Modulus operator can be applied to which of these

Answer: a Explanation: None. 5. Which of these statements are incorrect? a) Assignment operators are more efficiently implemented by Java run-time system than their equivalent long forms b) Assignment operators run faster than their equivalent long forms c) Assignment operators can be used only with numeric and character data type d) None of the mentioned View Answer

Computational operation In modulo operation returns the Given two positive numbers a and n, a modulo n (often abbreviated as a mod n) is the remainder of the a by n, where a is the n is the For example, the expression "5 mod 2" would evaluate to 1, because 5 divided by 2 has a Although typically performed with a and n both being n is 0 to n − 1 inclusive ( a mod 1 is always 0; a mod 0 is undefined, possibly resulting in a When exactly one of a or n is negative, the naive definition breaks down, and programming languages differ in how these values are defined. Variants of the definition [ ] In least positive residue, the smallest non-negative integer that belongs to that class (i.e., the remainder of the In nearly all computing systems, the quotient q and the remainder r of a divided by n satisfy the following conditions: ( 1) However, this still leaves a sign ambiguity if the remainder is non-zero: two possible choices for the remainder occur, one negative and the other positive, and two possible choices for the quotient occur. In number theory, the positive remainder is always chosen, but in computing, programming languages choose depending on the language and the signs of a or n. n or a is negative (see the table under a modulo 0 is undefined in most systems, although some do define it as a. Quotient ( q) and remainder ( r) as functions of dividend ( a), using truncated division Many implementations use truncated division, for which the quotient is defined by q = [ a n ]...

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A mod B = ( A + K ⋅ B ) mod B A \text K K . It is true that 5 divided by 3 gives you a remainder of 2 However, it is NOT true that -5 divided by 3 gives you a remainder of 2. It gives you a remainder of 1. We can do some long division to prove it: First we'll do the simple 5/3 _1_R2 3 / 5 -3 (Since 1 * 3 = 3) --- 2 We can check our result: 5 = 1 * 3 + 2 Now we can do -5/3 _-2_R1 3 / -5 -(-6) (Since -2 * 3= -6) --- 1 We can check our result: -5 = -2 * 3 + 1 The -2 seems strange since we might think 3 goes into 5, -1 times. But this would make our remainder negative (which is not allowed) Let's see what would happen, if we allowed negative remainders (we don't, but computers do!) it would look like this: _-1_R-2 3 / -5 -(-3) (Since -1 * 3= -3) --- -2 We can check our result: -5 = -1 * 3 + (-2) Notice that if we go 2 steps counter clockwise on the modular circle for 3 that we end up at 1. Notice that if we add one multiple of 3 to -2 we end up at 1. Again, we don't allow negative remainders, but it may give you some intuition for what is going on, and for how congruence modulo works later. Hope this makes sense Johann Carl Friedrich Gauss is usually attributed with the invention/discovery of modular arithmetic. In 1796 he did some work that advanced the field, and in 1801 published the book Disquisitiones Arithmeticae which, amongst other things, introduced congruence modulo and the ≡ symbol. So he is the person that laid out the modern approach to modular arithmetic that we ...