We have your learning goals covered with Spring and Summer courses available. Enroll today!

Contests & Programs AMC and other contests, summer programs, etc.
AMC and other contests, summer programs, etc.
3 M G
BBookmark  VNew Topic kLocked
Contests & Programs AMC and other contests, summer programs, etc.
AMC and other contests, summer programs, etc.
3 M G
BBookmark  VNew Topic kLocked
G
Topic
First Poster
Last Poster
k a March Highlights and 2025 AoPS Online Class Information
jlacosta   0
Mar 2, 2025
March is the month for State MATHCOUNTS competitions! Kudos to everyone who participated in their local chapter competitions and best of luck to all going to State! Join us on March 11th for a Math Jam devoted to our favorite Chapter competition problems! Are you interested in training for MATHCOUNTS? Be sure to check out our AMC 8/MATHCOUNTS Basics and Advanced courses.

Are you ready to level up with Olympiad training? Registration is open with early bird pricing available for our WOOT programs: MathWOOT (Levels 1 and 2), CodeWOOT, PhysicsWOOT, and ChemWOOT. What is WOOT? WOOT stands for Worldwide Online Olympiad Training and is a 7-month high school math Olympiad preparation and testing program that brings together many of the best students from around the world to learn Olympiad problem solving skills. Classes begin in September!

Do you have plans this summer? There are so many options to fit your schedule and goals whether attending a summer camp or taking online classes, it can be a great break from the routine of the school year. Check out our summer courses at AoPS Online, or if you want a math or language arts class that doesn’t have homework, but is an enriching summer experience, our AoPS Virtual Campus summer camps may be just the ticket! We are expanding our locations for our AoPS Academies across the country with 15 locations so far and new campuses opening in Saratoga CA, Johns Creek GA, and the Upper West Side NY. Check out this page for summer camp information.

Be sure to mark your calendars for the following events:
[list][*]March 5th (Wednesday), 4:30pm PT/7:30pm ET, HCSSiM Math Jam 2025. Amber Verser, Assistant Director of the Hampshire College Summer Studies in Mathematics, will host an information session about HCSSiM, a summer program for high school students.
[*]March 6th (Thursday), 4:00pm PT/7:00pm ET, Free Webinar on Math Competitions from elementary through high school. Join us for an enlightening session that demystifies the world of math competitions and helps you make informed decisions about your contest journey.
[*]March 11th (Tuesday), 4:30pm PT/7:30pm ET, 2025 MATHCOUNTS Chapter Discussion MATH JAM. AoPS instructors will discuss some of their favorite problems from the MATHCOUNTS Chapter Competition. All are welcome!
[*]March 13th (Thursday), 4:00pm PT/7:00pm ET, Free Webinar about Summer Camps at the Virtual Campus. Transform your summer into an unforgettable learning adventure! From elementary through high school, we offer dynamic summer camps featuring topics in mathematics, language arts, and competition preparation - all designed to fit your schedule and ignite your passion for learning.[/list]
Our full course list for upcoming classes is below:
All classes run 7:30pm-8:45pm ET/4:30pm - 5:45pm PT unless otherwise noted.

Introductory: Grades 5-10

Prealgebra 1 Self-Paced

Prealgebra 1
Sunday, Mar 2 - Jun 22
Friday, Mar 28 - Jul 18
Sunday, Apr 13 - Aug 10
Tuesday, May 13 - Aug 26
Thursday, May 29 - Sep 11
Sunday, Jun 15 - Oct 12
Monday, Jun 30 - Oct 20
Wednesday, Jul 16 - Oct 29

Prealgebra 2 Self-Paced

Prealgebra 2
Tuesday, Mar 25 - Jul 8
Sunday, Apr 13 - Aug 10
Wednesday, May 7 - Aug 20
Monday, Jun 2 - Sep 22
Sunday, Jun 29 - Oct 26
Friday, Jul 25 - Nov 21


Introduction to Algebra A Self-Paced

Introduction to Algebra A
Sunday, Mar 23 - Jul 20
Monday, Apr 7 - Jul 28
Sunday, May 11 - Sep 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Wednesday, May 14 - Aug 27
Friday, May 30 - Sep 26
Monday, Jun 2 - Sep 22
Sunday, Jun 15 - Oct 12
Thursday, Jun 26 - Oct 9
Tuesday, Jul 15 - Oct 28

Introduction to Counting & Probability Self-Paced

Introduction to Counting & Probability
Sunday, Mar 16 - Jun 8
Wednesday, Apr 16 - Jul 2
Thursday, May 15 - Jul 31
Sunday, Jun 1 - Aug 24
Thursday, Jun 12 - Aug 28
Wednesday, Jul 9 - Sep 24
Sunday, Jul 27 - Oct 19

Introduction to Number Theory
Monday, Mar 17 - Jun 9
Thursday, Apr 17 - Jul 3
Friday, May 9 - Aug 1
Wednesday, May 21 - Aug 6
Monday, Jun 9 - Aug 25
Sunday, Jun 15 - Sep 14
Tuesday, Jul 15 - Sep 30

Introduction to Algebra B Self-Paced

Introduction to Algebra B
Sunday, Mar 2 - Jun 22
Wednesday, Apr 16 - Jul 30
Tuesday, May 6 - Aug 19
Wednesday, Jun 4 - Sep 17
Sunday, Jun 22 - Oct 19
Friday, Jul 18 - Nov 14

Introduction to Geometry
Tuesday, Mar 4 - Aug 12
Sunday, Mar 23 - Sep 21
Wednesday, Apr 23 - Oct 1
Sunday, May 11 - Nov 9
Tuesday, May 20 - Oct 28
Monday, Jun 16 - Dec 8
Friday, Jun 20 - Jan 9
Sunday, Jun 29 - Jan 11
Monday, Jul 14 - Jan 19

