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
k i Adding contests to the Contest Collections
dcouchman   1
N Apr 5, 2023 by v_Enhance
Want to help AoPS remain a valuable Olympiad resource? Help us add contests to AoPS's Contest Collections.

Find instructions and a list of contests to add here: https://artofproblemsolving.com/community/c40244h1064480_contests_to_add
1 reply
dcouchman
Sep 9, 2019
v_Enhance
Apr 5, 2023
k i Zero tolerance
ZetaX   49
N May 4, 2019 by NoDealsHere
Source: Use your common sense! (enough is enough)
Some users don't want to learn, some other simply ignore advises.
But please follow the following guideline:


To make it short: ALWAYS USE YOUR COMMON SENSE IF POSTING!
If you don't have common sense, don't post.


More specifically:

For new threads:


a) Good, meaningful title:
The title has to say what the problem is about in best way possible.
If that title occured already, it's definitely bad. And contest names aren't good either.
That's in fact a requirement for being able to search old problems.

Examples:
Bad titles:
- "Hard"/"Medium"/"Easy" (if you find it so cool how hard/easy it is, tell it in the post and use a title that tells us the problem)
- "Number Theory" (hey guy, guess why this forum's named that way¿ and is it the only such problem on earth¿)
- "Fibonacci" (there are millions of Fibonacci problems out there, all posted and named the same...)
- "Chinese TST 2003" (does this say anything about the problem¿)
Good titles:
- "On divisors of a³+2b³+4c³-6abc"
- "Number of solutions to x²+y²=6z²"
- "Fibonacci numbers are never squares"


b) Use search function:
Before posting a "new" problem spend at least two, better five, minutes to look if this problem was posted before. If it was, don't repost it. If you have anything important to say on topic, post it in one of the older threads.
If the thread is locked cause of this, use search function.

Update (by Amir Hossein). The best way to search for two keywords in AoPS is to input
[code]+"first keyword" +"second keyword"[/code]
so that any post containing both strings "first word" and "second form".


c) Good problem statement:
Some recent really bad post was:
[quote]$lim_{n\to 1}^{+\infty}\frac{1}{n}-lnn$[/quote]
It contains no question and no answer.
If you do this, too, you are on the best way to get your thread deleted. Write everything clearly, define where your variables come from (and define the "natural" numbers if used). Additionally read your post at least twice before submitting. After you sent it, read it again and use the Edit-Button if necessary to correct errors.


For answers to already existing threads:


d) Of any interest and with content:
Don't post things that are more trivial than completely obvious. For example, if the question is to solve $x^{3}+y^{3}=z^{3}$, do not answer with "$x=y=z=0$ is a solution" only. Either you post any kind of proof or at least something unexpected (like "$x=1337, y=481, z=42$ is the smallest solution). Someone that does not see that $x=y=z=0$ is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
Similar, posting "I have solved this problem" but not posting anything else is not welcome; it even looks that you just want to show off what a genius you are.

e) Well written and checked answers:
Like c) for new threads, check your solutions at least twice for mistakes. And after sending, read it again and use the Edit-Button if necessary to correct errors.



To repeat it: ALWAYS USE YOUR COMMON SENSE IF POSTING!


Everything definitely out of range of common sense will be locked or deleted (exept for new users having less than about 42 posts, they are newbies and need/get some time to learn).

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
AMC- IMO preparation
asyaela.   11
N 34 minutes ago by hashbrown2009
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?
11 replies
asyaela.
6 hours ago
hashbrown2009
34 minutes ago
2025 ROSS Program
scls140511   15
N 40 minutes ago by akliu
Since the application has ended, are we now free to discuss the problems and stats? How do you think this year's problems are?
15 replies
+1 w
scls140511
Yesterday at 2:36 AM
akliu
40 minutes ago
Line Perpendicular to Euler Line
tastymath75025   55
N an hour ago by ohiorizzler1434
Source: USA TSTST 2017 Problem 1, by Ray Li
Let $ABC$ be a triangle with circumcircle $\Gamma$, circumcenter $O$, and orthocenter $H$. Assume that $AB\neq AC$ and that $\angle A \neq 90^{\circ}$. Let $M$ and $N$ be the midpoints of sides $AB$ and $AC$, respectively, and let $E$ and $F$ be the feet of the altitudes from $B$ and $C$ in $\triangle ABC$, respectively. Let $P$ be the intersection of line $MN$ with the tangent line to $\Gamma$ at $A$. Let $Q$ be the intersection point, other than $A$, of $\Gamma$ with the circumcircle of $\triangle AEF$. Let $R$ be the intersection of lines $AQ$ and $EF$. Prove that $PR\perp OH$.

