN300 2015-02-26 20:28:16
(w^(2015)-1)/(w-1)

DPatrick 2015-02-26 20:28:29
We could...but is there something we could do first that would be really really helpful?

DPatrick 2015-02-26 20:28:42
Hint: 2016 is a multiple of 12!

DivideBy0 2015-02-26 20:29:08
add w^2015

mathwrath 2015-02-26 20:29:20
1+w+w^2+...+w^11 = 0

TheStrangeCharm 2015-02-26 20:29:20
we know that $1 + \omega + ... + \omega ^11 = 0$

Waley 2015-02-26 20:29:25
add w^2015 and subtract it away

happiface 2015-02-26 20:29:25
subtract and add w^2015

DPatrick 2015-02-26 20:29:36
We can add a $w^{2015}$ term to the geometric series, and subtract another one off in that "extra" term. That is,
\[
s - ws = 1 + w + w^2 + \cdots + w^{2014} + w^{2015} - 2016w^{2015}.
\]
How does that help immensely?

ninjataco 2015-02-26 20:29:59
the sum of the first 2016 terms is 0

trumpeter 2015-02-26 20:29:59
that geo series equals 0

bharatputra 2015-02-26 20:29:59
the geometric series just becomes 0

celery2 2015-02-26 20:29:59
the front part is 0

Satyaprakash2009rta 2015-02-26 20:29:59
sum of geometric series is 0

DPatrick 2015-02-26 20:30:08
Now the geometric series part just sums to 0! If you prefer the formula, the sum is $\dfrac{1-w^{2016}}{1-w},$ and since 2016 is a multiple of 12, we have $w^{2016} = 1$, so this is just 0.

DPatrick 2015-02-26 20:30:20
So now we have simply $s - ws = -2016w^{2015}$, and hence
\[
s = \frac{2016w^{2015}}{w-1}.
\]

DPatrick 2015-02-26 20:30:30
How do we compute $|s|$ from this?

MSTang 2015-02-26 20:30:56
it equals 2016/magnitude(w-1)

DPatrick 2015-02-26 20:31:00
Magnitude is multiplicative, so
\[
|s| = \left| \frac{2016w^{2015}}{w-1} \right| = \frac{2016\,|w|^{2015}}{|w-1|}.
\]

dhruv 2015-02-26 20:31:08
|w|=1

DPatrick 2015-02-26 20:31:12
And we know $|w| = 1$, so we have just $|s| = \dfrac{2016}{|w-1|}.$

DPatrick 2015-02-26 20:31:20
And how do we compute $|w-1|$?

MSTang 2015-02-26 20:31:47
$w = \sqrt{3}/2 + i/2$

sparkles257 2015-02-26 20:31:47
using w and finding mag normally

DPatrick 2015-02-26 20:31:53
One idea is that we know $w = \dfrac{\sqrt3}{2} + \dfrac12i$, so we can just compute it directly:
\[
w-1 = \left(\frac{\sqrt3}{2}-1\right) + \frac12i,
\]
so
\[
|w-1|^2 = \left(\frac{\sqrt3}{2}-1\right)^2 + \left(\frac12\right)^2.
\]

DPatrick 2015-02-26 20:32:12
This computes to:
\[
|w-1|^2 = \frac{(\sqrt3-2)^2+1}{4} = \frac{8-4\sqrt3}{4} = 2-\sqrt3.
\]

DPatrick 2015-02-26 20:32:22
So $|w-1| = \sqrt{2-\sqrt3}.$ How do we compute this?

brian22 2015-02-26 20:33:11

MSTang 2015-02-26 20:33:11
$(\sqrt3-1)^2 = 4-2\sqrt3$

DPatrick 2015-02-26 20:33:21
Indeed, one clever way to continue is to recognize that
\[
2 - \sqrt3 = \frac{4 - 2\sqrt3}{2} = \frac{(\sqrt3 - 1)^2}{2},
\]
so that $\sqrt{2-\sqrt{3}} = \dfrac{\sqrt3 - 1}{\sqrt2}.$

DPatrick 2015-02-26 20:33:45
But that's kind of magical.