Intermediate: Grades 8-12

Intermediate Algebra
Sunday, Mar 16 - Sep 14
Tuesday, Mar 25 - Sep 2
Monday, Apr 21 - Oct 13
Sunday, Jun 1 - Nov 23
Tuesday, Jun 10 - Nov 18
Wednesday, Jun 25 - Dec 10
Sunday, Jul 13 - Jan 18
Thursday, Jul 24 - Jan 22

Intermediate Counting & Probability
Sunday, Mar 23 - Aug 3
Wednesday, May 21 - Sep 17
Sunday, Jun 22 - Nov 2

Intermediate Number Theory
Friday, Apr 11 - Jun 27
Sunday, Jun 1 - Aug 24
Wednesday, Jun 18 - Sep 3

Precalculus
Sunday, Mar 16 - Aug 24
Wednesday, Apr 9 - Sep 3
Friday, May 16 - Oct 24
Sunday, Jun 1 - Nov 9
Monday, Jun 30 - Dec 8

Advanced: Grades 9-12

Olympiad Geometry
Wednesday, Mar 5 - May 21
Tuesday, Jun 10 - Aug 26

Calculus
Sunday, Mar 30 - Oct 5
Tuesday, May 27 - Nov 11
Wednesday, Jun 25 - Dec 17

Group Theory
Thursday, Jun 12 - Sep 11

Contest Preparation: Grades 6-12

MATHCOUNTS/AMC 8 Basics
Sunday, Mar 23 - Jun 15
Wednesday, Apr 16 - Jul 2
Friday, May 23 - Aug 15
Monday, Jun 2 - Aug 18
Thursday, Jun 12 - Aug 28
Sunday, Jun 22 - Sep 21
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)

MATHCOUNTS/AMC 8 Advanced
Friday, Apr 11 - Jun 27
Sunday, May 11 - Aug 10
Tuesday, May 27 - Aug 12
Wednesday, Jun 11 - Aug 27
Sunday, Jun 22 - Sep 21
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)

AMC 10 Problem Series
Tuesday, Mar 4 - May 20
Monday, Mar 31 - Jun 23
Friday, May 9 - Aug 1
Sunday, Jun 1 - Aug 24
Thursday, Jun 12 - Aug 28
Tuesday, Jun 17 - Sep 2
Sunday, Jun 22 - Sep 21 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Monday, Jun 23 - Sep 15
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)

AMC 10 Final Fives
Sunday, May 11 - Jun 8
Tuesday, May 27 - Jun 17
Monday, Jun 30 - Jul 21

AMC 12 Problem Series
Tuesday, May 27 - Aug 12
Thursday, Jun 12 - Aug 28
Sunday, Jun 22 - Sep 21
Wednesday, Aug 6 - Oct 22

AMC 12 Final Fives
Sunday, May 18 - Jun 15

F=ma Problem Series
Wednesday, Jun 11 - Aug 27

WOOT Programs
Visit the pages linked for full schedule details for each of these programs!


MathWOOT Level 1
MathWOOT Level 2
ChemWOOT
CodeWOOT
PhysicsWOOT

Programming

Introduction to Programming with Python
Monday, Mar 24 - Jun 16
Thursday, May 22 - Aug 7
Sunday, Jun 15 - Sep 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Tuesday, Jun 17 - Sep 2
Monday, Jun 30 - Sep 22

Intermediate Programming with Python
Sunday, Jun 1 - Aug 24
Monday, Jun 30 - Sep 22

USACO Bronze Problem Series
Tuesday, May 13 - Jul 29
Sunday, Jun 22 - Sep 1

Physics

Introduction to Physics
Sunday, Mar 30 - Jun 22
Wednesday, May 21 - Aug 6
Sunday, Jun 15 - Sep 14
Monday, Jun 23 - Sep 15

Physics 1: Mechanics
Tuesday, Mar 25 - Sep 2
Thursday, May 22 - Oct 30
Monday, Jun 23 - Dec 15

Relativity
Sat & Sun, Apr 26 - Apr 27 (4:00 - 7:00 pm ET/1:00 - 4:00pm PT)
Mon, Tue, Wed & Thurs, Jun 23 - Jun 26 (meets every day of the week!)
0 replies
jlacosta
Mar 2, 2025
0 replies
AMC- IMO preparation
asyaela.   9
N 21 minutes ago by Schintalpati
I'm a ninth grader, and I recently attempted the AMC 12, getting 18 questions correct and leaving 7 empty. I started working on Olympiad math in November and currently dedicate about two hours per day to preparation. I'm feeling a bit demotivated, but if it's possible for me to reach IMO level, I'd be willing to put in more time. How realistic is it for me to get there, and how much study would it typically take?
9 replies
+3 w
asyaela.
3 hours ago
Schintalpati
21 minutes ago
Tennessee Math Tournament (TMT) Online 2025
TennesseeMathTournament   29
N 27 minutes ago by NashvilleSC
Hello everyone! We are excited to announce a new competition, the Tennessee Math Tournament, created by the Tennessee Math Coalition! Anyone can participate in the virtual competition for free.

The testing window is from March 22nd to April 5th, 2025. Virtual competitors may participate in the competition at any time during that window.

The virtual competition consists of three rounds: Individual, Bullet, and Team. The Individual Round is 60 minutes long and consists of 30 questions (AMC 10 level). The Bullet Round is 20 minutes long and consists of 80 questions (Mathcounts Chapter level). The Team Round is 30 minutes long and consists of 16 questions (AMC 12 level). Virtual competitors may compete in teams of four, or choose to not participate in the team round.

To register and see more information, click here!

