Search results
Results from the WOW.Com Content Network
Answer: 7 × 1 + 6 × 10 + 5 × 9 + 4 × 12 + 3 × 3 + 2 × 4 + 1 × 1 = 178 mod 13 = 9 Remainder = 9 A recursive method can be derived using the fact that = and that =. This implies that a number is divisible by 13 iff removing the first digit and subtracting 3 times that digit from the new first digit yields a number divisible by 13.
d() is the number of positive divisors of n, including 1 and n itself; σ() is the sum of the positive divisors of n, including 1 and n itselfs() is the sum of the proper divisors of n, including 1 but not n itself; that is, s(n) = σ(n) − n
Some sequences have alternate names: 4n+1 are Pythagorean primes, 4n+3 are the integer Gaussian primes, and 6n+5 are the Eisenstein primes (with 2 omitted). The classes 10 n + d ( d = 1, 3, 7, 9) are primes ending in the decimal digit d .
This is denoted as 20 / 5 = 4, or 20 / 5 = 4. [2] In the example, 20 is the dividend, 5 is the divisor, and 4 is the quotient. Unlike the other basic operations, when dividing natural numbers there is sometimes a remainder that will not go evenly into the dividend; for example, 10 / 3 leaves a remainder of 1, as 10 is not a multiple of 3.
Get AOL Mail for FREE! Manage your email like never before with travel, photo & document views. Personalize your inbox with themes & tabs. You've Got Mail!
The search engine that helps you find exactly what you're looking for. Find the most relevant information, video, images, and answers from all across the Web.
For example, there are six divisors of 4; they are 1, 2, 4, −1, −2, and −4, but only the positive ones (1, 2, and 4) would usually be mentioned. 1 and −1 divide (are divisors of) every integer. Every integer (and its negation) is a divisor of itself. Integers divisible by 2 are called even, and integers not divisible by 2 are called odd ...
Every prime number p divides a Fibonacci number that can be determined by the value of p modulo 5. If p is congruent to 1 or 4 modulo 5, then p divides F p−1, and if p is congruent to 2 or 3 modulo 5, then, p divides F p+1. The remaining case is that p = 5, and in this case p divides F p.