DPatrick 2015-02-26 20:33:56
Another approach, that avoids having to compute a square-root via magic, is to draw a picture:

DPatrick 2015-02-26 20:34:03

DPatrick 2015-02-26 20:34:17
The picture above is how we geometrically determine the location of $w-1$. The length $|w-1|$ is the length of the black segment in the picture.

DPatrick 2015-02-26 20:34:34
Notice it's the base of an isosceles triangle with legs of length 1 and vertex angle of $30^\circ.$

DPatrick 2015-02-26 20:34:47
So trig tells us...

mathtastic 2015-02-26 20:34:54
yeah just compute with $\sin 15^\circ$

DPatrick 2015-02-26 20:35:07
So by dropping the altitude from the vertex to the base, we see that the length is $2\sin 15^\circ.$

DPatrick 2015-02-26 20:35:15
And how do we compute that?

MSTang 2015-02-26 20:35:34
sin(45-30)

az_phx_brandon_jiang 2015-02-26 20:35:34
subtraction cause half angle is bad

hnkevin42 2015-02-26 20:35:34
sine subtraction formula

DrMath 2015-02-26 20:35:34
$\sin{(45-30)}$

Waley 2015-02-26 20:35:44
sin15=sin(45-30) use angle subtraction law

sillyd 2015-02-26 20:35:44
sin(45-30)

DPatrick 2015-02-26 20:35:47
We can use the difference-angle formula for sine:
\[
\sin(45^\circ - 30^\circ) = \sin 45^\circ \cos 30^\circ - \sin 30^\circ \cos 45^\circ.
\]

DPatrick 2015-02-26 20:35:53
Thus
\[
\sin 15^\circ = \frac{\sqrt3 - 1}{2\sqrt2},
\]
and $|w-1| = 2\sin 15^\circ = \dfrac{\sqrt3 - 1}{\sqrt2},$ just as before.

DPatrick 2015-02-26 20:36:17
Anyway, to finish, our answer is
\[
|s| = \frac{2016}{|w-1|} = \frac{2016\sqrt{2}}{\sqrt3 - 1}.
\]

sparkles257 2015-02-26 20:36:48
rationalize!

acegikmoqsuwy2000 2015-02-26 20:36:48
rationalize denominator

sharmag 2015-02-26 20:36:48
rationalize denominator?

eswa2000 2015-02-26 20:36:48
rationalize the denom

ninjataco 2015-02-26 20:36:48
multiply numerator and denominator by (sqrt3 + 1)

happiface 2015-02-26 20:36:48
they want us to rationalize more apparently

DPatrick 2015-02-26 20:36:59
We need square roots in our answer and no denominators.

DPatrick 2015-02-26 20:37:04
We rationalize the denominator by multiplying through by $\sqrt3 + 1$:
\[
|s| = \frac{2016\sqrt2(\sqrt3+1)}{(\sqrt3-1)(\sqrt3+1)} = \frac{2016(\sqrt6 + \sqrt2)}{2} = 1008\sqrt6 + 1008\sqrt2.
\]

alex7 2015-02-26 20:37:26
that equals 1008+6+1008+2=2024

ingridzhang97 2015-02-26 20:37:26
so answer is B) 2024

Darn 2015-02-26 20:37:26
1008+6+1008+2=B

DPatrick 2015-02-26 20:37:28
Thus our final answer is 1008+6+1008+2 = 2024. Answer (B).

droid347 2015-02-26 20:37:52
Can we do 10B #19 now pleaseee?

DPatrick 2015-02-26 20:38:05
Yeah, we've got a little more time, so let's do #19, which happened to be the same on both contests.

