Cracking Passwords: Calculating Outcomes with 3 Characters

  • Thread starter Thread starter Raza
  • Start date Start date
AI Thread Summary
The discussion focuses on calculating the number of possible outcomes for passwords of varying lengths and character types. For a 3-digit password, there are 1,000 combinations (10^3). If the password consists of 3 characters that can be either numbers or letters, the total outcomes are calculated as (26+10)^3. The conversation also touches on the complexity of outcomes when restrictions are applied, such as the format of license plates. Finally, there is a clarification on calculating outcomes for a 12-character password using either 36^12 or 62^12, depending on case sensitivity.
Raza
Messages
203
Reaction score
0
Hi,
I have recently cracked a password for a zip file using brute force which got me thinking. Let's say that the password is only 3 digits, how many outcome could it possibly have?
I know it's 999 but is there a equation that I can use to figure this out?
or let's say that it contains only 3 alphabets, how many outcome could it possibly have, or let's say that the password's length is three characters and it could have either alphabets or numbers, what are the possible outcome?

Please give me brief explanation and some equation.

Thanks
 
Mathematics news on Phys.org
in general, if you have a items with b places to but them, the number of outcomes is a^b.

for 3 digits there are actually 10^3= 1000 different choices (which can easily be seen by considering it is 000=choice 1, 001 = choice 2,...999= choice 1000).

if a password is 3 characters and can have alphabets or numbers, it is (26+10)^3.

these questions become harder if you start putting restrictions on order.
 
For each position, you know how many possible options there are, so you just multiply out to get the total number of possible results.

Like, if you have a single number position, and you are working in base 10, then there are 10 possible results, 0-9.

If you have a 2-digit number, then each position (the 1's postion and the 10's position) has 10 possibilities, so you get a total of 10 * 10 = 100 possible results (0-99).

If you have a capital character position, A-Z, that position has 26 possible outcomes. Two capital letter positions have 26 * 26 possible outcomes, etc.

So as a final example, in California, the standard license plates are formatted like this:

<1 number> <3 chars> <3 numbers>

So the total number of possible plates in that format would be 10 * 26^3 * 10^3.

Makes sense?


EDIT -- matticus beat me to it!
 
Thanks, I got it.
It's simply amazing how math works.
Right now, I am cracking another zip password, which is 12 characters, alphabets and numbers, so it would be (10+26)12,right?
 
Raza said:
Thanks, I got it.
It's simply amazing how math works.
Right now, I am cracking another zip password, which is 12 characters, alphabets and numbers, so it would be (10+26)12,right?

So now you're asking for help hacking passwords... :rolleyes:

No, sorry, your answer is all wrong. It should be the two's-complement of that number to be correct. :bugeye:
 
Raza said:
Right now, I am cracking another zip password, which is 12 characters, alphabets and numbers, so it would be (10+26)12,right?

36^12 if all letters are known to be lowercase (or case-insensitive); 62^12 if letters can be of any case.
 
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Thread 'Imaginary Pythagorus'
I posted this in the Lame Math thread, but it's got me thinking. Is there any validity to this? Or is it really just a mathematical trick? Naively, I see that i2 + plus 12 does equal zero2. But does this have a meaning? I know one can treat the imaginary number line as just another axis like the reals, but does that mean this does represent a triangle in the complex plane with a hypotenuse of length zero? Ibix offered a rendering of the diagram using what I assume is matrix* notation...
Back
Top