Which of the following is a composite number?
53
19
39
47
Answers
Answer:
39 is the answer
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Answer:
A composite number n is a positive integer n>1 which is not prime (i.e., which has factors other than 1 and itself). The first few composite numbers (sometimes called "composites" for short) are 4, 6, 8, 9, 10, 12, 14, 15, 16, ... (OEIS A002808), whose prime decompositions are summarized in the following table. Note that the number 1 is a special case which is considered to be neither composite nor prime.
n prime factorization n prime factorization
4 2^2 20 2^25
6 2·3 21 3·7
8 2^3 22 2·11
9 3^2 24 2^3·3
10 2·5 25 5^2
12 2^23 26 2·13
14 2·7 27 3^3
15 3·5 28 2^27
16 2^4 30 2·3·5
18 2·3^2 32 2^5
The nth composite number c_n can be generated using the Wolfram Language code
Composite[n_Integer] :=
FixedPoint[n + PrimePi[#] + 1&, n]
The Dirichlet generating function of the characteristic function of the composite numbers c_n is given by
sum_(n=1)^(infty)([n in {c_k}_(k=1)^infty])/(n^s) = sum_(n=1)^(infty)1/(c_n^s)
(1)
= 1/(4^s)+1/(6^s)+1/(8^s)+1/(9^s)+...
(2)
= zeta(s)-1-P(s),
(3)
where zeta(s) is the Riemann zeta function, P(s) is the prime zeta function, and [S] is an Iverson bracket.
There are an infinite number of composite numbers.
The composite number problem asks if there exist positive integers m and n such that N=mn.
A composite number C can always be written as a product in at least two ways (since 1·C is always possible). Call these two products
C=ab=cd,
(4)
then it is obviously the case that c|ab (c divides ab). Set
c=mn,
(5)
where m is the part of c which divides a, and n is the part of c which divides b. Then there are p and q such that
a = mp
(6)
b = nq.
(7)
Solving ab=cd for d gives
d=(ab)/c=((mp)(nq))/(mn)=pq.
(8)
It then follows that
S = a^2+b^2+c^2+d^2
(9)
= m^2p^2+n^2q^2+m^2n^2+p^2q^2
(10)
= (m^2+q^2)(n^2+p^2).
(11)
It therefore follows that a^2+b^2+c^2+d^2 is never prime! In fact, the more general result that
S=a^k+b^k+c^k+d^k
(12)
is never prime for k an integer >=0 also holds (Honsberger 1991).