If you have any questions, please email connect@tnmathcoalition.org or reply to this thread!
29 replies
TennesseeMathTournament
Mar 9, 2025
NashvilleSC
27 minutes ago
AIME score for college apps
Happyllamaalways   75
N 37 minutes ago by hashbrown2009
What good colleges do I have a chance of getting into with an 11 on AIME? (Any chances for Princeton)

Also idk if this has weight but I had the highest AIME score in my school.
75 replies
Happyllamaalways
Mar 13, 2025
hashbrown2009
37 minutes ago
AMC 8 discussion
Jaxman8   42
N an hour ago by mpcnotnpc
Discuss the AMC 8 below!
42 replies
Jaxman8
Jan 29, 2025
mpcnotnpc
an hour ago
No more topics!
MOCK AMC I RESULTS
Lucky707   21
N Jan 31, 2005 by pilot
Here are the results with a question-by-question analysis.

NOTE: If you have not taken the test and would like to, please see the other thread for the questions as this thread has the questions ALONG with the answers.

TOTAL NUMBER OF CONTESTANTS: 15

MOCK AMC I-
CREATED BY: Lucky707 and Singthesorrow
DATE ADMINISTERED: January 30, 2005

Rules:
(Standard AMC Rules)
1. You have 75 minutes to complete the test.
2. Only non-programmable calculators are allowed.
3. 6 points for a correct answer, 2.5 points for a question left unanswered and 0 points for an incorrect answer so it is in your best interest NOT to guess.
4. All other standard AMC Rules apply... We've had plenty of Mock AMCs so you should know the rules by now.
5. At 5:15, you must submit the test. If you do not, it will not count. I will allow a few extra minutes for slow modems, etc. but please submit the test at 5:15.
6. My AOL is hezgotzbeenz if anyone has any questions during the test.

1. The I.Q. of a Martian varies directly with the square of the number of eyes it has. If a Martian with 5 eyes has an I.Q. of 75, find the IQ of a Martian with 8 eyes.
(A) 120
(B) 128
(C) 144
(D) 192
(E) 200

Generally well done, those who missed this question did so because they didn't read it and multiplied 8 by 15 to get 120.
ANSWER: D
AVERAGE SCORE: 5.2 (13 correct/15 attempted)


2. The number of cubic feet in the volume of a cube is the same as the number of square inches of its surface area. What is the length of an edge of it expressed in feet? (NOTE: 12 inches = 1 foot)
(A) 6
(B) 36
(C) 72
(D) 108
(E) 864

Also fairly well done. Those who missed this one did not convert at some point down the line.
ANSWER: E
AVERAGE SCORE: 4.8 (12/15)


3. The first two terms of a geometric progression are $\sqrt{3}$ and the fourth root of three. What is the fifth term?
(A) $\frac{1}{\sqrt{3}}$
(B) $\sqrt{3}$
(C) fourth root of 3
(D) fourth root of $\frac{1}{3}$
(E) tenth root of 3

This was somewhat of a trick question. Those who were tricked by it chose E but if we look at the common ratio and work it out as usual, we get the correct answer of A.
CORRECT ANSWER: A
AVERAGE SCORE: 4.8 (12/15)


4. A leaky faucet drips at a steady rate. If we measure from the instant the first drop hits the bottom, to the instant the last drop falls, it takes 36 seconds for 9 drops to fall. How many seconds would it take for 12 drops to fall?
(A) 42 seconds
(B) 44 seconds
(C) 48 seconds
(D) 49.5 seconds
(E) 54 seconds

This question was nasty. So many places where one could make an error. Here is the solution:
We measure from the instant the first drop falls to the instant the last drop falls. 36 seconds for 9 drops to fall means 36/8 seconds between drops hitting the ground. That's the first place one can make a mistake. The other place is multiplying by 12 instead of 11. The answer is 49.5
ANSWER: D
AVERAGE SCORE: 3.0 (7/14)


5. A triangle has sides of length 15, 20 and 25 units. It is inscribed in a circle. What is the circle's radius?
(A) $21/2$
(B) $43/3$
(C) $64/7$
(D) $25/2$
(E) $25$

This question was very well done. After realizing that the triangle is a right triangle all one needs to do is divide the hypotenuse by 2 to get the answer.
ANSWER: D
AVERAGE SCORE: 6.0 (15/15)


6. How many zeroes are at the end of the product of the first fifty positive integers?
(A) 5
(B) 12
(C) 17
(D) 35
(E) 47

I don't think the average would be this high in real life for this type of question but I was sure that AOPSers could steamroll through this question... and steamroll they did!
ANSWER: B
AVERAGE SCORE: 6.0 (15/15)


7. If $100! = (100)(99)(98)...(2)(1)$, then what is the largest value of $x$ such that $2^x$ divides $100!$ evenly?
(A) 50
(B) 97
(C) 98
(D) 99
(E) 100

See comment above. 6 and 7 were quite similar actually. Nobody made an addition error so everyone got this right.
ANSWER: B
AVERAGE SCORE: 6.0 (15/15)


8. I have a red shirt, a blue shirt and a green shirt. I also have blue jeans, black jeans and khakis. How many possible combinations can I make if I absolutely hate wearing the blue shirt with the blue jeans?
(A) 5
(B) 8
(C) 14
(D) 16
(E) 17

This one was a gimme.
ANSWER: B
AVERAGE SCORE: 6.0 (15/15)


9. Let $x$ be the smallest perfect square that has 140 as a divisor. Then
(a) $0 < x < 1000$
(b) $1000 \leq x < 2000$
(c) $2000 \leq x < 3000$
(d) $3000 \leq x < 4000$
(e) $4000 \leq x < 5000$

Wow... there were a lot of prime factorization questions early on. Realizing that 140 = 7*5*2^2.... we find that 7^2*5^2*2^2 will be a perfect square. And that value is 4900, making the answer E.
ANSWER: E
AVERAGE SCORE: 6.0 (15/15)


10. What is the sum of the first twenty powers of 2 (2^1 + 2^2 + 2^3 ... + 2^20)?
(A) $2^{19} - 2$
(B) $2^{20} - 2$
(C) $2^{21} + 2$
(D) $2^{21} - 2$
(E) None of the above

