A number will have 12 factors if it satisfies any of the following prime factor configurations: P11 (P1)5*(P2) (P1)3*(P2)2 (P1)2*(P2)*(P3)
p2 - 2p + 2 can be factored as (p - 1)(p - 1)which can be written as ( p - 1)2.
It depends on whether 1 is considered a factor. If yes, then the answer to the question is no, since 1 and the number itself are the two factors implying that it is a prime. If 1 is not considered as a factor, then the square of any prime meets the requirements. If p is a prime, then p and p2 are the only two factors of p2.
P(x=n1,y=n2) = (n!/n1!*n2!*(n-n1-n2)) * p1^n1*p2^n2*(1-p1-p2) where n1,n2=0,1,2,....n n1+n2<=n
The proof is by contradiction: assume there is a finite number of prime numbers and get a contradiction by requiring a prime that is not one of the finite number of primes. Suppose there are only a finite number of prime numbers. Then there are n of them.; and they can all be listed as: p1, p2, ..., pn in order with there being no possible primes between p(r) and p(r+1) for all 0 < r < n. Consider the number m = p1 × p2 × ... × pn + 1 It is not divisible by any prime p1, p2, ..., pn as there is a remainder of 1. Thus either m is a prime number itself or there is some other prime p (greater than pn) which divides into m. Thus there is a prime which is not in the list p1, p2, ..., pn. But the list p1, p2, ..., pn is supposed to contain all the prime numbers. Thus the assumption that there is a finite number of primes is false; ie there are an infinite number of primes. QED.
By definition, prime number p has two factors, 1 and p. P2 has three factors, 1, p and p2 Therefore, p2 is composite.
No. Let p1 be a prime number. Let p2 be a multiple of p1 such that p2 = p1 * k. Then the factors of p2 are: 1, p1, k and p2. ==> p2 is not a prime number. Hence, a multiple of a prime number cannot be a prime number.
A number will have 12 factors if it satisfies any of the following prime factor configurations: P11 (P1)5*(P2) (P1)3*(P2)2 (P1)2*(P2)*(P3)
p2 - 2p + 2 can be factored as (p - 1)(p - 1)which can be written as ( p - 1)2.
If factors include 1 and the number itself then the any number (n) with only 4 factors has to be the product of two prime numbers, p1 and p2. The factors are then n, p1, p2, and 1. The largest 3-digit number with only 4 factors is 998. The factors are 998, 499, 2 and 1
3 and p2 are both factors of each part of the binomial, so divide it from each part and see what you have left. (3p4 / 3p2) - (12p2 / 3p2) (3p2)(p2-4)◄
p2 + 3p - 4What two factors of - 4 add up to 3 ??(p - 1)(p + 4)===============so,p = 1p = - 4
Let p1 and p2 be the two prime numbers. Because they are prime, their divisors are div(p1) = {1,p1} and div(p2) = {1,p2}. So GCD(p1,p2) = Greatest Common Divisor of p1 and p2 = p1 if p1 equals p2 1 if p1 is different from p2
The bond order of P2 is 2.
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p2+10d+7
p2 + 3p = p (p + 3)