Proposed by Ray Li
55 replies
tastymath75025
Jun 29, 2017
ohiorizzler1434
an hour ago
Foot from vertex to Euler line
cjquines0   31
N an hour ago by pUssydestroyer777
Source: 2016 IMO Shortlist G5
Let $D$ be the foot of perpendicular from $A$ to the Euler line (the line passing through the circumcentre and the orthocentre) of an acute scalene triangle $ABC$. A circle $\omega$ with centre $S$ passes through $A$ and $D$, and it intersects sides $AB$ and $AC$ at $X$ and $Y$ respectively. Let $P$ be the foot of altitude from $A$ to $BC$, and let $M$ be the midpoint of $BC$. Prove that the circumcentre of triangle $XSY$ is equidistant from $P$ and $M$.
31 replies
cjquines0
Jul 19, 2017
pUssydestroyer777
an hour ago
ABMC 2025 IN-PERSON Contest (April 5th)
ilovepizza2020   4
N an hour ago by mynameisjefff
The 9th annual Acton-Boxborough Math Competition (ABMC) is quickly approaching! This year's ABMC will be held in-person at RJ Grey Junior High School, Acton, MA, on April 5th, 2025. The competition includes individual rounds and a team round, in which teams of 2-4 students participate. Anyone in grade 8 or below is welcome! You must register to compete. For more information about registration and the tentative schedule, please consult our website: https://abmathcompetitions.org/2025-contest/.

We offer prizes not only to top competitors; several of our sponsor prizes and educational awards are raffled among all in-person participants. Additionally, there are separate prizes for the top-scoring elementary schoolers.


For more information, visit https://abmathcompetitions.org/, especially the 2025 Competition page.
For the mailing list, visit https://abmathcompetitions.org/contact/.

Best,
ABMC Coordinators
4 replies
ilovepizza2020
2 hours ago
mynameisjefff
an hour ago
is this really supposed to be #13???
hgmium   3
N an hour ago by hashbrown2009
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_13

I managed to do all the other geo problems for that year besides this one
misplaced?
3 replies
hgmium
2 hours ago
hashbrown2009
an hour ago
Inequality => square
Rushil   12
N 2 hours ago by ohiorizzler1434
Source: INMO 1998 Problem 4
Suppose $ABCD$ is a cyclic quadrilateral inscribed in a circle of radius one unit. If $AB \cdot BC \cdot CD \cdot DA \geq 4$, prove that $ABCD$ is a square.
12 replies
Rushil
Oct 7, 2005
ohiorizzler1434
2 hours ago
p + q + r + s = 9 and p^2 + q^2 + r^2 + s^2 = 21
who   28
N 2 hours ago by asdf334
Source: IMO Shortlist 2005 problem A3
Four real numbers $ p$, $ q$, $ r$, $ s$ satisfy $ p+q+r+s = 9$ and $ p^{2}+q^{2}+r^{2}+s^{2}= 21$. Prove that there exists a permutation $ \left(a,b,c,d\right)$ of $ \left(p,q,r,s\right)$ such that $ ab-cd \geq 2$.
28 replies
who
Jul 8, 2006
asdf334
2 hours ago
H not needed
dchenmathcounts   44
N 3 hours ago by Ilikeminecraft
Source: USEMO 2019/1
Let $ABCD$ be a cyclic quadrilateral. A circle centered at $O$ passes through $B$ and $D$ and meets lines $BA$ and $BC$ again at points $E$ and $F$ (distinct from $A,B,C$). Let $H$ denote the orthocenter of triangle $DEF.$ Prove that if lines $AC,$ $DO,$ $EF$ are concurrent, then triangle $ABC$ and $EHF$ are similar.

Robin Son
44 replies
dchenmathcounts
May 23, 2020
Ilikeminecraft
3 hours ago
IZHO 2017 Functional equations
user01   51
N 3 hours ago by lksb
Source: IZHO 2017 Day 1 Problem 2
Find all functions $f:R \rightarrow R$ such that $$(x+y^2)f(yf(x))=xyf(y^2+f(x))$$, where $x,y \in \mathbb{R}$
51 replies
user01
Jan 14, 2017
lksb
3 hours ago
chat gpt
fuv870   2
N 3 hours ago by fuv870
The chat gpt alreadly knows how to solve the problem of IMO USAMO and AMC?
2 replies
fuv870
3 hours ago
fuv870
3 hours ago
Inequality with wx + xy + yz + zw = 1
Fermat -Euler   23
N 3 hours ago by hgomamogh
Source: IMO ShortList 1990, Problem 24 (THA 2)
Let $ w, x, y, z$ are non-negative reals such that $ wx + xy + yz + zw = 1$.
Show that $ \frac {w^3}{x + y + z} + \frac {x^3}{w + y + z} + \frac {y^3}{w + x + z} + \frac {z^3}{w + x + y}\geq \frac {1}{3}$.
23 replies
Fermat -Euler
Nov 2, 2005
hgomamogh
3 hours ago
Waiting for a dm saying me again "old geometry"
drmzjoseph   0
3 hours ago
Source: Idk easy
Given $ABCD$ a tangencial quadrilateral that is not a rhombus, let $a,b,c,d$ be lengths of tangents from $A,B,C,D$ to the incircle of the quadrilateral which center is $I$. Let $M,N$ be the midpoints of $AC,BD$ resp. Prove that
\[ \frac{MI}{IN}=\frac{a+c}{b+d} \]
0 replies
drmzjoseph
3 hours ago
0 replies
Finally hard NT on UKR MO from NT master
mshtand1   2
N 4 hours ago by IAmTheHazard
Source: Ukrainian Mathematical Olympiad 2025. Day 1, Problem 11.4
A pair of positive integer numbers \((a, b)\) is given. It turns out that for every positive integer number \(n\), for which the numbers \((n - a)(n + b)\) and \(n^2 - ab\) are positive, they have the same number of divisors. Is it necessarily true that \(a = b\)?