This question can easily be done by induction... or if you want to stay honest (which is not a good idea during the AMC)... geometric series.
ANSWER: D
AVERAGE SCORE: 5.2 (13/15)


11. $9ax+7y=12$ and $4x+2by=9$ represent the same line. $a$ and $b$ can both be expressed as fractions in lowest terms. What is the numerator of $a$ plus the denominator of $b$ equal to?
(A) 212
(B) 144
(C) 72
(D) 48
(E) 24

Generally those who attempted this question got it right. One person got it wrong and I'm assuming it's because he forgot to reduce.
ANSWER: E
AVERAGE SCORE: 5.1 (12/13)

12. A square of side length 8 has its corners cut off to make a regular octagon. Find the area of the octagon to the nearest whole number.
(A) 32
(B) 48
(C) 53
(D) 55
(E) 59

Anyone who attempted this question got it right. The trick was not to be confused and make the cut along the midpoints or else it would be a hexagon. Luckily, nobody fell for that trick.
ANSWER: C
AVERAGE SCORE: 5.5 (13/13)


13. A piece of string is cut in two at a random point. What is the probability that the larger piece is at least $x$ times as large as the shorter piece? (where $x \geq 1$)
(A) $\frac{2}{1+\sqrt{5}}$
(B) $\frac{x^2+x-1}{x}$
(C) $\frac{2}{x+1}$
(D) $\frac{3}{x+2}$
(E) None of the Above

This is the type of question thta is far easier when it is a multiple choice question than when it is a written question. After a little bit of trial and error, one eventually arrives at the correct answer.
CORRECT ANSWER: C
AVERAGE SCORE: 4.5 (10/12)


14. Let $S$ denote the sum of all distinct four-digit numbers that contain only 1,2,3,4 or 5, each at most once. Then
(A) $S \leq 200,000$
(B) $200,000 < S \leq 300,000$
(C) $300,000 < S \leq 400,000$
(D) $400,000 < S \leq 500,000$
(E) $500,000 < S$

A disguised combinatorics problem here. The answer is C but the number itself is so very close to D that it is easy to make a mistake if using the wrong method. Luckily, nobody made THAT mistake (Even though few got it wrong).
ANSWER: C
AVERAGE SCORE: 3.9 (8/11)


15. How many digits are in $a^b$ if $b=\sqrt{9530}$ and $\log a$ ~ $2.8356$
(A) 277
(B) 278
(C) 282
(D) 283
(E) 284

Once again, a question AOPS knows how to do but I can't say the same about the rest of the mathematical world. If you know how to do this, which almost everyone did, then you will get the correct answer (unless you forget to take the floor, which nobody did!)
ANSWER: A
AVERAGE SCORE: 5.5 (13/13)


16. If $\frac{1}{(a)(b+1)} + \frac{1}{(b)(a+1)} = \frac{1}{(a+1)(b+1)}$, what is the value of $\frac{1}{a} + \frac{1}{b}$?
(A) $-1$
(B) $1$
(C) $2 \sqrt{5}$
(D) $-2 \sqrt{5}$
(E) Undefined

This was a nice question. Anyone who attempted it got it right but it wasn't exactly easy.
ANSWER: A
AVERAGE SCORE: 4.8 (10/10)


17. Let the parabola $x^2-11x+28$ meet the $x$-axis at two distinct points, $A$ and $B$. Let the vertex of said parabola be $C$. The area of triangle $ABC$ can be expressed as $\frac{f}{g}$, where $f$ and $g$ are relatively prime. $f$ and $g$ are both...
(A) Prime (but not in the form $2^n-1$)
(B) Perfect Squares
(C) Perfect Cubes
(D) In the form $2^n-1$
(E) None of the Above

A question dealing with co-ordinates. After finding the vertex of the parabola, it becomes pretty straightforward. The answer is 27/8, which corresponds to C.
ANSWER: C
AVERAGE SCORE: 5.0 (12/14)


18. Let $x$ be the number of positive integral factors of $90^9$. The sum of the digits of $x$ equals
(A) 7
(B) 10
(C) 13
(D) 16
(E) None of the Above

ANSWER: B (I believe it was 1900)
AVERAGE SCORE: 5.3 (12/12)


19. Bette visits her friend Keith and the nreturns home by the same route. She always walks 2 km/h when going uphill, 6 km/h when going downhill, and 3 km/h when on level ground. If her total walking time is 6 hours, then the total distance she walks, in km, is
(A) 9
(B) 12
(C) 18
(D) 22
(E) 36

I like this question. One must realize that she will walk the same distance uphill and downhill. If one finds her average speed uphill and downhill, it will be 3 km/h... the same as on level ground! Thus, in 6 hours, she walks 18 km.
ANSWER: C
AVERAGE SCORE: 3.7 (8/12)


20. If $JKLMN$ is a regular pentagon with side length $a$ and diagonal length $b$, what is the value of $\frac{b}{a} - \frac{a}{b}$?
(A) $\frac{3}{4}$
(B) 1
(C) $\frac{4}{3}$
(D) 2
(E) Depends on $a$

One could approach this using Ptolemy's Theorem, trigonometry or just having seen it before. E was intended to throw some people off but wasn't that effective.
ANSWER: B
AVERAGE SCORE: 4.7 (10/11)


21. Let $P$ be a point outside circle $O$. Draw two tangents from $P$ such that they intersect $O$ at $A$ and $B$. If $AP=BP=7$ and the larger of the two arcs $AB$ is twice as large as the smaller of the two arcs, find the length of $AB$.
(A) $6$
(B) $7 \sqrt{3}$
(C) $\frac{11 \sqrt{3}}{2}$
(D) $7$
(E) None of the Above

I came up with this geometry question myself after looking at a random theorem.
ANSWER: D
AVERAGE SCORE: 4.3 (9/11)


