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k a April Highlights and 2025 AoPS Online Class Information
jlacosta   0
Apr 2, 2025
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0 replies
jlacosta
Apr 2, 2025
0 replies
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2025 MATHCOUNTS State Hub
SirAppel   488
N 9 minutes ago by DhruvJha
Previous Years' "Hubs": (2022) (2023) (2024)Please Read

Now that it's April and we're allowed to discuss ...
[list=disc]
[*] CA: 43 (45 44 43 43 43 42 42 41 41 41)
[*] NJ: 43 (45 44 44 43 39 42 40 40 39 38) *
[*] NY: 42 (43 42 42 42 41 40)
[*] TX: 42 (43 43 43 42 42 40 40 38 38 38)
[*] MA: 41 (45 43 42 41)
[*] WA: 41 (41 45 42 41 41 41 41 41 41 40) *
[*]VA: 40 (41 40 40 40)
[*] FL: 39 (42 41 40 39 38 37 37)
[*] IN: 39 (41 40 40 39 36 35 35 35 34 34)
[*] NC: 39 (42 42 41 39)
[*] IL: 38 (41 40 39 38 38 38)
[*] OR: 38 (44 40? 38 38)
[*] PA: 38 (41 40 40 38 38 37 36 36 34 34) *
[*] MD: 37 (43 39 39 37 37 37)
[*] AZ: 36 (40? 39? 39 36)
[*] CT: 36 (44 38 38 36 35 35 34 34 34 33 33)
[*] MI: 36 (39 41 41 36 37 37 36 36 36 36) *
[*] MN: 36 (40 36 36 36 35 35 35 34)
[*] CO: 35 (41 37 37 35 35 35 ?? 31 31 30) *
[*] GA: 35 (38 37 36 35 34 34 34 34 34 33)
[*] OH: 35 (41 37 36 35)
[*] AR: 34 (46 45 35 34 33 31 31 31 29 29)
[*] NV: 34 (41 38 ?? 34)
[*] WI: 34 (40 37 37 34 35 30 28 29 29 29) *
[*] HI: 32 (35 34 32 32)
[*] NH: 31 (42 35 33 31 30)
[*] DE: 30 (34 33 32 30 30 29 28 27 26? 24)
[*] SC: 30 (33 33 31 30)
[*] IA: 29 (33 30 31 29 29 29 29 29 29 29 29 29) *
[*] NE: 28 (34 30 28 28 27 27 26 26 25 25)
[*] SD: 22 (30 29 24 22 22 22 21 21 20 20)
[/list]
Cutoffs Unknown

* means that CDR is official in that state.

Notes

For those asking about the removal of the tiers, I'd like to quote Jason himself:
[quote=peace09]
learn from my mistakes
[/quote]

Help contribute by sharing your state's cutoffs!
488 replies
SirAppel
Apr 1, 2025
DhruvJha
9 minutes ago
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I think I regressed at math
PaperMath   49
N 32 minutes ago by sadas123
I found the slip of paper a few days ago that I think I wrote when I was in kindergarten. It is just a sequence of numbers and you have to find the next number, the pattern is $1,2,5,40,1280,?$. I couldn't solve this and was wondering if any of you can find the pattern
49 replies
1 viewing
PaperMath
Mar 8, 2025
sadas123
32 minutes ago
J
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two solutions
τρικλινο   8
N 41 minutes ago by sadas123
in a book:CORE MATHS for A-LEVEL ON PAGE 41 i found the following


1st solution


$x^2-5x=0$



$ x(x-5)=0$



hence x=0 or x=5



2nd solution



$x^2-5x=0$

$x-5=0$ dividing by x



hence the solution x=0 has been lost



is the above correct?
8 replies
τρικλινο
Yesterday at 6:20 PM
sadas123
41 minutes ago
J
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What Are The Chances?
IbrahimNadeem   59
N an hour ago by sadas123
Hello, I'm curious to have honest advice on how far I can make it (by 11th-12th grade-ish);

If I have:

- Started AMC 8 study in 6th grade
- Started AMC 10 study in 7th grade
- Started practicing harder & went from 60 to around 100 on AMC 10 (on practice tests with official conditions)
- Started AMC 12 study in 8th grade
- Currently (fall of 8th grade) getting ~120 on AMC 10/12 & 7-10 while practicing AIME

At this rate, what are the chances of me making the USA(J)MO, for example, by ~11th grade?