Proposed by Oleksii Masalitin
2 replies
mshtand1
Mar 13, 2025
IAmTheHazard
4 hours ago
Official Mock AMC Solutions Thread
samus   20
N Mar 21, 2005 by kaycubs
Since people may be wondering about the answers to some problems, why don't we put all of the detailed solutions on this thread. Some rules
-let's try to post in the general order of questions
-try to explain in as much detail as possible
-If anyone has any questions, lets post in another thread.
-Try to quote the question.

Here's questions 1-4:
1. A space diagonal of a polyhedron is a line that connects two vertices of the polyhedron that do not lie on the same face. How many space diagonals does a cube have?

(A) 0 (B) 2 (C) 4 (D) 8 (E) 28

Answer: Let us call the cube ABCDEFGH, ABCD being the face closer to us with A: top left corner, B: top right corner, C: bottom right corner, D:bottom left corner; EFGH being the face farther away from us with E: top left corner, F: top right corner, G: bottom right corner, H:bottom left corner. The space diagnolls have to be on opposite faces, left/right positions, and up/down positions. So the space diagnols are AG,BH, EC, and FD. There are 4, so the answer is: (C) 4.

2. Find the units digit of the product of the first 2004 odd primes.

(A) 1 (B) 3 (C) 5 (D) 7 (E) 9

Answer: The first odd prime numbers are 3,5,7,11,17..... The product of the first 3 prime #'s is 105, first 4 primes is 735, and so on. Therefore, from this pattern, we determine that the units dgit of the first 2004 odd primes will be 5. Another way to look at it: 5 times any number of odd #'s (such a 2003 more) will always have a unit digit of 5. Therefore the answer is: (C) 5.

3. The area of the triangle bounded by the lines y=x, y=-x, and y=6 is

(A) 12 (B) 12 :sqrt:2 (C) 24 (D) 24 :sqrt:2 (E) 36

Answer: The three intersection points of these lines are (0,0),(-6,6), and (6,6). This forms a triangle of height 6 (from a point (x,6) to (x,0)) and length (base) of 12 ((6,6)-(-6,6)=(12,0)). The area of a triangle is bh/2, so the area of this triangle is 12*6/2=36. Therefore the answer is: (E) 36.

4. How many odd positive integers between 1000 and 2000, inclusive, have all digits different?

(A) 224 (B) 280 (C) 504 (D) 512 (E) 729

Answer: This is a combanatorics/counting problem. There are 4 digits:_ _ _ _. We know that the thousands digit can only be 1 (2000 isn't odd). There are 4 possibilities for the units digit: one of the odd numbers 3,5,7,9 since 1 has already been used and we need all the digits different. The tens digit can have only 8 possibities from the 10 digits; 1 can't be used since it is used in the thousands digit, and another digit can't be used since it has already been used in the units place. The hundreds digit has 7 possibilities: 1 can't be used since it is used in the thousands digit, and another digit can't be used since it has already been used in the units place, and yet another digit can't be used since it has been used in the tens digit. We already know there is 1 possibility for the thousands digit: the number 1. Therefore the total number of possibilities for this number are 1 * 7 * 8 * 4 = 224. The answer is therefore: (A) 224.




Again: lets keep questions in order and POST YOUR ANSWERS!
20 replies
samus
Jul 24, 2004
kaycubs
Mar 21, 2005
Official Mock AMC Solutions Thread
G H J
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
samus
155 posts
#1 • 2 Y
Y by Adventure10, Mango247
Since people may be wondering about the answers to some problems, why don't we put all of the detailed solutions on this thread. Some rules
-let's try to post in the general order of questions
-try to explain in as much detail as possible
-If anyone has any questions, lets post in another thread.
-Try to quote the question.

Here's questions 1-4:
1. A space diagonal of a polyhedron is a line that connects two vertices of the polyhedron that do not lie on the same face. How many space diagonals does a cube have?