22. The system of equations
$4y+gx=34$
$\frac{g^4x^2-16y^2}{g^2x+4y} = -10$
has how many solutions where $g$, $x$ and $y$ are all integers?
(A) $0$
(B) $1$
(C) $2$
(D) $4$
(E) None of the Above

I presented this question for my math class and wow was it hard. This is actually a spinoff of an old Cayley question but I like my version better. The only solution is (2,3,7)
ANSWER: B
AVERAGE SCORE: 3.0 (5/9)


23. How many arrangements of the letters of the word "INDEPENDENT" have no adjacent E's?
(A) $\frac{(2)(7!)}{5!}$
(B) $\frac{11!}{9!-2!}$
(C) $\frac{11!}{3!}$
(D) $\frac{11!}{3!2!}$
(E) None of the Above

I swear I made up options B, C and D off the top of my head. Question was well done after people figured out that I was randomly putting factorials wherever I wanted.
ANSWER: E
AVERAGE SCORE: 4.4 (9/10)


24. If $f(x-y)=f(x)f(y)$ for all $x$ and $y$, and $f(x)$ never equals $0$, then $f(3)$ equals...
(A) $-3$
(B) $3$
(C) $9$
(D) :pm: 1
(E) None of the above

This one was a killer. The only person to get this correct was towersfreak2006. Most people determined f(x)^2=1 correctly but then mistakenly thought f(x) = :pm: 1

But f(x) is a function. How can f(x) = :pm: 1

The correct method involves setting y = x/2 and working from there. Try it yourself. The correct answer is simply 1.

ANSWER: E
AVERAGE SCORE: 1.2 (1/10)


25. Pieces of paper numbered from $1$ to $100$ are each placed in one of three hats such that there is at least one piece of paper in each hat. In how many ways can this be done such that if two hats are selected and a card is taken from each, then the knowledge of their sum alone is always sufficient to identify the third hat?
(A) $6$
(B) $12$
(C) $18$
(D) $24$
(E) None of the Above

Nobody attempted this question. I put it on here to simulate an extremely difficult question 25 and to prevent any perfects. I felt that if you can get perfect on AMC10, you don't really need practice for it.

SOURCE OF THIS QUESTION: IMO 2000 B1 ... yeah, don't feel upset if you didn't solve it... it was only supposed to be a simulation. Nobody got anywhere near perfect anyway.

And for whitehorseking88, here's what the question is asking: Let's say you had 3 hats and you knew the sum of the numbers in all 3. Let's say the sums were A, B and C respectively. Now if you randomly pick two pieces of paper out of any two hats and are told their sum, you are able to identify which hat has sum A, which hat has sum B and which hat has sum C. Now how many possible sums A, B and C exist for this to work?

(HINT: Try sums like 1 and 2...)

ANSWER: I won't tell ya... IMO 2000 if you really need to know. kalva.demon.co.uk
AVERAGE SCORE: 2.5 (0 correct/ 0 attempted)


Results up soon.
21 replies
Lucky707
Jan 29, 2005
pilot
Jan 31, 2005
MOCK AMC I RESULTS
G H J
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Lucky707
113 posts
#1 • 2 Y
Y by Adventure10, Mango247
Here are the results with a question-by-question analysis.

NOTE: If you have not taken the test and would like to, please see the other thread for the questions as this thread has the questions ALONG with the answers.

TOTAL NUMBER OF CONTESTANTS: 15

MOCK AMC I-
CREATED BY: Lucky707 and Singthesorrow
DATE ADMINISTERED: January 30, 2005

Rules:
(Standard AMC Rules)
1. You have 75 minutes to complete the test.
2. Only non-programmable calculators are allowed.
3. 6 points for a correct answer, 2.5 points for a question left unanswered and 0 points for an incorrect answer so it is in your best interest NOT to guess.
4. All other standard AMC Rules apply... We've had plenty of Mock AMCs so you should know the rules by now.
5. At 5:15, you must submit the test. If you do not, it will not count. I will allow a few extra minutes for slow modems, etc. but please submit the test at 5:15.
6. My AOL is hezgotzbeenz if anyone has any questions during the test.

1. The I.Q. of a Martian varies directly with the square of the number of eyes it has. If a Martian with 5 eyes has an I.Q. of 75, find the IQ of a Martian with 8 eyes.
(A) 120
(B) 128
(C) 144
(D) 192
(E) 200

Generally well done, those who missed this question did so because they didn't read it and multiplied 8 by 15 to get 120.
ANSWER: D
AVERAGE SCORE: 5.2 (13 correct/15 attempted)


2. The number of cubic feet in the volume of a cube is the same as the number of square inches of its surface area. What is the length of an edge of it expressed in feet? (NOTE: 12 inches = 1 foot)
(A) 6
(B) 36
(C) 72
(D) 108
(E) 864

Also fairly well done. Those who missed this one did not convert at some point down the line.
ANSWER: E
AVERAGE SCORE: 4.8 (12/15)


3. The first two terms of a geometric progression are $\sqrt{3}$ and the fourth root of three. What is the fifth term?
(A) $\frac{1}{\sqrt{3}}$
(B) $\sqrt{3}$
(C) fourth root of 3
(D) fourth root of $\frac{1}{3}$
(E) tenth root of 3

This was somewhat of a trick question. Those who were tricked by it chose E but if we look at the common ratio and work it out as usual, we get the correct answer of A.
CORRECT ANSWER: A
AVERAGE SCORE: 4.8 (12/15)


4. A leaky faucet drips at a steady rate. If we measure from the instant the first drop hits the bottom, to the instant the last drop falls, it takes 36 seconds for 9 drops to fall. How many seconds would it take for 12 drops to fall?
(A) 42 seconds
(B) 44 seconds
(C) 48 seconds
(D) 49.5 seconds
(E) 54 seconds