DPatrick 2015-02-26 20:38:11
19. In $\triangle ABC$, $\angle C = 90^\circ$ and $AB = 12$. Squares $ABXY$ and $ACWZ$ are constructed outside of the triangle. The points $X$, $Y$, $Z$, and $W$ lie on a circle. What is the perimeter of the triangle?
$\phantom{19B}$
(A) $12 + 9\sqrt3$ (B) $18 + 6\sqrt3$ (C) $12 + 12\sqrt2$ (D) $30$ (E) $32$

DPatrick 2015-02-26 20:38:31
Here we can sketch a reasonable picture:

DPatrick 2015-02-26 20:38:36

DPatrick 2015-02-26 20:38:49
The other condition is that $X$, $Y$, $Z$, and $W$ all lie on a circle. Where does the center of that circle have to be?

AMN300 2015-02-26 20:39:17
Midpoint of AB

sparkles257 2015-02-26 20:39:17
center of circle is midpt.of AB

noobynoob 2015-02-26 20:39:17
midpoint of AB

EulerMacaroni 2015-02-26 20:39:17
The midpoint of $\overline{AB}$

BobaFett101 2015-02-26 20:39:17
midpoint of AB

gurev 2015-02-26 20:39:17
midpoint of AB

Deathranger999 2015-02-26 20:39:17
The midpoint of the hypotenuse.

brian22 2015-02-26 20:39:17
bisectory of $WZ$ intersected with bisector of $XY$

vinayak-kumar 2015-02-26 20:39:17
Intersection of perpendicular bisctors

DPatrick 2015-02-26 20:39:28
Any circle containing $X$ and $Y$ has its center on the perpendicular bisector of $\overline{XY}$. This is the same as the perpendicular bisector of $\overline{AB}$.

DPatrick 2015-02-26 20:39:36
Any circle containing $Z$ and $W$ has its center on the perpendicular bisector of $\overline{WZ}$. This is the same as the perpendicular bisector of $\overline{AC}$.

DPatrick 2015-02-26 20:40:01
But the perpendicular bisector of $\overline{AC}$ is parallel to $\overline{CB}$, so it creates a triangle similar to $ABC$ but half the size.

DPatrick 2015-02-26 20:40:19
Thus, it interescts $\overline{AB}$ at the midpoint of $\overline{AB}$! Let's call that point $O$. That's the center of the circle.

DPatrick 2015-02-26 20:40:31
From this, can we determine the radius of the circle?

mathperson9 2015-02-26 20:40:55
Which is 6\sqrt{5} from X, Y, Z, and W

DivideBy0 2015-02-26 20:40:55
pythag!

CountDown 2015-02-26 20:40:55
yes, it's 6sqrt5

EulerMacaroni 2015-02-26 20:40:55
$6\sqrt{5}$

mathperson9 2015-02-26 20:40:55
OZ is 6\Sqrt{5}

DPatrick 2015-02-26 20:40:58
Sure: we can just look at triangle $OBY$:

DPatrick 2015-02-26 20:41:03

DPatrick 2015-02-26 20:41:17
We have $OB = 6$ and $BY = 12$, so by the Pythagorean Theorem $OY = \sqrt{6^2 + 12^2} = 6\sqrt5$ is the radius.

Darn 2015-02-26 20:41:32
Then find distance to Z or W!

DrMath 2015-02-26 20:41:35
so we can use that to make an equation for $OZ$

DPatrick 2015-02-26 20:41:45
Right. We can compute $OZ = 6\sqrt5$ in a similar fashion. But we'll need to name some lengths.

DPatrick 2015-02-26 20:41:51
Let $BC = a$ and $AC = b$, and consider $OPZ$ as shown below:

DPatrick 2015-02-26 20:41:55

DPatrick 2015-02-26 20:42:04
What are $OP$ and $ZP$?