(A) 0 (B) 2 (C) 4 (D) 8 (E) 28

Answer: Let us call the cube ABCDEFGH, ABCD being the face closer to us with A: top left corner, B: top right corner, C: bottom right corner, D:bottom left corner; EFGH being the face farther away from us with E: top left corner, F: top right corner, G: bottom right corner, H:bottom left corner. The space diagnolls have to be on opposite faces, left/right positions, and up/down positions. So the space diagnols are AG,BH, EC, and FD. There are 4, so the answer is: (C) 4.

2. Find the units digit of the product of the first 2004 odd primes.

(A) 1 (B) 3 (C) 5 (D) 7 (E) 9

Answer: The first odd prime numbers are 3,5,7,11,17..... The product of the first 3 prime #'s is 105, first 4 primes is 735, and so on. Therefore, from this pattern, we determine that the units dgit of the first 2004 odd primes will be 5. Another way to look at it: 5 times any number of odd #'s (such a 2003 more) will always have a unit digit of 5. Therefore the answer is: (C) 5.

3. The area of the triangle bounded by the lines y=x, y=-x, and y=6 is

(A) 12 (B) 12 :sqrt:2 (C) 24 (D) 24 :sqrt:2 (E) 36

Answer: The three intersection points of these lines are (0,0),(-6,6), and (6,6). This forms a triangle of height 6 (from a point (x,6) to (x,0)) and length (base) of 12 ((6,6)-(-6,6)=(12,0)). The area of a triangle is bh/2, so the area of this triangle is 12*6/2=36. Therefore the answer is: (E) 36.

4. How many odd positive integers between 1000 and 2000, inclusive, have all digits different?

(A) 224 (B) 280 (C) 504 (D) 512 (E) 729

Answer: This is a combanatorics/counting problem. There are 4 digits:_ _ _ _. We know that the thousands digit can only be 1 (2000 isn't odd). There are 4 possibilities for the units digit: one of the odd numbers 3,5,7,9 since 1 has already been used and we need all the digits different. The tens digit can have only 8 possibities from the 10 digits; 1 can't be used since it is used in the thousands digit, and another digit can't be used since it has already been used in the units place. The hundreds digit has 7 possibilities: 1 can't be used since it is used in the thousands digit, and another digit can't be used since it has already been used in the units place, and yet another digit can't be used since it has been used in the tens digit. We already know there is 1 possibility for the thousands digit: the number 1. Therefore the total number of possibilities for this number are 1 * 7 * 8 * 4 = 224. The answer is therefore: (A) 224.




Again: lets keep questions in order and POST YOUR ANSWERS!
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
beta
3001 posts
#2 • 4 Y
Y by Adventure10, Adventure10, Adventure10, Mango247
Don't we already have a discussion thread??
5. For how many integer values of x will there be a non-degenerate triangle with distinct side lengths 11, 15, and x?

(A) 18 (B) 19 (C) 20 (D) 21 (E) 22

Solution: By Triangle Inequality, 11+15>x, or 26>x. By Triangle Inequality again, 11+x>15, x>15-11=4. Therefore 26>x>4. The total number of possiblity for x is 26-4-1=21. However, we included 11 and 15 in the 21 values of x. Thus 21-19=2, or B.

6. How many right triangles with integer side lengths are such that their side lengths are in a nonconstant geometric sequence?

(A) 0 (B) 1 (C) 2 (D) 4 (E) Infinitely many

Solution: Call the sides a b and c. Since they are in geom. progression they can be labeled like this: a, ar, and ar^2 respectively. For this to be a right triangle a^2+(ar)^2=(ar^2)^2 must be true.

Factor an a^2 from the left sides and divide by a^2 on both sides and we are left with 1+r^2=r^4. To find r we use quad form. and take the square root of it(when using quad form we find r^2 so we need to take the sq. root). You can find this and figure out it is irrational. But the sides must be integers. With integer sides and an irrational ratio there are no solutions.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
nr1337
1213 posts
#3 • 2 Y
Y by Adventure10, Mango247
Yay number 5!

5. For how many integer values of x will there be a non-degenerate triangle with distinct side lengths 11, 15, and x?

(A) 18 (B) 19 (C) 20 (D) 21 (E) 22

First we look for x such that 11 + x > 15 and x <= 15. Thus x > 4 from our first inequality, so the minimum x is 5. Next we look for 11+15 > x. So 26 > x, and the maximum x is 25. There are 25-5+1 = 21 numbers between 5 and 25 inclusive BUT WAIT! The problem said DISTINCT side lengths! Thus 11 and 15 don't work for x, so we subtract 2. 21 - 2 = 19, thus our answer is B.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
JSRosen3
335 posts
#4 • 2 Y
Y by Adventure10, Mango247
nr1337 wrote:
BUT WAIT! The problem said DISTINCT side lengths!