This question was nasty. So many places where one could make an error. Here is the solution:
We measure from the instant the first drop falls to the instant the last drop falls. 36 seconds for 9 drops to fall means 36/8 seconds between drops hitting the ground. That's the first place one can make a mistake. The other place is multiplying by 12 instead of 11. The answer is 49.5
ANSWER: D
AVERAGE SCORE: 3.0 (7/14)


5. A triangle has sides of length 15, 20 and 25 units. It is inscribed in a circle. What is the circle's radius?
(A) $21/2$
(B) $43/3$
(C) $64/7$
(D) $25/2$
(E) $25$

This question was very well done. After realizing that the triangle is a right triangle all one needs to do is divide the hypotenuse by 2 to get the answer.
ANSWER: D
AVERAGE SCORE: 6.0 (15/15)


6. How many zeroes are at the end of the product of the first fifty positive integers?
(A) 5
(B) 12
(C) 17
(D) 35
(E) 47

I don't think the average would be this high in real life for this type of question but I was sure that AOPSers could steamroll through this question... and steamroll they did!
ANSWER: B
AVERAGE SCORE: 6.0 (15/15)


7. If $100! = (100)(99)(98)...(2)(1)$, then what is the largest value of $x$ such that $2^x$ divides $100!$ evenly?
(A) 50
(B) 97
(C) 98
(D) 99
(E) 100

See comment above. 6 and 7 were quite similar actually. Nobody made an addition error so everyone got this right.
ANSWER: B
AVERAGE SCORE: 6.0 (15/15)


8. I have a red shirt, a blue shirt and a green shirt. I also have blue jeans, black jeans and khakis. How many possible combinations can I make if I absolutely hate wearing the blue shirt with the blue jeans?
(A) 5
(B) 8
(C) 14
(D) 16
(E) 17

This one was a gimme.
ANSWER: B
AVERAGE SCORE: 6.0 (15/15)


9. Let $x$ be the smallest perfect square that has 140 as a divisor. Then
(a) $0 < x < 1000$
(b) $1000 \leq x < 2000$
(c) $2000 \leq x < 3000$
(d) $3000 \leq x < 4000$
(e) $4000 \leq x < 5000$

Wow... there were a lot of prime factorization questions early on. Realizing that 140 = 7*5*2^2.... we find that 7^2*5^2*2^2 will be a perfect square. And that value is 4900, making the answer E.
ANSWER: E
AVERAGE SCORE: 6.0 (15/15)


10. What is the sum of the first twenty powers of 2 (2^1 + 2^2 + 2^3 ... + 2^20)?
(A) $2^{19} - 2$
(B) $2^{20} - 2$
(C) $2^{21} + 2$
(D) $2^{21} - 2$
(E) None of the above

This question can easily be done by induction... or if you want to stay honest (which is not a good idea during the AMC)... geometric series.
ANSWER: D
AVERAGE SCORE: 5.2 (13/15)


11. $9ax+7y=12$ and $4x+2by=9$ represent the same line. $a$ and $b$ can both be expressed as fractions in lowest terms. What is the numerator of $a$ plus the denominator of $b$ equal to?
(A) 212
(B) 144
(C) 72
(D) 48
(E) 24

Generally those who attempted this question got it right. One person got it wrong and I'm assuming it's because he forgot to reduce.
ANSWER: E
AVERAGE SCORE: 5.1 (12/13)

12. A square of side length 8 has its corners cut off to make a regular octagon. Find the area of the octagon to the nearest whole number.
(A) 32
(B) 48
(C) 53
(D) 55
(E) 59

Anyone who attempted this question got it right. The trick was not to be confused and make the cut along the midpoints or else it would be a hexagon. Luckily, nobody fell for that trick.
ANSWER: C
AVERAGE SCORE: 5.5 (13/13)


13. A piece of string is cut in two at a random point. What is the probability that the larger piece is at least $x$ times as large as the shorter piece? (where $x \geq 1$)
(A) $\frac{2}{1+\sqrt{5}}$
(B) $\frac{x^2+x-1}{x}$
(C) $\frac{2}{x+1}$
(D) $\frac{3}{x+2}$
(E) None of the Above

This is the type of question thta is far easier when it is a multiple choice question than when it is a written question. After a little bit of trial and error, one eventually arrives at the correct answer.
CORRECT ANSWER: C
AVERAGE SCORE: 4.5 (10/12)


14. Let $S$ denote the sum of all distinct four-digit numbers that contain only 1,2,3,4 or 5, each at most once. Then
(A) $S \leq 200,000$
(B) $200,000 < S \leq 300,000$
(C) $300,000 < S \leq 400,000$
(D) $400,000 < S \leq 500,000$
(E) $500,000 < S$

A disguised combinatorics problem here. The answer is C but the number itself is so very close to D that it is easy to make a mistake if using the wrong method. Luckily, nobody made THAT mistake (Even though few got it wrong).
ANSWER: C
AVERAGE SCORE: 3.9 (8/11)


15. How many digits are in $a^b$ if $b=\sqrt{9530}$ and $\log a$ ~ $2.8356$
(A) 277
(B) 278
(C) 282
(D) 283
(E) 284

Once again, a question AOPS knows how to do but I can't say the same about the rest of the mathematical world. If you know how to do this, which almost everyone did, then you will get the correct answer (unless you forget to take the floor, which nobody did!)
ANSWER: A
AVERAGE SCORE: 5.5 (13/13)


16. If $\frac{1}{(a)(b+1)} + \frac{1}{(b)(a+1)} = \frac{1}{(a+1)(b+1)}$, what is the value of $\frac{1}{a} + \frac{1}{b}$?
(A) $-1$
(B) $1$
(C) $2 \sqrt{5}$
(D) $-2 \sqrt{5}$
(E) Undefined

This was a nice question. Anyone who attempted it got it right but it wasn't exactly easy.
ANSWER: A
AVERAGE SCORE: 4.8 (10/10)


