Stay ahead of learning milestones! Enroll in a class over the summer!

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k a April Highlights and 2025 AoPS Online Class Information
jlacosta   0
Apr 2, 2025
Spring is in full swing and summer is right around the corner, what are your plans? At AoPS Online our schedule has new classes starting now through July, so be sure to keep your skills sharp and be prepared for the Fall school year! Check out the schedule of upcoming classes below.

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Looking for summer camps in math and language arts? Be sure to check out the video-based summer camps offered at the Virtual Campus that are 2- to 4-weeks in duration. 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)!

Be sure to mark your calendars for the following events:
[list][*]April 3rd (Webinar), 4pm PT/7:00pm ET, Learning with AoPS: Perspectives from a Parent, Math Camp Instructor, and University Professor
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April 9th (Webinar), 4:00pm PT/7:00pm ET, Learn about Video-based Summer Camps at the Virtual Campus
[*]April 10th (Math Jam), 4:30pm PT/7:30pm ET, 2025 MathILy and MathILy-Er Math Jam: Multibackwards Numbers
[*]April 22nd (Webinar), 4:00pm PT/7:00pm ET, Competitive Programming at AoPS (USACO).[/list]
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0 replies
jlacosta
Apr 2, 2025
0 replies
9 Was the 2025 AMC 8 harder or easier than last year?
Sunshine_Paradise   189
N 8 minutes ago by K1mchi_
Also what will be the DHR?
189 replies
Sunshine_Paradise
Jan 30, 2025
K1mchi_
8 minutes ago
Confusion about 2 counting strategies
Spacepandamath13   7
N 10 minutes ago by K1mchi_
What is the difference between PIE (Principle of Inclusion and Exclusion) and Complementary Counting?
7 replies
Spacepandamath13
Mar 23, 2025
K1mchi_
10 minutes ago
The daily problem!
Leeoz   108
N 18 minutes ago by AbhayAttarde01
Every day, I will try to post a new problem for you all to solve! If you want to post a daily problem, you can! :)

Please hide solutions and answers, hints are fine though! :)

Problems usually get harder throughout the week, so Sunday is the easiest and Saturday is the hardest!

Past Problems!
108 replies
Leeoz
Mar 21, 2025
AbhayAttarde01
18 minutes ago
Is there a point in me continuing AoPS?
julll55   26
N an hour ago by jxie2022
Okay so for context, I am in 8th grade and I'm bad at math, well at least on the competitive level like AMC and stuff. When I was in 7th grade I did some practice AMC8 tests and got like 15-17.

I've just been recently wondering if there is a point in me continuing AoPS. It's not like I'm ever really going to be competitive at math at this rate, and I can't exactly devote more time to it because I have too many other activities I have to do. And since I'm going into 9th grade next year, I have to start taking the AMC10 to be able to go into further math competitions like the AIME. Plus, my school system doesn't do anything related to a math team (I live in Alabama).

So I'm not sure if I should continue learning math from AoPS, because I feel like I could just devote the 12-15 hours I spend on math each week to something else that I am actually good at. The only thing that is sort of stopping me is how much I've already invested into AoPS. I've done Beast Academy, pre-algebra through int algebra. intro c&p, into geo, intro number theory, amc8 and mathcounts.

To be honest, I just don't think I'm talented or even wanting to spend like 20+ hours a week on math just to be able to qualify for these types of competitions. I know that it can be good as an extracurricular and all, but I don't know if it's worth it to grind math. Because in school I do fine, I have like 100 in math, so it's not like I need extra tutoring.
26 replies
julll55
Feb 24, 2025
jxie2022
an hour ago
No more topics!
Something Horrible-a Challenge
Xueshuxue   24
N Apr 7, 2025 by Solocraftsolo
Hello, I was wondering if it's possible to make 8 with the numbers 5, 3, 5, and 7 under the following rules:
-You can only use 5 twice, 3 once, and 7 once.
-You must use all the numbers.
You can stack numbers to form larger numbers (example: I could take 3 and 5 and turn it into 35 or 53, or use 7, 3, and 5 to make 375.)
-You are allowed to use parentheses.
(Also, I already found out that no 3 digital numbers will work for the solution.)
24 replies
Xueshuxue
Mar 28, 2025
Solocraftsolo
Apr 7, 2025
Something Horrible-a Challenge
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Xueshuxue
45 posts
#1
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Hello, I was wondering if it's possible to make 8 with the numbers 5, 3, 5, and 7 under the following rules:
-You can only use 5 twice, 3 once, and 7 once.
-You must use all the numbers.
You can stack numbers to form larger numbers (example: I could take 3 and 5 and turn it into 35 or 53, or use 7, 3, and 5 to make 375.)
-You are allowed to use parentheses.
(Also, I already found out that no 3 digital numbers will work for the solution.)
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Inaaya
260 posts
#2
Y by
(5x3)-7 !!!
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sanaops9
822 posts
#3
Y by
you need to use both fives rbo

you didn't mention what functions to use so $\lfloor{\dfrac{55+3}{7}\rfloor}$
This post has been edited 1 time. Last edited by sanaops9, Mar 28, 2025, 7:28 PM
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derekwang2048
1219 posts
#4
Y by
if you're allowed to use more than just the basic operations, this is simple
This post has been edited 2 times. Last edited by derekwang2048, Mar 28, 2025, 7:29 PM
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sanaops9
822 posts
#5
Y by
@above, you can use \lfloor and \rfloor for left floor and right floor, respectively.

