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For example, the expression "5 mod 2" evaluates to 1, because 5 divided by 2 has a quotient of 2 and a remainder of 1, while "9 mod 3" would evaluate to 0, because 9 divided by 3 has a quotient of 3 and a remainder of 0. Although typically performed with a and n both being integers, many computing systems now allow other types of numeric operands.
Time-keeping on this clock uses arithmetic modulo 12. Adding 4 hours to 9 o'clock gives 1 o'clock, since 13 is congruent to 1 modulo 12. In mathematics, modular arithmetic is a system of arithmetic for integers, where numbers "wrap around" when reaching a certain value, called the modulus.
His proposal was the R-36O (8K69 by GRAU index and SS-9 Mod 3 Scarp by NATO report) which was approved for development by Soviet officials on April 16, 1962. [11] [3] [4] Yangel used an ICBM design of his own, the R-36 (NATO designation SS-9 Scarp), as a base for the R-36O. [9]
Modulo is a mathematical jargon that was introduced into mathematics in the book Disquisitiones Arithmeticae by Carl Friedrich Gauss in 1801. [3] Given the integers a, b and n, the expression "a ≡ b (mod n)", pronounced "a is congruent to b modulo n", means that a − b is an integer multiple of n, or equivalently, a and b both share the same remainder when divided by n.
If the digit 9 is ignored when summing the digits, the effect is to "cast out" one more 9 to give the result 12. More generally, when casting out nines by summing digits, any set of digits which add up to 9, or a multiple of 9, can be ignored. In the number 3264, for example, the digits 3 and 6 sum to 9.
Single nuclear warhead of 20 megatons TNT (NATO codename SS-9 Mod 1). [18] Single nuclear warhead of 8.3 megatons TNT (NATO codename SS-9 Mod 2). [18] The first launch of an R-36 took place on September 28, 1963, and ended ignominiously when the missile lost thrust one second after liftoff and fell back onto the pad, causing a fire. [19]
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For each of them, compute the remainder by 4 (the second largest modulus) until getting a number congruent to 3 modulo 4. Then one can proceed by adding 20 = 5 × 4 at each step, and computing only the remainders by 3. This gives 4 mod 4 → 0. Continue 4 + 5 = 9 mod 4 →1. Continue 9 + 5 = 14 mod 4 → 2. Continue 14 + 5 = 19 mod 4 → 3.