EulerMacaroni 2015-02-26 20:42:31
OP is $b+a/2$

donot 2015-02-26 20:42:31
$b+\frac{a}{2}$

bharatputra 2015-02-26 20:42:31
b+a/2 and b/2

TheStrangeCharm 2015-02-26 20:42:31
b + a\2 and b/2

akaashp11 2015-02-26 20:42:31
$OP = a/2 + b$

ingridzhang97 2015-02-26 20:42:31
OP=b+a/2, ZP=b/2

DominicanAOPSer 2015-02-26 20:42:31
OP = b + a/2

CountDown 2015-02-26 20:42:31
b/2, b + a/2

DPatrick 2015-02-26 20:42:38
Right. $OP = \frac{a}{2} + b$ and $PZ = \frac{b}{2}$.

DPatrick 2015-02-26 20:42:48
So the Pythagorean Theorem on $OPZ$ gives us
\[
\left(\frac{a}{2}+b\right)^2 + \left(\frac{b}{2}\right)^2 = (6\sqrt5)^2.
\]

EulerMacaroni 2015-02-26 20:43:09
but multiply through by 4 here so its not ugly

DPatrick 2015-02-26 20:43:16
Good idea. Multiplying through by 4 gives
\[
(a+2b)^2 + b^2 = 720,
\]
or $a^2 + 4ab + 5b^2 = 720.$

happiface 2015-02-26 20:43:26
But we also have that $a^2 + b^2 = 144$ from the triangle $ABC$.

droid347 2015-02-26 20:43:26
but $a^2+b^2=144$

DPatrick 2015-02-26 20:43:33
Right, they're the sides of the original triangle, so $a^2 + b^2 = 144.$

DPatrick 2015-02-26 20:43:53
Subtracting this from our other equation gives $4ab + 4b^2 = 720-144 = 576.$

thequantumguy 2015-02-26 20:44:22
divide by 4

DPatrick 2015-02-26 20:44:25
And now dividing by 4 again gives $ab + b^2 = 144.$

DPatrick 2015-02-26 20:44:32
But we also just had $a^2 + b^2 = 144.$

noobynoob 2015-02-26 20:44:44
a=b

EulerMacaroni 2015-02-26 20:44:44
$ab=a^2$

RJ810 2015-02-26 20:44:44
a=b

AMN300 2015-02-26 20:44:44
Hence a=b !

NewbieGamer 2015-02-26 20:44:44
a=b

DPatrick 2015-02-26 20:44:53
So we must have $ab = a^2$, or $b = a$.

DPatrick 2015-02-26 20:44:59
So the original right triangle $ABC$ is isosceles. How do we finish the problem?

dli00105 2015-02-26 20:45:16
a=b=6\sqrt{2}

sparkles257 2015-02-26 20:45:16
a= 6sqrt(2) = b

uniqueearth 2015-02-26 20:45:16
a=b=6sqrt2

AMN300 2015-02-26 20:45:16
So a=b=6 sqrt 2 by 45-45-90, so we have an answer of C after adding

MathLearner01 2015-02-26 20:45:24
12+12sqrt2

alex7 2015-02-26 20:45:24
12+6sqrt2*2

acegikmoqsuwy2000 2015-02-26 20:45:24
12+6sqrt2+6sqrt2 = 12+12sqrt2 so (C)

DPatrick 2015-02-26 20:45:27
If $AB = 12$ and $ABC$ is right isosceles, then $AC = BC = 6\sqrt2$. So its perimeter is $12 + 12\sqrt2$, answer (C).

DPatrick 2015-02-26 20:45:59
Let's quickly do both #20's before calling it a night. One is sorta interesting, and one is really not.

DPatrick 2015-02-26 20:46:07
First, the 10B #20:

DPatrick 2015-02-26 20:46:11
20. Erin the ant starts at a given corner of a cube and crawls along exactly 7 edges in such a way that she visits every corner exactly once and then finds that she is unable to return along an edge to her starting point. How many paths are there meeting these conditions?
$\phantom{20B}$
(A) 6 (B) 9 (C) 12 (D) 18 (E) 24

DPatrick 2015-02-26 20:46:41
Heh. Erin the ant in #20, and Aaron the ant in #24. I just noticed that.

