Aug 29, 2009 · Much easier if u use bananas and a group of people. Say you have 1 banana and group of 6 people, this you would express: 1 mod 6 / 1 % 6 / 1 modulo 6.

Jul 8, 2013 · That said, your intuition was that it could be -2 and not 5. Actually, in modular arithmetic, -2 = 5 (mod 7) because it exists k in Z such that 7k - 2 = 5. You may not have …

When your modulus n is a power of 2, then x mod n will count up in binary from 0 to n-1, back to 0, to n-1, etc; for modulus n that looks like binary 01xxxxx, x mod n will cycle through every of …

Nov 17, 2015 · The difference between mod and rem is subtle, but important. (-1 mod 2) would normally give 1. More specifically given two integers, X and Y, the operation (X mod Y) tends …

May 9, 2010 · Obviously, sequence of 2^n mod 11 will be cyclical. 2^0 mod 11 = 1 2^1 mod 11 = 2 2^2 mod 11 = 4 2^3 mod 11 = 8 2^4 mod 11 = 5 2^5 mod 11 = 10 2^6 mod 11 = 9 2^7 mod …

Dec 7, 2011 · If you don't see any fraction then the mod is 0 like 50 / 5 = 10 (mod is 0). The remainder fraction is shown in reduced form, so 60 / 8 will result in 7 1/2. Remainder is 1/2 …

Instead, you can use mod_authnz_sspi which is compatible with Apache 2.4 for Windows (x86 and x64). This module used to be available from ApacheHaus.net but their website is down …

Feb 1, 2010 · In the step where we calculate 5^1 mod 221, 5^2 mod 221, etc. we note that 5^(2^k) = 5^(2^(k-1)) * 5^(2^(k-1)) because 2^k = 2^(k-1) + 2^(k-1) so that we can first compute 5^1 …

Jun 19, 2010 · There is only a simple way to find modulo of 2^i numbers using bitwise. There is an ingenious way to solve Mersenne cases as per the link such as n % 3, n % 7... There are …

Apr 18, 2009 · The identity you're looking for is x mod N = (x mod 2^q)- c*floor(x/2^q), given that N = 2^q + c and c is any integer (but typically ±1). You may want to read section 9.2.3: "Moduli …