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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

Are you interested in working towards MATHCOUNTS and don’t know where to start? We have you covered! If you have taken Prealgebra, then you are ready for MATHCOUNTS/AMC 8 Basics. Already aiming for State or National MATHCOUNTS and harder AMC 8 problems? Then our MATHCOUNTS/AMC 8 Advanced course is for you.

Summer camps are starting next month at the Virtual Campus in math and language arts that are 2 - to 4 - weeks in duration. Spaces are still available - don’t miss your chance to have an enriching summer experience. There are middle and high school competition math camps as well as Math Beasts camps that review key topics coupled with fun explorations covering areas such as graph theory (Math Beasts Camp 6), cryptography (Math Beasts Camp 7-8), and topology (Math Beasts Camp 8-9)!

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[*]May 20th, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 1 Math Jam, Problems 1 to 4, join the Canada/USA Mathcamp staff for this exciting Math Jam, where they discuss solutions to Problems 1 to 4 of the 2025 Mathcamp Qualifying Quiz!
[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
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0 replies
jlacosta
May 1, 2025
0 replies
what number have you memorized perfect squares
ellenssim   3
N an hour ago by whwlqkd
Up to what number have you memorized perfect squares, and how often does it help you in solving problems?
3 replies
ellenssim
5 hours ago
whwlqkd
an hour ago
Challenge: Make every number to 100 using 4 fours
CJB19   266
N 2 hours ago by Leeoz
I've seen this attempted a lot but I want to see if the AoPS community can actually do it. Using ONLY 4 fours and math operations, make as many numbers as you can. Try to go in order. I'll start:
$$(4-4)*4*4=0$$$$4-4+4/4=1$$$$4/4+4/4=2$$$$(4+4+4)/4=3$$$$4+(4-4)*4=4$$$$4+4^{4-4}=5$$$$4!/4+4-4=6$$$$4+4-4/4=7$$$$4+4+4-4=8$$
266 replies
CJB19
May 15, 2025
Leeoz
2 hours ago
Last challenge problems in the books
ysn613   10
N 4 hours ago by mdk2013
Algebra
It is known that $\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}\dots=\frac{\pi^2}{6}$ Given this fact, determine the exact value of $$\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}\dots.$$(Source: Mandelbrot)
Counting and Probability
A $3\times3\times3$ wooden cube is painted on all six faces, then cut into 27 unit cubes. One unit cube is randomly selected and rolled. After it is rolled, $5$ out of the $6$ faces are visible. What is the probability that exactly one of the five visible faces is painted? (Source: MATHCOUNTS)
Number Theory(This technically isn't the last problem but the last chapter doesn't have challenge problems)
The integer p is a 50-digit prime number. When its square is divided by 120, the remainder is not 1. What is the remainder?
I didn't include geometry because I haven't taken it yet, feel free to post it
Answer these problems and post what you think is the order of difficulty
10 replies
ysn613
Monday at 7:04 PM
mdk2013
4 hours ago
What's $(-1)^0?$
Vulch   12
N 4 hours ago by Li0nking
What's $(-1)^0?$

(It may be a silly question,but still I want to know it's value)
12 replies
Vulch
Oct 26, 2024
Li0nking
4 hours ago
No more topics!
Doubt on a math problem
AVY2024   19
N May 7, 2025 by sadas123
Solve for x and y given that xy=923, x+y=84
19 replies
AVY2024
Apr 8, 2025
sadas123
May 7, 2025
Doubt on a math problem
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AVY2024
29 posts
#1
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Solve for x and y given that xy=923, x+y=84
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fruitmonster97
2504 posts
#2
Y by
We have $x=42-a$ and $y=42+a$ for some $a.$ We have $42^2-a^2=923,$ so $a=29.$ Thus, $(x,y)=\boxed{(13,71)}.$
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sp0rtman00000
2 posts
#3
Y by
Yous must solve the following second degree equation: t^2-84t+932=0
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CatsRule222
2 posts
#4
Y by
I agree with sp0rtman00000;
It will be easier to change it to a second degree polynomial and then use the quadratic formula, which is: (-b+-Sqrt(b^2-4ac))/2a
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fruitmonster97
2504 posts
#5 • 1 Y
Y by DDCN_2011
CatsRule222 wrote:
I agree with sp0rtman00000;
It will be easier to change it to a second degree polynomial and then use the quadratic formula, which is: (-b+-Sqrt(b^2-4ac))/2a

That's what I did, except instead of using the quadratic formula or factoring I used Po-Shen Loh's recently(ish) found method which combines vieta's and some common sense.
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jasperE3
11385 posts
#6 • 1 Y
Y by Solocraftsolo
fruitmonster97 wrote:
We have $x=42-a$ and $y=42+a$ for some $a.$ We have $42^2-a^2=923,$ so $a=29.$ Thus, $(x,y)=\boxed{(13,71)}.$