Darn 2015-02-26 20:47:14
Notice that by parity, we can only have the diagonally opposite corner as our end of the path

fluffyanimal 2015-02-26 20:47:14
There are 6 ways to do the first 2 steps which are all the same

DPatrick 2015-02-26 20:47:32
That's a really key observation for this problem: the first two moves don't matter.

DPatrick 2015-02-26 20:47:39
All 3 choices for the first move are symmetric.

DPatrick 2015-02-26 20:47:54
And then for the second move, there are 2 choices, but both lead to a corner that's opposite (across a face) from the starting point.

DPatrick 2015-02-26 20:48:01
So the first $3 \cdot 2 = 6$ possible pairs of first moves all lead to symmetric positions.

DPatrick 2015-02-26 20:48:14
Here's a picture with the first two moves shown in blue:

DPatrick 2015-02-26 20:48:19

DPatrick 2015-02-26 20:48:30
We just need to count how many ways there are to finish, and multiply by 6.

Bob_Smith 2015-02-26 20:48:53
And there's only one way!

celery2 2015-02-26 20:48:53
one way to finish

DPatrick 2015-02-26 20:49:06
Indeed, from here, all of Erin's moved are forced! Let's see why...

DPatrick 2015-02-26 20:49:17
We have to go left now: otherwise we won't reach that vertex until the end, but that's not allowed because "she is unable to return along an edge to her starting point":

DPatrick 2015-02-26 20:49:30

DPatrick 2015-02-26 20:49:41
And now the next move is forced:

DPatrick 2015-02-26 20:49:46

uniqueearth 2015-02-26 20:50:08
you have to go down

acegikmoqsuwy2000 2015-02-26 20:50:08
and we now go down

AMN300 2015-02-26 20:50:08
now there's only 1 good way

DPatrick 2015-02-26 20:50:17
And now we have to move around the back face...but we have to go counterclockwise, because if we go clockwise then we'll end at the corner adjacent to Start, which is not allowed.

DPatrick 2015-02-26 20:50:22

DPatrick 2015-02-26 20:50:37
So after the first two moves, all the rest of the moves are forced.

crosby87 2015-02-26 20:50:50
answer is 6 $A$

SimonSun 2015-02-26 20:50:50
so A 6

akaashp11 2015-02-26 20:50:50
A

celery2 2015-02-26 20:50:50
so the answer is 1x6=6 (A)

DPatrick 2015-02-26 20:50:51
So there are only 6 paths. Answer (A).

DPatrick 2015-02-26 20:51:12
That was sort of interesting I guess.

DPatrick 2015-02-26 20:51:36
12B #20 was not (in my opinion), but let's do it anyway...

DPatrick 2015-02-26 20:51:40
20. For every positive integer $n$, let $\text{mod}_5(n)$ be the remainder obtained when $n$ is divided by 5. Define a function $f : \{0,1,2,3,\ldots\} \times \{0,1,2,3,4\} \rightarrow \{0,1,2,3,4\}$ recursively as follows:
\[
f(i,j) = \left\{
\begin{array}{ll}
\text{mod}_5(j+1) & \text{if } i=0 \text{ and } 0 \le j \le 4, \\
f(i-1,1) & \text{if } i \ge 1 \text{ and } j=0, \text{ and} \\
f(i-1,f(i,j-1)) & \text{if } i \ge 1 \text{ and } 1 \le j \le 4.
\end{array}
\right.
\]
What is $f(2015,2)$?
$\phantom{12B:20}$
(A) 0 (B) 1 (C) 2 (D) 3 (E) 4

DPatrick 2015-02-26 20:51:57
Wow.

Darn 2015-02-26 20:52:27
That looks ugly, so list out a table of values?

DrMath 2015-02-26 20:52:27
ok lets play around...

DPatrick 2015-02-26 20:52:36
Yeah, let's just try to make a little chart of this function for small values of $i$, and see if we notice any patterns.