Please be completely honest and don't hold back; This can be useful to see if I have the need to practice harder.
59 replies
IbrahimNadeem
Oct 31, 2021
sadas123
an hour ago
J
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Index of Coincidence of a ciphertext with respect to itself.
fortenforge   0
Oct 18, 2009
If we are comparing a text to itself, we basically are mathematically finding the probability that if we choose $ 2$ characters from the text, the characters will be the same.
Here is the formula:
$ \displaystyle\sum_{i=1}^{c}\frac{n_i(n_i - 1)}{N(N-1)}$.
where $ c$ is the number of characters in the alphabet, $ n_i$ is the number of times the $ i$th of the alphabet appears in the plaintext, and $ N$ is the number of letters in the plaintext.
Let us try to derive this formula. Probability is defined as the number of ways you get what you want divided by the total number of possibilities. How many ways are there to choose any $ 2$ letters from a group of $ N$ letters? It is of course, $ \dbinom{N}{2}$ which is equal to $ \frac{N!}{2!(N-2)!} = \frac{N(N-1)}{2}$. This is the denominator. To calculate the numerator, we first calculate the number of ways to pick $ 2$ a's from our plaintext and add that to the number of ways to pick $ 2$ b's from our plaintext, and so on. If the number of a's in our plaintext was $ n_i$, then the number of ways to pick $ 2$ a's is $ \dbinom{n_i}{2}$, this is equal to $ \frac{n_i(n_i-1)}{2}$ as we have shown before. This numerator and denominator gives us $ \displaystyle\sum_{i=1}^{c}\frac{n_i(n_i - 1)/2}{N(N-1)/2}$, the $ /2$'s cancel giving us our desired formula:

$ \boxed{\displaystyle\sum_{i=1}^{c}\frac{n_i(n_i - 1)}{N(N-1)}}$.
0 replies
fortenforge
Oct 18, 2009
0 replies
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Why frequency analysis does not work
fortenforge   0
Aug 30, 2009
Frequency Analysis works because there is a one to one correspondence between the plaintext alphabet and the ciphertext alphabet. If frequency analysis is going to work, the letter $ p$ should ALWAYS be encrypted as the letter $ c$. In a Vigenere cipher this does not occur. Depending of $ p$'s position in the plaintext, $ p$ could be encrypted as one of several letters. If the keyword has length $ 5$, then $ p$ could be encrypted as $ c_1,c_2,c_3,c_4,c_5$. This is not a one to one correspondence so frequency analysis does not work.

Let us say that the frequency of $ p$ in normal English was $ i$. If the key word was of length $ 1$, the frequency of $ c$ in the cipher text would be $ i$ as well. But if the key word was of length $ 2$, then the frequency of $ c_1 = i/2$ and the frequency of $ c_2 = i/2$. Basically frequency analysis works if there is one alphabet that corresponds to another alphabet in a 1 to 1 correspondence.
To find a method for cryptanalysis we need to be more creative.
0 replies
fortenforge
Aug 30, 2009
0 replies
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Mathematics of the Vigenere Cipher
fortenforge   0
Aug 21, 2009
Ok, so I lied. I said that the next post was going to be about why frequency analysis fails on the Vigenere cipher but I decided to talk about how to mathematically define the cipher.

Let us say that we have already translated our plaintext into numbers (A = 0, B = 1, C = 2, ...). Let us say that the numbers are $ p_0, p_1, p_2, \cdots p_n$.

Let us say that we have chosen a key of length $ x$ and the letters of our key transformed into numbers are $ k_0, k_1, \cdots k_x$.

To encrypt plaintext number $ p_i$ we use $ k_j$ where $ j \equiv i \pmod{x}$. This accounts for the fact that the key is repeated over each letter of the plaintext. We use mod $ x$ because $ x$ is the length of the keyword.

Call $ c_i$ the corresponding ciphertext number to $ p_i$.

$ c_i \equiv p_i + k_j \pmod{26}$ where $ k_j \equiv i \pmod{x}$.

The first part is just the Caesar Cipher mathematically. The only difference is that as $ p_i$ changes $ k_j$ changes as well. This is what makes the Vigenere cipher a much better code.

Let's take an example:

plaintext: BLITZKRIEG
Numerical equivalent: 1 11 8 19 25 10 17 8 4 6

keyword: WAR
Numerical equivalent: 22 0 17

For the $ 0$th number of our plaintext, $ 1$, to find the equivalent key letter we take $ 0 \pmod{3} = 0$. So we take the $ 0$th keyword number which is $ 22$.

For the $ 5$th number of our plaintext, $ 10$ to find the equivalent key letter we take $ 5 \pmod{3} = 2$. So we take the $ 2$nd keyword number which is $ 17$.

We would continue this process for all the letters there by finding each letters keyword equivalent.

Notice that the first number in our list is being considered as our 0th number and the 2nd number is being considered as the 1st number to make the math work. In cryptography this is normal.

