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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.
The constants R mod N and R 3 mod N can be generated as REDC(R 2 mod N) and as REDC((R 2 mod N)(R 2 mod N)). The fundamental operation is to compute REDC of a product. When standalone REDC is needed, it can be computed as REDC of a product with 1 mod N. The only place where a direct reduction modulo N is necessary is in the precomputation of R ...
The most direct method of calculating a modular exponent is to calculate b e directly, then to take this number modulo m.Consider trying to compute c, given b = 4, e = 13, and m = 497:
However, the linear congruence 4x ≡ 6 (mod 10) has two solutions, namely, x = 4 and x = 9. The gcd(4, 10) = 2 and 2 does not divide 5, but does divide 6. Since gcd(3, 10) = 1, the linear congruence 3x ≡ 1 (mod 10) will have solutions, that is, modular multiplicative inverses of 3 modulo 10 will exist. In fact, 7 satisfies this congruence (i ...
If the time is 7:00 now, then 8 hours later it will be 3:00. Simple addition would result in 7 + 8 = 15, but 15:00 reads as 3:00 on the clock face because clocks "wrap around" every 12 hours and the hour number starts again at zero when it reaches 12. We say that 15 is congruent to 3 modulo 12, written 15 ≡ 3 (mod 12), so that 7 + 8 ≡ 3 ...
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.
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For any integer n, n ≡ 1 (mod 2) if and only if 3n + 1 ≡ 4 (mod 6). Equivalently, n − 1 / 3 ≡ 1 (mod 2) if and only if n ≡ 4 (mod 6) . Conjecturally, this inverse relation forms a tree except for the 1–2–4 loop (the inverse of the 4–2–1 loop of the unaltered function f defined in the Statement of the problem section of ...