Homework Statement , relevant equations, and the attempt at a solution are all in the attached file.
I was reading through Invitation to Discrete Mathematics and attempted to solve an exercise that involved a proof. I've typeset everything in LaTeX and made a PDF out of it so that it does not...
Dream case scenario I would get that
\text{Given that } S \text{ is an array of } m \text{ small nonnegative integers and } \alpha \text{ is a prime larger than the largest integer in } S
\sum_{i=0}^{m-1}\left(\alpha^{m-(i+1)}\times S_{1_i} \right) \ne...
This is a self-assigned question. Not homework.
I may have the right answer, but would like some reviewing. It came to me while reading on a CS topic, this did not come from a math textbook, otherwise it would be slightly more in context.
I did not use any formula (so 2 would be empty) and 1...
"formula known to you earlier for certain numbers". I know about 3 and 41 by heart.
I do not know if it is a property, I just happen to remember it.
n : 3 ** n % 100
0 : 1
1 : 3
2 : 9
3 : 27
4 : 81
5 : 43
6 : 29
7 : 87
8 : 61
9 : 83
10 : 49
11 : 47
12 : 41
13 : 23
14 : 69
15 : 7
16 : 21
17 : 63...
Your English is so bad that I can't even understand if you are getting it or not. Add a space after your punctuation.
If the remainder of the division by 100 was 43 I would finish with 4 + 3 = 7.
As the remainder is 3, I end up with 0 + 3 = 3. Get it? It is modulo 100 and not modulo 10.
"I want to do it myself and possibly learn something" way:
27 ^ {27} = \left( 3^3 \right) ^ {27} = 3 ^ {81}
Now finish it. I find easier to work with a power of 3, you may think differently, though.
I provide two solutions below. I wouldn't peek if I were you, though.
I had never written the proof myself. Inspired by what jedishrfu wrote, my attempt follows.
To avoid the problem of showing unnecessary \LaTeX, the proof is hidden under spoiler tags.
Making a keylogger run in Windows is really easy.
Still, if he typed the URL manually (or at least part of it), the keylogger output parser should retrieve it automatically and allow whoever has it to have some fun.