You can see that this method works by verifying it here:

WARWARWARW
BLITZKRIEG

The math method and the visual method match up, the 0th plaintext letter corresponds to the 0th key letter and the 5th plaintext letter corresponds to the 2nd key letter.

If we wanted to encrypt the 0th letter mathematically we would find:

$ 1 + 22 \pmod{26} \equiv 23$. So our 0th ciphertext number is 23.

If we wanted to encrypt the 5th letter mathematically we would find:

$ 10 + 17 \pmod{26} \equiv 1$. So our 5th ciphertext number is 1.

We would continue the process for all the letters.

Again when doing it mathematically and doing it without math you get the same ciphertext:

XLZPZBNIVC.

Learning what a cipher is mathematically is not much useful if you are decrypting a message by hand, but it is enormously useful if you are trying to program a cipher on a computer.
0 replies
fortenforge
Aug 21, 2009
0 replies
J
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Number of Possible Keys for Substitution cipher.
fortenforge   0
Jul 6, 2009
We know that the Monoalphabetic Substitution Cipher should have a lot more keys than the Caesar Cipher, but how many more?

Ptext: ABCDEFGHIJKLMNOPQRSTUVWXYZ
Ctext: ??????????????????????????

Let's look at the first "?" under the "A" in the ciphertext.
How many choices do we have for that "?". Well, we have $ 26$ choices because we can choose any letter of the alphabet. Let us say we chose "R".

Ptext: ABCDEFGHIJKLMNOPQRSTUVWXYZ
Ctext: R?????????????????????????

How many choices do we have for the next "?". We can choose any letter of the alphabet except "R", because we have already chosen that for the plaintext letter "A". So we have $ 25$ choices. Let's say we chose "E". Now for the next question mark we can't chose the letter "R" or "E" so we have $ 24$ choices. By now you should see the pattern, we have $ 26 \cdot 25 \cdot 24 \cdot 23 \ldots$ choices. That is equivalent to $ 26$ factorial.

$ 26! \approx 400000000000000000000000000$.

This is a lot of keys. Much much more keys than a Caesar Cipher. Unfortunately, with today's computers this is not that many keys. But this makes it impossible to try to crack a monoalphabetic substitution cipher using brute force by hand. There is however another method to crack this code...
0 replies
fortenforge
Jul 6, 2009
0 replies
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Mathematics of the Caesar Cipher
fortenforge   0
Jun 21, 2009
We can write the algorithm for the Caesar cipher in terms of math.

$ k$ is the key. $ p$ is the letter being encrypted and $ c$ is the encrypted letter. The variables p and c are used to represent the letter being encrypted because in cryptography we refer to the original message as the 'plaintext' and the encrypted message as the 'ciphertext'.

We know $ k$, represents a number because it is the key. But $ p$ and $ c$ are actually letters. We need to convert them into numbers. This is very simple. Represent A by 0, B by 1, C by 2 ... Z by 25.

When encrypting a message we are shifting it by $ k$ letters. In terms of numbers we are just adding $ k$ to $ p$ to get $ c$.

$ c = p + k$.

There is one problem with this. If $ p = 25$ and $ k = 1$ then $ c = 26$ which is a number we cannot convert to a letter. This problem occurs because of the 'wrapping around' from Z to A. To fix this we can use modular arithmetic. If you don't know what this is try googling it. We will almost always be working in mod 26 because there are 26 letters in the alphabet. Our new equation would be:

$ c \equiv p + k \text{ }(\text{mod } 26)$

This is how to encrypt a message. To decrypt a message instead of adding $ k$ we should subtract it.

$ c\equiv p - k \text{ }(\text{mod } 26)$
0 replies
fortenforge
Jun 21, 2009
0 replies
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No more topics!
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After Mathcounts
Existing_Human1   15
N Apr 6, 2025 by jb2015007
Hello Community!

I am officially done with my mathcounts career, as I have officially failed state, and I am now left an aloof blob reminiscing about the good old days.

So ... I was wondering if any of you have fun competitions I can do to relive the glory days of mathcounts. Obviously, their are the AMCs but I'm looking for something more team/travel based and one that preferably has a CDR.

Please specify if the competition is team based and if it has a cdr, and also when it takes place

Thank you in advance!
15 replies
Existing_Human1
Apr 5, 2025
jb2015007
Apr 6, 2025
J
After Mathcounts
G H J
Reveal topic tags
MATHCOUNTS
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G H BBookmark kLocked kLocked NReply
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Existing_Human1
208 posts
Existing_Human1
#1 h  Apr 5, 2025, 6:21 PM • 2 Y
Y by PikaPika999, RocketScientist
Hello Community!

I am officially done with my mathcounts career, as I have officially failed state, and I am now left an aloof blob reminiscing about the good old days.

