modulus

At least in MySQL v5.1.73 (CentOS 6.6), MOD() function returns a bogus result... unless someone can explain how this is actually correct. mysql> select MOD(-385.4784399 ,1440); +-------------------------+ | MOD(-385.4784399 ,1440) | +-------------------------+ | -385.4784399 | +-------------------------+ 1 row in set (0.01 sec) Is this the best alternative? mysql> select -385.478439885319 - 1440 * FLOOR(-385.478439885319/1440); +----------------------------------------------------------+ | -385.478439885319 - 1440 * FLOOR(-385.478439885319/1440) | +---------------------------------------------

I have to find c, c = (a*b) mod m a,b,c,m are 32-bit integers. But (a*b) can be more than 32 bits. I am trying to figure out a way to compute c, without using long or any data type > 32 bit. Any ideas? What if m is a prime number, can things be simplified? Note: based on a few comments, c = ((a mod m) * (b mod m)) mod m, but in my case even this multiplication will overflow I happen to have this 256 by 128 bit division code with a random-based test and I'm sure you can reduce it trivially to 128 by 64 bit division and then to 64 by 32: #include <limits.h> #include <stdio.h> #include <stdlib.h>

I'd searching many websites trying to understand how does RSA works. I have a modulus "A89F25A56FA6DA258C8CA8B40427D927B4A1EB4D7EA326BBB12F97DED70AE5E4480FC9C5E8A972177110A1CC318D06D2F8F5C4844AC5FA79A4DC470BB11ED635699C17081B90F1B984F12E92C1C529276D8AF8EC7F28492097D8CD5BECEA16FE4088F6CFAB4A1B42328A1B996F9278B0B7E3311CA5EF856C2F888474B83612A82E4E00D0CD4069A6783140433D50725F" and exponent "03" and i have to decrypt information formated in hex bytes. My questions are: How do i create a public key? Once i have the public key do i have to encode in base64 or the public key is ready to decrypt?

This seems to be the #1 thing that is asked when dealing with Remainder/Mod, and I'm kind of hitting a wall with it. I'm teaching myself to program with a textbook and a chuck of C code. Seeing as I don't really have an instructor to say, "No, no. It actually works like this", I thought I'd try my hand here. I haven't found a conclusive answer to the mathematical part of this, though. So... I'm under the impression that this is a pretty rare occurrence, but I'd still like to know what it is that happens underneath the shiny compiling. Plus, this textbook would like for me to supply all values

I need a way to calculate: (g^u * y^v) mod p in Java. I've found this algorithm for calculating (g^u) mod p: int modulo(int a,int b,int c) { long x=1 long y=a; while(b > 0){ if(b%2 == 1){ x=(x*y)%c; } y = (y*y)%c; // squaring the base b /= 2; } return (int) x%c; } and it works great, but I can't seem to find a way to do this for (g^u * y^v) mod p as my math skills are lackluster. To put it in context, it's for a java implementation of a "reduced" DSA - the verifying part requires this to be solved. Assuming that the two factors will not overflow, I believe you can simplify an expression like

How to calculate modulus of the form (a*b)%c? i want to calculate modulus of multiplication of two int numbers where they are almost in the stage of overflow... here c is also int (a * b) % c == ((a % c) * (b % c)) % c What about ((a % c) * (b % c)) % c ? Depending on your architecture this could be faster or slower than casting to a bigger type. You may cast a and c to long long , so the multiplication won't overflow. ((long long)a * (long long)b) % c 来源： https://stackoverflow.com/questions/5915808/how-to-calculate-modulus-of-the-form-abc

I want to create a list of string that resemble the column letters in Microsoft Excel. For example, after 26 columns, the next columns become AA , AB , AC , etc. I have tried using the modulus operator, but I just end up with AA , BB , CC , etc... import string passes_through_alphabet = 0 for num, col in enumerate([_ for _ in range(40)]): if num % 26 == 0: passes_through_alphabet += 1 excel_col = string.ascii_uppercase[num%26] * passes_through_alphabet print(num, excel_col) 0 A 1 B 2 C 3 D ... 22 W 23 X 24 Y 25 Z 26 AA 27 BB 28 CC ... You can use itertools.product for this: >>> import

Using this code to retrieve the public key bytes... var pubKey = AppDomain.CurrentDomain.DomainManager.EntryAssembly .GetName().GetPublicKey(); What is this common structure at the start (first 32 bytes) of the key? It's not ASN.1 and it might not be variable. I can google it and get repeats. // 00 24 00 00 04 80 00 00 94 00 00 00 06 02 00 00 00 24 00 00 52 53 41 31 Is it all reversed or just part of it (e.g. the modulus at the end)? 52 53 41 31 is a string of RSA1 . My key's modulus is 1024 bit, so I was looking for something that described the length. 0x0400 ( 00 04 B.E.) would be 1024 (bits

I want to find a way to know if an integer is divided by 3 or 7 without using division, because it is very slow in MIPS assembly. I have done a lot of research but found nothing. There's a method described by Granlund & Montgomery that requires the modular / multiplicative inverse of the (odd) divisor modulo 2**b . (Some parts of this paper have been improved recently ) The divisors: (d) = 3, 7 (odd numbers) are an easy case. Assuming 32 bit (unsigned) arithmetic, the inverses modulo 2**32 yield 2863311531 (0xAAAAAAAB) and 3067833783 (0xB6DB6DB7) respectively. There's an online calculator here