DPatrick 2015-02-26 20:52:46
$i=0$ is pretty straightforward: we just add 1 to $j$ and take the value mod 5.

DPatrick 2015-02-26 20:52:52
\[
\begin{array}{c|c}
j & i=0 \\ \hline
0 & 1 \\
1 & 2 \\
2 & 3 \\
3 & 4 \\
4 & 0
\end{array}
\]

DPatrick 2015-02-26 20:53:02
Let's now look at $i=1$. First, I'll just put in the instructions from the function:

DPatrick 2015-02-26 20:53:06
\[
\begin{array}{c|c|c}
j & i=0 & i=1 \\ \hline
0 & 1 & f(0,1) \\
1 & 2 & f(0,f(1,0)) \\
2 & 3 & f(0,f(1,1)) \\
3 & 4 & f(0,f(1,2)) \\
4 & 0 & f(0,f(1,3))
\end{array}
\]

DPatrick 2015-02-26 20:53:30
$f(0,j)$ is just $j+1$ (mod 5), so that's write it that way:

DPatrick 2015-02-26 20:53:34
\[
\begin{array}{c|c|c}
j & i=0 & i=1 \\ \hline
0 & 1 & 2 \\
1 & 2 & f(1,0)+1 \\
2 & 3 & f(1,1)+1 \\
3 & 4 & f(1,2)+1 \\
4 & 0 & f(1,3)+1
\end{array}
\]

Darn 2015-02-26 20:53:49
2,3,4,0,1

celery2 2015-02-26 20:53:49
2,3,4,0,1

Darn 2015-02-26 20:53:49
Keep adding 1

brian22 2015-02-26 20:53:49

DPatrick 2015-02-26 20:53:50
Now it's clear: we just add 1 to the number above.

DPatrick 2015-02-26 20:53:56
\[
\begin{array}{c|c|c}
j & i=0 & i=1 \\ \hline
0 & 1 & 2 \\
1 & 2 & 3 \\
2 & 3 & 4 \\
3 & 4 & 0 \\
4 & 0 & 1
\end{array}
\]

DPatrick 2015-02-26 20:54:10
So it looks like $f(1,j) = j+2$ (mod 5).

DPatrick 2015-02-26 20:54:23
Hmmm...maybe we guess that $f(2,j) = j+3$ (mod 5).

DPatrick 2015-02-26 20:54:33
Let's see...

DPatrick 2015-02-26 20:54:37
\[
\begin{array}{c|c|c|c}
j & i=0 & i=1 & i=2 \\ \hline
0 & 1 & 2 & 3\\
1 & 2 & 3 & f(2,0)+2\\
2 & 3 & 4 & f(2,1)+2\\
3 & 4 & 0 & f(2,2)+2\\
4 & 0 & 1 & f(2,3)+2
\end{array}
\]

DPatrick 2015-02-26 20:54:52
So each number is 2 plus the number above (mod 5). That works out to:

DPatrick 2015-02-26 20:54:57
\[
\begin{array}{c|c|c|c}
j & i=0 & i=1 & i=2 \\ \hline
0 & 1 & 2 & 3\\
1 & 2 & 3 & 0\\
2 & 3 & 4 & 2\\
3 & 4 & 0 & 4\\
4 & 0 & 1 & 1
\end{array}
\]

DPatrick 2015-02-26 20:55:14
Now the function appears to be $f(2,j) = 2(j-1)$.

DPatrick 2015-02-26 20:55:30
So let's move on to $i=3$:

DPatrick 2015-02-26 20:55:35
\[
\begin{array}{c|c|c|c|c}
j & i=0 & i=1 & i=2 & i=3 \\ \hline
0 & 1 & 2 & 3 & 0\\
1 & 2 & 3 & 0 & 2(f(3,0)-1) \\
2 & 3 & 4 & 2 & 2(f(3,1)-1) \\
3 & 4 & 0 & 4 & 2(f(3,2)-1) \\
4 & 0 & 1 & 1 & 2(f(3,3)-1) \\
\end{array}
\]

DPatrick 2015-02-26 20:55:53
In other words, we subtract 1 from the number above, and then double it, then take mod 5.