So ... I was wondering if any of you have fun competitions I can do to relive the glory days of mathcounts. Obviously, their are the AMCs but I'm looking for something more team/travel based and one that preferably has a CDR.

Please specify if the competition is team based and if it has a cdr, and also when it takes place

Thank you in advance!
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Dream9
82 posts
Dream9
#2 h  Apr 5, 2025, 6:23 PM • 1 Y
Y by PikaPika999
BmMT. Berkely mini math tornement. it's team based and registration ends today. The competition is online and in-person on June 7th
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franklin2013
239 posts
franklin2013
#3 h  Apr 5, 2025, 7:12 PM • 1 Y
Y by PikaPika999
Yes! BmMT is definitely a great choice. in fact I'm competing in the online tournament!
Z K Y
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Andyluo
916 posts
Andyluo
#4 h  Apr 5, 2025, 7:23 PM • 1 Y
Y by PikaPika999
here's some math stuff I'm going to do in the future

USAMTS (if can oly qual)
SMT Online
Purple Comet
MC Nationals
ARML
(Bmmt is probably good)
AMC 10/12
AIME
USAJMO trust
Ohio Math League (state stuff)
HMMT November
HMMT Februrary
CAMP stuff?
MOP!?!?!?!??!?!??!?!?!? (not happening)
This post has been edited 2 times. Last edited by Andyluo, Apr 5, 2025, 7:28 PM
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huajun78
60 posts
huajun78
#5 h  Apr 5, 2025, 8:34 PM • 1 Y
Y by PikaPika999
there are a lot of mu alpha theta competitions
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jkim0656
714 posts
jkim0656
#6 h  Apr 5, 2025, 8:36 PM • 1 Y
Y by PikaPika999
give me some info on BmMt pls
im intrested maybe for next year
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jkim0656
714 posts
jkim0656
#7 h  Apr 5, 2025, 8:36 PM • 1 Y
Y by PikaPika999
also my newbie brain does not know what HMMT is
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huajun78
60 posts
huajun78
#8 h  Apr 5, 2025, 8:39 PM • 1 Y
Y by PikaPika999
Harvard MIT Math Tournament
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DDCN_2011
352 posts
DDCN_2011
#9 h  Apr 5, 2025, 9:19 PM • 1 Y
Y by PikaPika999
im js doing amc 10 i also failed mc D:
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RocketScientist
336 posts
RocketScientist
#10 h  Apr 6, 2025, 1:46 AM • 1 Y
Y by PikaPika999
relatable post :(
from what i've heard, college comps like hmmt, pumac or cmimc are really fun
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Existing_Human1
208 posts
Existing_Human1
#11 h  Apr 6, 2025, 2:08 AM • 1 Y
Y by PikaPika999
Anything with CDR? I need to cope with the fact that FTW isn't useful

Also, can you put the registration dates? (Sorry, I'm just lazy)
This post has been edited 1 time. Last edited by Existing_Human1, Apr 6, 2025, 2:08 AM
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mathprodigy2011
286 posts
mathprodigy2011
#12 h  Apr 6, 2025, 2:31 AM • 1 Y
Y by PikaPika999
Existing_Human1 wrote:
Hello Community!

I am officially done with my mathcounts career, as I have officially failed state, and I am now left an aloof blob reminiscing about the good old days.

So ... I was wondering if any of you have fun competitions I can do to relive the glory days of mathcounts. Obviously, their are the AMCs but I'm looking for something more team/travel based and one that preferably has a CDR.

Please specify if the competition is team based and if it has a cdr, and also when it takes place

Thank you in advance!

ARML is a cool alternative
Z K Y
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jb2015007
1845 posts
jb2015007
#13 h  Apr 6, 2025, 2:56 AM • 1 Y
Y by PikaPika999
franklin2013 wrote:
Yes! BmMT is definitely a great choice. in fact I'm competing in the online tournament!

same! do u already have a team?
if not PM me!
Z K Y
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Existing_Human1
208 posts
Existing_Human1
#14 h  Apr 6, 2025, 3:17 AM • 1 Y
Y by PikaPika999
For BMMT I'll be out of middle school in june, do I still count as an 8th grader?
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franklin2013
239 posts
franklin2013
#15 h  Apr 6, 2025, 3:59 PM • 1 Y
Y by PikaPika999
jb2015007 wrote:
franklin2013 wrote:
Yes! BmMT is definitely a great choice. in fact I'm competing in the online tournament!

same! do u already have a team?
if not PM me!

I already have a team with 4 people.
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jb2015007
1845 posts
jb2015007
#16 h  Apr 6, 2025, 4:02 PM • 1 Y
Y by PikaPika999
Oh ok nvm then.
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