<Evil cackle> :twisted:

Only 6 people got this right!!! :(
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
joml88
6343 posts
#5 • 2 Y
Y by Adventure10, Mango247
7. 7. A box contains 11 marbles, numbered 1, 2, 3, ... , 11. If 6 marbles are drawn simultaneously at random, and the probability that the sum of the numbers drawn is odd is D/Q, with D and Q relatively prime positive integers, find D + Q.

(A) 3 (B) 17 (C) 331 (D) 346 (E) 349


First note that:
Even+Even=even, Even+Odd=odd, and Odd+odd=even and that we have 6 odds and 5 evens in our set of numbers.

We are going to try to find the number of times that it makes it odd over the total number of times. You can choose 6 from a group of 11 in \[{11\choose 6}=462\]

Now to find how many ways it will be even we break it into cases.

1st case- 6 evens... this doesn't create an odd

2nd-5 evens and 1 odd...this creates odd numbers. There is one way to get 5 evens and 6 to get an odd making 6 total ways.

3rd-4 evens 2 odd... none

4th-3 even 3 odds... this works. 10 ways to choose 3 odds and 20 for the evens, 200 ways

5th-2 evens 4 odds...none.

6th-1 even 5 odds... works. 5 ways for the evens and 6 for the odds 30.

So we have $\frac{200+30+6}{462}=\frac{118}{231}$.

D+Q=349 (E)
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
beta
3001 posts
#6 • 2 Y
Y by Adventure10, Mango247
7. A box contains 11 marbles, numbered 1, 2, 3, ... , 11. If 6 marbles are drawn simultaneously at random, and the probability that the sum of the numbers drawn is odd is
\frac{D}{Q}
, with D and Q relatively prime positive integers, find D + Q.

(A) 3 (B) 17 (C) 331 (D) 346 (E) 349

Solution: There are 6 odd numbers and 5 even numbers. If the sum of the six numbers chosen is odd, then we have 3 cases:
1)5 odd and 1 even
2) 3 odd and 3 even
3) 1 odd and 5 even.
In case 1), the total number of ways to get the result is (6C5)(5C1)
In case 2), the total number of ways to get the result is (6C3)(5C3)
In case 3), the total number of ways to get the result is (6C1)(5C5)
Thus, the total number of way of getting the sum of 6 numbers odd is
(6C5)(5C1)+(6C3)(5C3)+(6C1)(5C5)=236
The total number of ways to draw 6 marbles simultaneously is 11C6
The probability is 236/11C6=118/231. 118+231=349, the answer is E.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
beta
3001 posts
#7 • 2 Y
Y by Adventure10, Mango247
8. Find the number of integers
k
such that
k^3 + k^2 + k + 7
is divisible by
k+1
.

(A) 2 (B) 4 (C) 6 (D) 8 (E) Infinitely many

Solution:$k^3+k^2+k+7$ must be divisible by $k+1$, so divide it to get $\displaystyle k^2+1+\frac{6}{k+1}$. If it's divisible, that will have to be an integer, meaning $k+1|6$. From this, we see $\pm(k+1)\in \{1,2,3,6\}$ So there are 8 possible values, 1, 2, 5, 0, -2, -3, -4, -7 for k, and the answer is d.

9. Circle A is inscribed in an isosceles right triangle and circle B is circumscribed about the same triangle. The ratio of the area of circle A to the area of circle B is

(A)
\frac{1}{2}
(B)
3 - \sqrt{2}
(C)
3 + \sqrt{2}
(D)
3 - 2\sqrt{2}
(E)
3 + 2\sqrt{2}


Solution: Without Loss of generality let the length of the leg of right triangle be 1. Thus the length of the hypotenuse is $\sqrt2$.Since the center of the circumscribed circle of a right triangle is the midpoint of the hypotenuse of the right triangle. The length of the circumradius is $\frac{\sqrt2}{2}$. Now note that the area of the triangle is $1(1)/2=\frac{1}{2}=rs$, where r is the inradius, and s is the semiperimeter. The semiperimeter=$\frac{1+1+\sqrt2}{2}$. Thus the inradius$=\frac{1}{2}/ \frac{2+\sqrt2}{2}=\frac{1}{2+\sqrt2}$. Inradius: Circumradius=$\frac{1}{2+\sqrt2}:\frac{\sqrt2}{2}=\frac{\sqrt2}{2+\sqrt2}=\sqrt 2-1$. Ratio of area=ratio of radius^2
=$(\sqrt 2-1)^2=3-2\sqrt2$, or D
This post has been edited 1 time. Last edited by beta, Jul 25, 2004, 2:33 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
kool_dudy
303 posts
#8 • 2 Y
Y by Adventure10, Mango247
Thx to all solution posters.