DPatrick 2015-02-26 20:56:04
This gives

DPatrick 2015-02-26 20:56:08
\[
\begin{array}{c|c|c|c|c}
j & i=0 & i=1 & i=2 & i=3 \\ \hline
0 & 1 & 2 & 3 & 0\\
1 & 2 & 3 & 0 & 3 \\
2 & 3 & 4 & 2 & 4 \\
3 & 4 & 0 & 4 & 1 \\
4 & 0 & 1 & 1 & 0 \\
\end{array}
\]

RJ810 2015-02-26 20:56:23
There is no 2!

DPatrick 2015-02-26 20:56:30
Yes. Note that we've lost 2 as an output of the function, and that because of the recursive nature of the function, we can never get 2 as a value again. That allows us to eliminate (C).

Darn 2015-02-26 20:56:52
maybe it will continue to eliminate terms?

DPatrick 2015-02-26 20:56:58
Let's hope so, and soon!

DPatrick 2015-02-26 20:57:03
Let's continue:

DPatrick 2015-02-26 20:57:06
\[
\begin{array}{c|c|c|c|c|c}
j & i=0 & i=1 & i=2 & i=3 & i=4 \\ \hline
0 & 1 & 2 & 3 & 0 & 3\\
1 & 2 & 3 & 0 & 3 & f(3,f(4,0)) \\
2 & 3 & 4 & 2 & 4 & f(3,f(4,1)) \\
3 & 4 & 0 & 4 & 1 & f(3,f(4,2)) \\
4 & 0 & 1 & 1 & 0 & f(3,f(4,3)) \\
\end{array}
\]

DPatrick 2015-02-26 20:57:29
Let's just grind through them...

DPatrick 2015-02-26 20:57:41
$f(4,0) = 3$, so $f(4,1) = f(3,3) = 1$.

DPatrick 2015-02-26 20:57:50
Then $f(4,2) = f(3,1) = 3$.

DPatrick 2015-02-26 20:57:56
Then $f(4,3) = f(3,3) = 1$.

DPatrick 2015-02-26 20:58:01
And finally $f(4,4) = f(3,1) = 3$.

DPatrick 2015-02-26 20:58:09
\[
\begin{array}{c|c|c|c|c|c}
j & i=0 & i=1 & i=2 & i=3 & i=4 \\ \hline
0 & 1 & 2 & 3 & 0 & 3\\
1 & 2 & 3 & 0 & 3 & 1 \\
2 & 3 & 4 & 2 & 4 & 3 \\
3 & 4 & 0 & 4 & 1 & 1 \\
4 & 0 & 1 & 1 & 0 & 3
\end{array}
\]

acegikmoqsuwy2000 2015-02-26 20:58:15
so 4 and 0 are gone

RJ810 2015-02-26 20:58:15
There is no 4 and no 0!

abean077 2015-02-26 20:58:15
4 and 0 are gone too

DPatrick 2015-02-26 20:58:20
Now the answer must be 1 or 3. 50-50 chance!

DPatrick 2015-02-26 20:58:30
Maybe $i=5$ will finally provide some clarity:

DPatrick 2015-02-26 20:58:35
\[
\begin{array}{c|c|c|c|c|c|c}
j & i=0 & i=1 & i=2 & i=3 & i=4 & i=5 \\ \hline
0 & 1 & 2 & 3 & 0 & 3 & 1\\
1 & 2 & 3 & 0 & 3 & 1 & f(4,f(5,0)) \\
2 & 3 & 4 & 2 & 4 & 3 & f(4,f(5,1)) \\
3 & 4 & 0 & 4 & 1 & 1 & f(4,f(5,2)) \\
4 & 0 & 1 & 1 & 0 & 3 & f(4,f(5,3))
\end{array}
\]

DPatrick 2015-02-26 20:58:46
Again, let's just grind through them.

