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for prime p>= 5
p = (6n+/-1)
p^2 + 2 = 36n^2+1 +/- 12n +2 = 36n^2+3 +/- 12n
divisible by 3 and p^2+2 >= 27 so 2nd factor >=9 so not a prime or composite
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What is the largest three digit number that has four factors?
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
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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)
Can a multiple of a prime number be prime?
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.
How do you caluculate the GCD of two prime numbers?
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
Is Composite numbers have only 2 factors?
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.
What is the mathematical proof to show that the number of prime numbers is infinite?
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.
What number has exactly 7 factor pairs?
If p is any prime number, then p12 would meet the requirements. The seven factor pairs would be: (p0, p12) (p1, p11) (p2, p10) (p3, p9) (p4, p8) (p5, p7) (p6, p6) It can be shown that, because 7 is a prime, a composite number cannot generate such a solution.
Why is the sum of two prime numbers greater than two never prime?
Consider prime numbers p1, p2 greater than 2. Since p1 and p2 are prime and greater than 2, they are both necessarily odd. Hence, they are of the form: p1 = 2k+1, p2 = 2j+1, where j and k are positive integers. Their sum is then: p1+p2 = (2k+1)+(2j+1) = 2k+2j+2 = 2 (k+j+1). So 2 divides their sum, hence the sum can't be prime
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There cannot be such a number. Suppose the number is P. Then 10% of P is P/10 Their multiple is P*P/10 = P2/10 If P2/10 = 726 then P2 = 7260 so that P = sqrt(7260) = 85.2 (approx) which is not an integer and so cannot be a prime.
What is the form of the Pythagorean triple generator?
If p and q are integers, then a = p2 - q2 b = 2pq, and c = p2 + q2 form a Pythagorean triple. Furthermore, if p and q are co-prime then the triple is primitive Pythagorean.
How do you get the black boat in the prime pier in pokemon black?
I have a feeling it is the boat u ride to the p2 lab for the genesect event.
Why have there been no direct proofs of the intinity of primes and naturals after 2000 years of study?
Sorry, but both of these have direct and relatively simple proofs. To prove the infinity of primes, let p1, p2, p3, ... pn be the first n primes. Form the product p1 x p2 x p3 ... x pn. Clearly it is divisible by all the prime numbers up to pn. Then add 1 to that sum. The new value cannot be divisible by any of the preceding primes so it must be a new prime number. Because you can do this for any value of n, the number of primes must be infinite. The proof of the irrationality of the square root of 2 is similarly easy and should be available in any high-school math book at the level of Algebra 2 or above.