This is what a getting started problem solver needs.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
joml88
6343 posts
#9 • 2 Y
Y by Adventure10, Mango247
10. A box contains 2 pennies, 4 nickels, and 6 dimes. Six coins are drawn randomly and without replacement. Given that the probability that the value of the coins drawn is at least 50 cents is m/n, with m and n relatively prime positive integers, find m + n.

(A) 961 (B) 1015 (C) 1051 (D) 1056 (E) None of these


The total number of ways to choose 6 is . We are left with finding how many ways to get 50 or over.

We can have
dddddd
dddddn
dddddp or
ddddnn
There is one way to get the first case.

For the second case you can chose the 5 d's in ways and the n in 4 ways(since order doesn't matter). So we have 6*4=24 ways for this case.

For case 3, the d's can be chosen in 6 again. The p can be chosen in 2 which means there are 12 for this case.

For the last case, the d's can be picked in ways and the n's in ${4\choose 2}=6$ ways. This makes 90 ways.

So the answer is 127/924. m+n= 1051 (C)

11. 11. Given that $i^2=-1$, for how many integers n is $(n+i)^4$ an integer?

(A) 0 (B) 1 (C) 2 (D) 3 (E) 4

Expanding the binomial we get
\[n^4+4n^3i+6n^2i^2+4ni^3+i^4=n^4+4n^3i-6n^2-4ni+1\]

Taking away the parts that we know are integers we are left with:
\[4n^3i-4ni=4ni(n+1)(n-1)\]
This is only an integer when it is 0. so $n\in \{-1,0,1\}$. (D)
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
JSRosen3
335 posts
#10 • 2 Y
Y by Adventure10, Mango247
Pork_Chop8 wrote:
5. I missed the distinct part. Great trick, evil problem writer. :lol:

Hey, I resent that... ;)

Only 6 people out of 29 got that one right.

Also, I think this thread was intended have the solutions in order of the questions...so you may want to edit your post, or move it to the other discussion thread where solutions are not being discussed in order.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
beta
3001 posts
#11 • 2 Y
Y by Adventure10, Mango247
12. A quadrilateral circumscribed about a circle has three of its side lengths 132, 162, and 201. What is the positive difference between the minimum and maximum lengths for the fourth side?

(A) 60 (B) 78 (C) 93 (D) 138 (E) 231

Solution:
Lemma: If a circle can be inscribed in a quadrilateral ABCD, then AB+CD=AD+BC, so AD=AB+CD-BC.
Proof: Let the circle be tangent at AB, BC, CD, AD at W, X, Y, Z respectively.
Thus AW=AZ, BW=BX, CY=CX, DY=DZ. Add all the equations up yields
AB+CD=BC+AD as desired.
Therefore, let AD be the fourth side, AD is at minimum when AB+CD=132+162=294, and BC=201. AD=294-201=93.
AD is at maximum when AB+CD=201+162=363, BC=132. AD=363-132=231.
Thus 231-93=138, or D.

13. For how many primes
p
is
p^4 + p^5
the square of an integer?

(A) 0 (B) 1 (C) 2 (D) 3 (E) Infinitely many

Solution: First we factor \[p^4(p+1)\].
So we need the square root of this to be an integer. $p^4$ will come out of the square so we are left with finding were $\sqrt{p+1}$ is an integer. In other words we want \[p+1=a^2\] where a is an integer.

Subtract by one on both sides and factor to get \[p=(a+1)(a-1)\] Wait a minute a prime is only divisible by one and itself. So this could only work when a-1=1. this makes p=3 and 3 is prime so we only have one solution, or B
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
beta
3001 posts
#12 • 2 Y
Y by Adventure10, Mango247
14. If
x
,
y
, and
z
are positive numbers satisfying
x + \frac{1}{y} = 4
,
y + \frac{1}{z} = 1
, and
z + \frac{1}{x} = \frac{7}{3}
, then find
xyz
.

(A)
\frac{2}{3}
(B)
1
(C)
\frac{4}{3}
(D)
2
(E)
\frac{7}{3}


Solution: $(x+\frac{1}{y})(y+\frac{1}{z}) (z+\frac{1}{x})$
$=xyz+x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{xyz}$
$=\frac{28}{3}$

Since
$x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$
$=(x+\frac{1}{y})+ (y+\frac{1}{z})+(z+\frac{1}{x})$
$=\frac{22}{3}$
Sub that in, and let $xyz=k$
$k+\frac{1}{k}+\frac{22}{3}=\frac{28}{3}$
$3k^2-6k+3=0$
$(k-1)^2=0$
$k=1$, or B.

15. If
\frac{\pi}{2} < \alpha < \frac{3\pi}{4}
,
\cos{\alpha} + \sin{\alpha} = \frac{1}{3}
, and
\left|\cos{2\alpha} + \sin{2\alpha}\right|= \frac{n + \sqrt{p}}{q}
, where
n
,
p
, and
q
are relatively prime positive integers and
p
is not divisible by the square of any prime, find
n+p+q
.