DPatrick 2015-02-26 20:59:01
$f(5,0)= 1$, so $f(5,1) = f(4,1) = 1$.

abean077 2015-02-26 20:59:09
They're all 1

Darn 2015-02-26 20:59:09
Just 1,1,1,1,1

DPatrick 2015-02-26 20:59:13
Aha---they're all 1! They're all $f(4,\text{the number above})$ and since that number above is always 1, they're all $f(4,1) = 1$.

DPatrick 2015-02-26 20:59:25
So $f(5,j) = 1$ for all $j$.

trumpeter 2015-02-26 20:59:31
from i=5 and on, the output will only be 1!

DPatrick 2015-02-26 20:59:45
Right, $f(i,j)$ for $i>5$ are all 1 too! Since $f(i,j) = f(i-1,\text{something})$, once $f(i,j)=1$ at $i=5$, it'll be the constant function 1 for all larger $i$.

dli00105 2015-02-26 20:59:57
answer (B)

acegikmoqsuwy2000 2015-02-26 20:59:57
So (B) 1

sharmag 2015-02-26 20:59:57
therefore the answer is B (1)

AMN300 2015-02-26 20:59:57
so since it's recursive, the answer is 1 B

NewbieGamer 2015-02-26 20:59:57
the answer is (B) 1

DPatrick 2015-02-26 21:00:00
Thus our final answer is (B) 1.

DPatrick 2015-02-26 21:00:06
Wow, that problem was boring. Sorry. I don't know of a clever way.

DPatrick 2015-02-26 21:00:39
Kinda sad to end on that problem, but it's 9 PM ET / 6 PM PT, so I'm afraid we're done for the evening! Thanks for coming!

DPatrick 2015-02-26 21:00:46
Please join us again on Saturday March 21 and Friday March 27 when we will be discussing the AIME I and II contests.

nosaj 2015-02-26 21:00:53
Will Richard be making videos?

anandiyer12 2015-02-26 21:00:53
will videos come out?

DPatrick 2015-02-26 21:01:04
I don't believe so -- he usually only makes videos for the A contests.

DPatrick 2015-02-26 21:01:28
Plus we're all super-busy with last-minute details to get ready for the new AoPS website launch this weekend!

DrMath 2015-02-26 21:01:48
what about the guy in the black hoodie that looks suspiciously like him?

DPatrick 2015-02-26 21:01:58
Maybe he has some free time, but he never responds to my friend requests.

awesomenesspanda 2015-02-26 21:02:12
So approximately when will the answer key come out?

DPatrick 2015-02-26 21:02:25
I would guess that the AMC will put up an answer key very soon. But I have no idea for sure.

rockyisi 2015-02-26 21:02:36
when will all the solutions come out?

DPatrick 2015-02-26 21:02:56
A copy of the solutions was sent to every school. The AMC never posts full solutions.

thequantumguy 2015-02-26 21:03:21
when will the cut off be posted

DPatrick 2015-02-26 21:03:23
I have no idea.

nosaj 2015-02-26 21:03:36
Thanks for putting in a lot of work on the new site!

hnkevin42 2015-02-26 21:03:36
Excited for the new website for AoPS!

DPatrick 2015-02-26 21:03:39
Thanks -- we are too!

DPatrick 2015-02-26 21:04:17
OK, I need to kick you all out of the classroom so that I can generate and post the transcript of the session. Have a good night!

AkhilY 2015-02-26 21:04:49
where can we find the transcript

DPatrick 2015-02-26 21:05:18
School -> MathJams, then the Transcripts link.

DPatrick 2015-02-26 21:05:28
(Not for another 5-10 minutes though. I'll also post the link on the Forum.)

RevL8ion 2015-02-26 21:06:51
Thanks again for the Math Jam!

2015 AMC 10/12 B Discussion

Facilitator: Dave Patrick