17. Let the parabola $x^2-11x+28$ meet the $x$-axis at two distinct points, $A$ and $B$. Let the vertex of said parabola be $C$. The area of triangle $ABC$ can be expressed as $\frac{f}{g}$, where $f$ and $g$ are relatively prime. $f$ and $g$ are both...
(A) Prime (but not in the form $2^n-1$)
(B) Perfect Squares
(C) Perfect Cubes
(D) In the form $2^n-1$
(E) None of the Above

A question dealing with co-ordinates. After finding the vertex of the parabola, it becomes pretty straightforward. The answer is 27/8, which corresponds to C.
ANSWER: C
AVERAGE SCORE: 5.0 (12/14)


18. Let $x$ be the number of positive integral factors of $90^9$. The sum of the digits of $x$ equals
(A) 7
(B) 10
(C) 13
(D) 16
(E) None of the Above

ANSWER: B (I believe it was 1900)
AVERAGE SCORE: 5.3 (12/12)


19. Bette visits her friend Keith and the nreturns home by the same route. She always walks 2 km/h when going uphill, 6 km/h when going downhill, and 3 km/h when on level ground. If her total walking time is 6 hours, then the total distance she walks, in km, is
(A) 9
(B) 12
(C) 18
(D) 22
(E) 36

I like this question. One must realize that she will walk the same distance uphill and downhill. If one finds her average speed uphill and downhill, it will be 3 km/h... the same as on level ground! Thus, in 6 hours, she walks 18 km.
ANSWER: C
AVERAGE SCORE: 3.7 (8/12)


20. If $JKLMN$ is a regular pentagon with side length $a$ and diagonal length $b$, what is the value of $\frac{b}{a} - \frac{a}{b}$?
(A) $\frac{3}{4}$
(B) 1
(C) $\frac{4}{3}$
(D) 2
(E) Depends on $a$

One could approach this using Ptolemy's Theorem, trigonometry or just having seen it before. E was intended to throw some people off but wasn't that effective.
ANSWER: B
AVERAGE SCORE: 4.7 (10/11)


21. Let $P$ be a point outside circle $O$. Draw two tangents from $P$ such that they intersect $O$ at $A$ and $B$. If $AP=BP=7$ and the larger of the two arcs $AB$ is twice as large as the smaller of the two arcs, find the length of $AB$.
(A) $6$
(B) $7 \sqrt{3}$
(C) $\frac{11 \sqrt{3}}{2}$
(D) $7$
(E) None of the Above

I came up with this geometry question myself after looking at a random theorem.
ANSWER: D
AVERAGE SCORE: 4.3 (9/11)


22. The system of equations
$4y+gx=34$
$\frac{g^4x^2-16y^2}{g^2x+4y} = -10$
has how many solutions where $g$, $x$ and $y$ are all integers?
(A) $0$
(B) $1$
(C) $2$
(D) $4$
(E) None of the Above

I presented this question for my math class and wow was it hard. This is actually a spinoff of an old Cayley question but I like my version better. The only solution is (2,3,7)
ANSWER: B
AVERAGE SCORE: 3.0 (5/9)


23. How many arrangements of the letters of the word "INDEPENDENT" have no adjacent E's?
(A) $\frac{(2)(7!)}{5!}$
(B) $\frac{11!}{9!-2!}$
(C) $\frac{11!}{3!}$
(D) $\frac{11!}{3!2!}$
(E) None of the Above

I swear I made up options B, C and D off the top of my head. Question was well done after people figured out that I was randomly putting factorials wherever I wanted.
ANSWER: E
AVERAGE SCORE: 4.4 (9/10)


24. If $f(x-y)=f(x)f(y)$ for all $x$ and $y$, and $f(x)$ never equals $0$, then $f(3)$ equals...
(A) $-3$
(B) $3$
(C) $9$
(D) :pm: 1
(E) None of the above

This one was a killer. The only person to get this correct was towersfreak2006. Most people determined f(x)^2=1 correctly but then mistakenly thought f(x) = :pm: 1

But f(x) is a function. How can f(x) = :pm: 1

The correct method involves setting y = x/2 and working from there. Try it yourself. The correct answer is simply 1.

ANSWER: E
AVERAGE SCORE: 1.2 (1/10)


25. Pieces of paper numbered from $1$ to $100$ are each placed in one of three hats such that there is at least one piece of paper in each hat. In how many ways can this be done such that if two hats are selected and a card is taken from each, then the knowledge of their sum alone is always sufficient to identify the third hat?
(A) $6$
(B) $12$
(C) $18$
(D) $24$
(E) None of the Above

Nobody attempted this question. I put it on here to simulate an extremely difficult question 25 and to prevent any perfects. I felt that if you can get perfect on AMC10, you don't really need practice for it.

SOURCE OF THIS QUESTION: IMO 2000 B1 ... yeah, don't feel upset if you didn't solve it... it was only supposed to be a simulation. Nobody got anywhere near perfect anyway.

And for whitehorseking88, here's what the question is asking: Let's say you had 3 hats and you knew the sum of the numbers in all 3. Let's say the sums were A, B and C respectively. Now if you randomly pick two pieces of paper out of any two hats and are told their sum, you are able to identify which hat has sum A, which hat has sum B and which hat has sum C. Now how many possible sums A, B and C exist for this to work?

(HINT: Try sums like 1 and 2...)

