Which is the only even prime number?

Answer 1

2 is the only even prime number, and it divides evenly into every even number. 2 is the smallest factorial prime and the only factorial that is prime. The first 30 prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, and 113 .

Prime numbers are numbers greater than 1 that are divisible by exactly two numbers: 1 and themselves. For example, the prime numbers through 20 are 2, 3, 5, 7, 11, 13, 17 and 19. As numbers grow larger, prime numbers become scarcer. If you examine the numbers from 1 to 100, you'll find that 25% of them are prime numbers. When you extend the list and check the numbers from 1 to 1000, the percentage drops dramatically to 16.8%. Only about 7% of the numbers from 1 to 1,000,000 are prime numbers.

Mathematicians believe (but have not conclusively proven) that every even number is the sum of two prime numbers except for the number 2.♥

Answer 2

2

Answer 3

2

Answer 4

2

Answer 5

2

Answer 6

2

Answer 7

all of us comprehend that is genuine because of the fact we can tutor it. What does it advise to be a sq. sort? It means that our sq. root is an integer (and consequently a rational sort - a sort which will nicely be expressed as a fragment of two integers a,b interior the kind a/b). although we can tutor that the sq. root of any top sort is irrational (that is, it won't be able to be expressed as a fragment of integers). enable's tutor it: enable p be a main, anticipate for contradiction that the sq. root of p is rational. If sqrt(p) is rational then there exist integers a,b such that sqrt(p)=a/b. additionally anticipate (without loss of generality) that gcd(a,b)=one million, that is that a and b are coprime and subsequently a/b is a fragment in lowest phrases. if that's the case then pI(a^2) (p divides (a^2)) which suggests pIa. If pIa then there exists some q interior the integers such that a=pq, then p(b^2)=(a^2)=(pq)^2=(p^2)(q^2) and consequently b^2=p(q^2). If b^2=p(q^2) then pI(b^2) which suggests pIb consequently p is a trouble-free component or the two a and b. although a and b are coprime and subsequently we've a contradiction. consequently the sq. root of any top p is irrational. If the sq. root of any top is irrational then no top is often a sq. sort.

Answer 8

2 because it is the only even number which is divisible only by 1 and itself.

Answer 9

maybe hes just young you dont have to call him dumb.the answer is 2 honey