(A) 34 (B) 35 (C) 36 (D) 37 (E) 38

Solution: $(\sin{\alpha}+\cos{\alpha})^2$
$=(\sin^2{\alpha}+\cos^2{\alpha})+2\sin{\alpha}\cos{\alpha}$
$=1+\sin{2\alpha}$
$=\frac{1}{9}$
Solve for $\sin{2\alpha}$
to get $\sin{2\alpha}=-\frac{8}{9}$
$\cos{2\alpha}=\pm \sqrt{1-(\frac{8}{9})^2}$
$=\pm \frac{\sqrt{17}}{9}$
Because of our restrictions, $\cos{2\alpha}$ is negative
$=-\frac{\sqrt{17}}{9}$
absolute value of $\cos{2\alpha}+\sin{2\alpha}=\frac{\sqrt{17}+8}{9}$
17+8+9=34, or A.

16. If
r_1
,
r_2
, and
r_3
are the solutions to the equation
x^3 - 4x^2 - 4x + 17 = 0
, find
r_1^3 + r_2^3 + r_3^3
.

(A) -68 (B) -4 (C) 0 (D) 61 (E) 68

Solution:First off, rename $ r_1=a; r_2=b; r_3=c $(so that you don't get all the r's confused). Since they are solutions to the equation, the equation $x^3-4x^2-4x+17=(x-a)(x-b)(x-c)$. Expanding the right side gives $x^3-(a+b+c)x^2+(ab+bc+ac)x-abc=0$. Therefore, $a+b+c=4; abc=-17$; $ab+bc+ac=-4$. Now take $a+b+c=4$ and multiply both sides by (a+b+c), giving $a^2+b^2+c^2+ab+ac+bc=16$. Replace $ab+ac+bc$ with -8, swing the -8 to the other side, then multiply both sides by $(a+b+c)$ again. Now you have (with a little bit of simplification):
$a^3+b^3+c^3+a(ab+ac)+b(ab+bc)+c(ac+bc)=96$
Change the equation we already had to ab+ac=-4-bc and put that into the equation we have, then repeat for the other two with the other variables. It now looks like this:
$a^3+b^3+c^3+a(-4-bc)+b(-4-ac)+c(-4-ab)=96$ which is equivalent to this:
$a^3+b^3+c^3-4(a+b+c)-3abc=96$ Substituting from our equations for a, b, and c, gives us $a^3+b^3+c^3-16+51=96$; therefore, $a^3+b^3+c^3=61$ and the answer is (D).

17. Find
\sum_{k=0}^{49} (-1)^k  {99 \choose 2k}
.

(A)
 
(B)
-2^{49}
(C)
2^{49}
(D)
-2^{50}
(E)
2^{50}


Solution:Note that
$(1-i)^{99}=1-{99 \choose 1}i-{99 \choose 2}+ {99 \choose 3}i+{99 \choose 4}.....$
$=\displaystyle\sum_{k=0}^{49} (-1)^k {99 \choose 2k}+i\sum_{k=0}^{49} (-1)^k {99 \choose {2k+1}}$
And $1=-(\sqrt2)(\cos\frac{3\pi}{4}), -1=-(\sqrt2)(\sin\frac{3\pi}{4})$
By DeMorvie's Theorem
$(1-i)^{99}=-((\sqrt2)(\cos\frac{3\pi}{4})+i(\sqrt2)(\sin\frac{3\pi}{4}))^{99}
=-(\sqrt2)^{99}(\cos\frac{(99)(3)\pi}{4}+i(\sin\frac{(99)(3)\pi}{4}))$
Thus
$\displaystyle\sum_{k=0}^{49} (-1)^k {99 \choose 2k}$
$=-(\sqrt2)^{99}(\cos\frac{297\pi}{4})$
$=-2^{49}(\sqrt2)(\cos\frac{\pi}{4})$
$=-2^{49}$, or B.

That took A LOT of Copy and Paste. :)
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
joml88
6343 posts
#13 • 2 Y
Y by Adventure10, Mango247
18. A rising number is a positive integer, each digit of which is larger than each of the digits to its left, and no digit of which is 0. For example, 12689 is a rising number. When all five-digit rising numbers are listed from smallest to largest, the 97th number in the list does not contain the digit

(A) 4 (B) 5 (C) 6 (D) 7 (E) 8



The smallest is 12345. With 4 as the second number we have 5 that work(i.e.12345,12346,12347,12348,12349).
If we keep the first 3 the same and keep increasing the second number by 1 we will get 4, 3, 2 ,and 1 that work for each one. Therefore there are 5+4+3+2+1=15 rising numbers of the form 123_ _ . Continuing in this way we find that there are 4+3+2+1=10 rising numbers of the form 124_ _. 6 with a 5 in the third, 3 with a 6 and 1 with a 7. This makes 35 of the form 12_ _ _. For 13_ _ _ there are 4+3+2+1 + 3+2+1 + 2+1 + 1= 20. For 14_ _ _ we have 10. For 15_ _ _ there are 4. For 16_ _ _ there is 1. This makes a grand total of 70.
For 2 as the first number we do the same thing. We find that 24589 is the 96 so 24678 is the 97. This doesn't contain a 5 so (B)
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
kool_dudy
303 posts
#14 • 2 Y
Y by Adventure10, Mango247
Joml, I understood your solution. Thx.