ANSWER: I won't tell ya... IMO 2000 if you really need to know. kalva.demon.co.uk
AVERAGE SCORE: 2.5 (0 correct/ 0 attempted)


Results up soon.
This post has been edited 1 time. Last edited by Lucky707, Jan 29, 2005, 11:25 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Lucky707
113 posts
#2 • 2 Y
Y by Adventure10, Mango247
Results -

TOTAL NUMBER OF CONTESTANTS: 15
AVERAGE SCORE: 116.43

AIME CUTOFF: 120 (Hey, why not? Sure... two people have scores of 119 but 8/15 make it... it works out so nicely)

FIRST: alan - 134.5
SECOND: towersfreak2006 - 133.5
*frankthetank - 133.5
FOURTH: Danbert - 128.5
FIFTH: *d343seven - 125
white_horse_king88 - 125
SEVENTH: keta - 124
EIGHTH: interesting_move - 122.5

Barely missed the cut - *rep123max (119) and *Yrael (119)

* - took it early

If you need to see where you rank, the other scores were (without names): 103.5, 100.5, 100.5, 92, 85.5

pkothari13 - unofficial score 122.5
This post has been edited 2 times. Last edited by Lucky707, Jan 30, 2005, 12:56 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
1234567890
1553 posts
#3 • 2 Y
Y by Adventure10, Mango247
i'd like to respectfully disagree with #4. It deals with gravity, and does factor that in. This would definitely not be an AMC problem, for test writers usually avoid all physics.
If you use kinematic equations, you will see the answer is none.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
h_s_potter2002
1306 posts
#4 • 2 Y
Y by Adventure10, Mango247
I hate stupid mistakes soooooooo much. I could have gotten 18 extra points. And one of the stupid mistakes was me writing down the wrong answer! I just hope I remember to check my work, and look over my answer on the real AMC this Tuesday.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
eryaman
1130 posts
#5 • 2 Y
Y by Adventure10, Mango247
shoot. I couldn't make it in time to take it :(
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
white_horse_king88
1779 posts
#6 • 2 Y
Y by Adventure10, Mango247
Actually, it wouldn't matter if gravity was there or not because he specifies the 9 drops from the point the first one hits to the last one falls. That's a period, so hypothetically, he already included gravity as a factor. Now, all you have to do is figure out the time from one drop to the next and multiply that by 11.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
pkothari13
709 posts
#7 • 2 Y
Y by Adventure10, Mango247
shoooooot! i did only the distance going one way on number 19.... yes.. stupid mistakes will soon kill us all...! GAH!
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Lucky707
113 posts
#8 • 2 Y
Y by Adventure10, Mango247
Erm yes... I gave you 115.5 before factoring in the INDEPENDENT... but when I added 6 for the question I forgot to take away the original 2.5 that I gave you. So I gave you 2.5 points extra. I caught it after making a spreadsheet of all the scores however.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
alan
228 posts
#9 • 2 Y
Y by Adventure10, Mango247
Rep123max wrote:
I'm extremely stupid.

well, at least you didnt read #10 as "What is the sum of 2 (2^1 + 2^2 + 2^3 ... + 2^20)"
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
keta
244 posts
#10 • 2 Y
Y by Adventure10, Mango247
Argh,.... Number 19 about the walking one really got me. I only calculated the distance one way like pkothari13 and foolishly choose A. $9$. Oh well....

Also, to everyone who took this, would you rank this mock AMC as harder or easier than the typical AMC 10? It seemed a little more difficult to me since there was alot of number theory and functional analysis not found on the AMC 10. (Let's not mention #25-Lucky707, how could you do this to us!!??? :lol: JK).
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
pkothari13
709 posts
#11 • 2 Y
Y by Adventure10, Mango247
alan wrote:
Rep123max wrote:
I'm extremely stupid.

well, at least you didnt read #10 as "What is the sum of 2 (2^1 + 2^2 + 2^3 ... + 2^20)"

omg! dude, thats what i thought all throughout the test... and then after spending like 5-10 minutes on it, i realized what it meant!
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
pkothari13
709 posts
#12 • 3 Y
Y by Adventure10, Adventure10, Mango247
hey Lucky707, do you know if i got number 25 right?
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
pkothari13
709 posts
#13 • 2 Y
Y by Adventure10, Mango247
lol oh yeah, how did all of you do that INDEPENDENT one.. I just guessed on that one hoping i'd get it right.... :D guess that didnt work out...
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Lucky707
113 posts
#14 • 2 Y
Y by Adventure10, Mango247
Yes you got #25 right.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
d343seven
237 posts
#15 • 2 Y
Y by Adventure10, Mango247
when i took it i had an answer for the INDEPENDENT one. what's the actual answer?
i don't wanna do it now, but my answer was something like 11! - something
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
The Original Pi Guy
172 posts
#16 • 2 Y
Y by Adventure10, Mango247
Could someone explain to me how you do number 15?
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
plokoon51
135 posts
#17 • 2 Y
Y by Adventure10, Mango247
Could someone explain to me how to come up with the answer to #22?
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
chenwb
67 posts
#18 • 3 Y
Y by Adventure10, Adventure10, Mango247
For the INDEPENDENT one, I had:

Lay down the 8 "non-E" letters first. There are $\frac{8}{3!2!3!}$ ways to do this (analogous to the number of possible words from rearranging MISSISSIPPI, if you've seen this before). Insert a wedge between each letter and at the ends. There are 9 wedges. Choosing 3 wedges to place the Es can occur in $\binom{9}{3}$ ways. The answer is then $\frac{8}{3!2!3!} \cdot \binom{9}{3}$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
plokoon51
135 posts
#19 • 2 Y
Y by Adventure10, Mango247
never mind on #22. STUPID MISTAKES!!! they will be the cause of the end of all civilization :P
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
chenwb
67 posts
#20 • 2 Y
Y by Adventure10, Mango247
For 15:

We take $[\log{a^b}]$, which will give us the greatest power of 10 less than $a^b$. To evaluate this, we simply rewrite as $b\log{a}$, and multiply out, leaving 276. Adding 1 for the extra digit (this is like how 1002, whose greatest power of 10 less than it is $10^3$, has 3+1=4 digits)., we have the final answer: 277 digits, or choice A.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
chenwb
67 posts
#21 • 2 Y
Y by Adventure10, Mango247
I was writing up 22...guess it won't be needed :P
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
pilot
152 posts
#22 • 2 Y
Y by Adventure10, Mango247
I'd like to see a solution for #22
Z K Y
N Quick Reply
G
H
=
a