$(x,y)=(71,13)$
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Soupboy0
492 posts
#7
Y by
$x = 1434$ by 1434 number theory lemma. Now, define a sigma number $\mathcal{J}$ from the set $\mathcal{S}$ and such that such that any number can now be expressed as ultra-complex, e.g. $a+bi+c\mathcal{J}$. Now, using the chicken jockey steve theorem, we find there are $2$ distinct solutions. These solutions actually alter the given value of $x$. The $2$ solutions are $4+37\sqrt{3}i+27\sqrt[3]{3} \mathcal{J}$ and $26.53+34\sqrt{2} i +14 \mathcal{J}.$ These $2$ solutions alter the quantum space limit, and change the values of $x$ to be $13$ and $71$. Now, using the newly derived quintic formula yields $y=13, 71$. Therefore, the $2$ solutions are $(13, 71)$ and $(71, 13)$
This post has been edited 1 time. Last edited by Soupboy0, Apr 11, 2025, 5:59 PM
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sadas123
1325 posts
#8
Y by
Soupboy0 wrote:
$x = 1434$ by 1434 number theory lemma. Now, define a sigma number $\mathcal{J}$ from the set $\mathcal{S}$ and such that such that any number can now be expressed as ultra-complex, e.g. $a+bi+c\mathcal{J}$. Now, using the chicken jockey steve theorem, we find there are $2$ distinct solutions. These solutions actually alter the given value of $x$. The $2$ solutions are $4+37\sqrt{3}i+27\sqrt[3]{3} \mathcal{J}$ and $26.53+34\sqrt{2} i +14 \mathcal{J}.$ These $2$ solutions alter the quantum space limit, and change the values of $x$ to be $13$ and $71$. Now, using the newly derived quintic formula yields $y=13, 71$. Therefore, the $2$ solutions are $(13, 71)$ and $(71, 13)$

I agree with you :D
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maxamc
585 posts
#9
Y by
Since $3|7$, $13|71$ (add $10$ to the left hand side and $1$ to the other, difference is $3 \cdot 3=9$, then Euclidian Algorithm), and $71|13$ (commutativity of the divides relation). Thus our answers are $(13,71)$ and all other permutations.
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evt917
2424 posts
#10
Y by
AVY2024 wrote:
Solve for x and y given that xy=923, x+y=84

me: bash
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Yiyj1
1267 posts
#11
Y by
evt917 wrote:
AVY2024 wrote:
Solve for x and y given that xy=923, x+y=84

me: bash

me: runs a program that tests every single pair (x,y)
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giratina3
556 posts
#12
Y by
Yiyj1 wrote:
evt917 wrote:
AVY2024 wrote:
Solve for x and y given that xy=923, x+y=84

me: bash

me: runs a program that tests every single pair (x,y)

me: crashes out
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derekwang2048
1228 posts
#13
Y by
wait is this not trivial by quadratic formula
if you're fine with squaring 84 and 4*923
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giratina3
556 posts
#14
Y by
derekwang2048 wrote:
wait is this not trivial by quadratic formula
if you're fine with squaring 84 and 4*923

This is why we don't use the quadratic formula whenever we see massive numbers.

You're incredibly prone to making an arithmetic mistake.
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Soupboy0
492 posts
#15
Y by
do you know what else is massive
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giratina3
556 posts
#17
Y by
Soupboy0 wrote:
do you know what else is massive

I don’t know. For some reason, I thought of hairstyles when I heard “massive”.
This post has been edited 1 time. Last edited by giratina3, May 6, 2025, 11:47 AM
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maxamc
585 posts
#18
Y by
giratina3 wrote:
derekwang2048 wrote:
wait is this not trivial by quadratic formula
if you're fine with squaring 84 and 4*923

This is why we don't use the quadratic formula whenever we see massive numbers.

You're incredibly prone to making an arithmetic mistake.

I always use the quadratic formula on huge numbers.
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LXC007
62 posts
#19
Y by
giratina3 wrote:
Soupboy0 wrote:
do you know what else is massive

I don’t know. For some reason, I thought of hairstyles when I heard “massive”.

I can hear the “Low taper fade” right now
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Capybara7017
441 posts
#20
Y by
jasperE3 wrote:
fruitmonster97 wrote:
We have $x=42-a$ and $y=42+a$ for some $a.$ We have $42^2-a^2=923,$ so $a=29.$ Thus, $(x,y)=\boxed{(13,71)}.$

$(x,y)=(71,13)$

they aren't wrong
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sadas123
1325 posts
#21
Y by
Capybara7017 wrote:
jasperE3 wrote:
fruitmonster97 wrote:
We have $x=42-a$ and $y=42+a$ for some $a.$ We have $42^2-a^2=923,$ so $a=29.$ Thus, $(x,y)=\boxed{(13,71)}.$

$(x,y)=(71,13)$

they aren't wrong

It doesn't matter which order you put in but just use subsitiution then difference of squares and then just use algebra and solve for the final answer.
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