I got one question though. How do you know that 24589 is the 96th number in that pattern?
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
confuted
711 posts
#15 • 2 Y
Y by Adventure10, Mango247
kool_dudy wrote:
Joml, I understood your solution. Thx.

I got one question though. How do you know that 24589 is the 96th number in that pattern?
Get as close as you can with the method he used up to 70, and then list the rest.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Ultimatemathgeek
111 posts
#16 • 2 Y
Y by Adventure10, Mango247
Thanks every one
More solutions please
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
beta
3001 posts
#17 • 2 Y
Y by Adventure10, Mango247
19.
\bigtriangleup ABC
has all side lengths distinct. Two altitudes of this triangle have lengths 4 and 12. If the length of the third altitude is also an integer, find its maximum value.

(A) 4 (B) 5 (C) 6 (D) 7 (E) None of these

Solution: Let the length of AB be a, BC be 1, AC be c. Let the altitude of length 4 and 12 be opposite to AB and BC respectively. Let the length of the third altitude be h. The area of ABC=base*altitude/2 $=\frac{4a}{2}=\frac{12}{2}=\frac{hc}{2}$. Multiply by 2 yields 4a=12=hc, thus a=3
Since AB, BC, AC must satisfy triangle inequality, c+1>3, c>2. Since hc=12,
h<6. The maximum integral value of h is 5, or (B).
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
beta
3001 posts
#18 • 2 Y
Y by Adventure10, Mango247
20. It is given that
(\log_z{3})(\log_x{3})+ (\log_y{3})(\log_x{3}) + (\log_y{3})(\log_z{3}) = 6(\log_z{3})(\log_x{3})(\log_y{3})
and
x<y<z
. If
x
,
y
, and
z
are in a geometric sequence with common ratio 2, and
x + y + z = \frac{m}{n}
, with m and n relatively prime positive integers, find
m + n
.

(A) 61 (B) 62 (C) 63 (D) 64 (E) 65

Solution: Note that $(\log_m{n})(\log_n{m})=1, \log_m{n}=\frac{1}{\log_n{m}}$. Divide both side of the original equation by $(\log_z{3})(\log_x{3})(\log_y{3})$ we have $\frac{1}{\log_y{3}}+\frac{1}{\log_z{3}}+\frac{1}{\log_x{3}}=6$, or $\log_3{y}+\log_3{z}+\log_3{x}=6$,
$\log_3{xyz}=6$, which yields $xyz=3^6=729$. Since x, y, z are in geometric squence of common ratio 2, y=2x, z=2y=4x. xyz=8x^3=729, solve for x yields $x=\frac{9}{2}$. $x+2x+4x=7x=\frac{63}{2}$. m+n=63+2=65, or (E).
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Ultimatemathgeek
111 posts
#19 • 2 Y
Y by Adventure10, Mango247
Where can I find the solutions for the rest of the test? Please, I need this.
I appreciate it
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
kaycubs
5 posts
#20 • 2 Y
Y by Adventure10, Mango247
22. Find the number of distinct pairs of integers (x, y) such that 0<x<y and sqrt{1984} =sqrt{x} +sqrt{y}

(A) 0 (B) 1 (C) 3 (D) 4 (E) 7



you have to factor out 1984 which is 2^6*31
root(1984) simplifies to 8root(31)



x has to be 31 and that multiplied by a perfect square for root(X) and root(y) to have a common factor of root 31

when x is 31*1, the other number is (8-root(1))squared*31 which is 49*31 and is 1519
(31,1519) is another solution

then you find the next perfect square

when x is 31*4, the other number is (8-root(4))squared*31 which is 36*31 and is 1116
(124,1116) is another solution

when x is 31*9, the other number is (8-root(9))squared*31 which is 25*31 and is 775
(279,775) is another solution

you can do this for the next perfect square until you realize that x and y will be the same in that

therefore:
(279,775),(124,1116), and (31,1519) are the 3 ordered pairs that work

the answer is (C). or 3

p.s. sorry for first response, i felt i had to correct myself
This post has been edited 4 times. Last edited by kaycubs, Mar 21, 2005, 3:14 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
kaycubs
5 posts
#21 • 2 Y
Y by Adventure10, Mango247
:( sorry, i didnt know calculators werent allowed, look back up for the non- calculator solution :P
Z K Y
N Quick Reply
G
H
=
a