EDIT: ig u figured it out
This post has been edited 2 times. Last edited by sanaops9, Mar 28, 2025, 7:31 PM
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vincentwant
1293 posts
#6 • 1 Y
Y by ARWonder
Click to reveal hidden text
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Solocraftsolo
2495 posts
#7
Y by
the answer is obviously

\[|P\left(\left\lfloor \sqrt{\sqrt{\left\lceil \sqrt{\left\lfloor \sqrt{\left\lceil \sqrt{\sqrt{\left\lfloor \sqrt{\sqrt{\left\lceil \sqrt{\sqrt{\sqrt{5537}}}\right\rceil !!}}\right\rfloor !}}\right\rceil !}\right\rfloor !}\right\rceil !}}\right\rfloor\right)|\]

where the P function is the power set and the | | means cardinality
This post has been edited 1 time. Last edited by Solocraftsolo, Mar 28, 2025, 8:51 PM
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Leeoz
174 posts
#8 • 2 Y
Y by Exponent11, TTNT
chat gpt cooked
chat gpt wrote:
To make 8 using only the numbers **3**, **5**, **5**, and **7**, one possible solution is:

\[
(5 + 5) - (7 - 3) = 8
\]
Explanation:
- First, add the two 5s: \(5 + 5 = 10\).
- Then, subtract the difference between 7 and 3: \(7 - 3 = 4\).
- Finally, subtract 4 from 10: \(10 - 4 = 8\).

This gives you 8!
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ChickensEatGrass
44 posts
#9
Y by
Leeoz wrote:
chat gpt cooked
chat gpt wrote:
To make 8 using only the numbers **3**, **5**, **5**, and **7**, one possible solution is:

\[
(5 + 5) - (7 - 3) = 8
\]
Explanation:
- First, add the two 5s: \(5 + 5 = 10\).
- Then, subtract the difference between 7 and 3: \(7 - 3 = 4\).
- Finally, subtract 4 from 10: \(10 - 4 = 8\).

This gives you 8!

what the heck
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huajun78
61 posts
#11
Y by
Leeoz wrote:
chat gpt cooked
chat gpt wrote:
To make 8 using only the numbers **3**, **5**, **5**, and **7**, one possible solution is:

\[
(5 + 5) - (7 - 3) = 8
\]
Explanation:
- First, add the two 5s: \(5 + 5 = 10\).
- Then, subtract the difference between 7 and 3: \(7 - 3 = 4\).
- Finally, subtract 4 from 10: \(10 - 4 = 8\).

This gives you 8!

$6=8$, proof by ChatGPT
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SirAppel
872 posts
#12
Y by
$(5+3)^{\lfloor \frac{7}{5} \rfloor}$
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cheltstudent
567 posts
#13
Y by
$lcm(7,5) - 3^{3}=8$
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Soupboy0
322 posts
#14 • 1 Y
Y by PuppyPenguinDolphin
$\frac{5!}{3!}-5-7$
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Owen314159
12 posts
#15
Y by
Soupboy0 wrote:
$\frac{5!}{3!}-5-7$

I think this is the most normal sol
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RandomMathGuy500
57 posts
#16
Y by
chat gpt wrote:
You can make 8 using the numbers 5, 5, 3, and 7 with the following equation:
$(7+3)-(5+5)=8$
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Total_Awesomeness
288 posts
#17
Y by
derekwang2048 wrote:
if you're allowed to use more than just the basic operations, this is simple

ur not allowed to use 8
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happymoose666
301 posts
#18
Y by
he is not using 8, it's an equal sign
This post has been edited 1 time. Last edited by happymoose666, Mar 29, 2025, 10:37 PM
Reason: yes
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Total_Awesomeness
288 posts
#19
Y by
no at the end
he says that it equals the digit 8, which is not allowed
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sanaops9
822 posts
#20
Y by
Owen314159 wrote:
Soupboy0 wrote:
$\frac{5!}{3!}-5-7$

I think this is the most normal sol
vincentwant wrote:
Click to reveal hidden text

No vincent's was prolly the best and intended sol
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pieMax2713
4176 posts
#21
Y by
$|\{5, 5, 7\}|^3$ where | | denotes cardinality
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K1mchi_
25 posts
#22 • 1 Y
Y by ChickensEatGrass
(7!/(5!+3!))/5
I just used lots of factorial
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WiseHawkCuteFriendly
580 posts
#23
Y by
chat gpt wrote:
$\frac{7+5}{3}+5=8$,
reasoning: $\frac{12}{3}+5=8$
$4+5=8$
This post has been edited 2 times. Last edited by WiseHawkCuteFriendly, Mar 31, 2025, 1:46 AM
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happymoose666
301 posts
#24
Y by
Total_Awesomeness wrote:
no at the end
he says that it equals the digit 8, which is not allowed

But he meant $\frac{55+\lfloor{\sqrt3}\rfloor}{7}$ is equal to 8, so he solved the problem, he is not using eight
This post has been edited 2 times. Last edited by happymoose666, Mar 31, 2025, 1:21 PM
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Demetri
1272 posts
#25
Y by
Soupboy0 wrote:
$\frac{5!}{3!}-5-7$

I think this is the most normal sol
This doesn't use factorials
Click to reveal hidden text
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Solocraftsolo
2495 posts
#26 • 1 Y
Y by Exponent11
huajun78 wrote:

$6=8$, proof by ChatGPT

proof by exhaustion
proof by contradiction
proof by chatGPT
This post has been edited 1 time. Last edited by Solocraftsolo, Apr 7, 2025, 4:10 PM
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