#include<iostream.h> #include<conio.h>
int main()
{
int i,n;
clrscr();
cout<<"PROGRAM TO CHECK IF THE NUMBER IS PRIME OR NOT:"<<endl;
cout<<"Enter a number:";
cin>>n;
for(int i=2;i<n;i++)
{
if(n%i==0)
cout<<"\nTHE NUMBER IS COMPOSITE"<<endl;
else
cout<<"\nTHE NUMBER IS PRIME"<<endl;
}
return 0;
}
It's even. All even numbers (except for 2) are composite.
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By learning how to program on C+.
Since there is an infinite set of prime numbers the answer would be infinity.
Any number of the form n = a*b*c*d*e*f where a, b, c, d, e and f are different prime numbers. n has 26 = 64 factors in total, of which 1 is the number 1 (neither prime nor composite), 6 are prime, and the remaining 57 are composite.
Product of a prime number and a composite number results in a composite number.Now consider the product of a composite number(a) and a prime number(b) is equal to c.i.e. c = a x bIt is clear that c is divisible by both a and b.Also c is divisible by itself and 1, this means that c has more than two factors.A number having more than two factors is composite, therefore product of a prime number and a composite number results in a composite number.
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3, 61 and 113 are prime; the rest composite.
1, which is 2 b/c all of the other even numbers are divisible by 2, therefore making it composite.
The answer depends on the value of C, which is unknown.
Assuming that a and c are both integers, 10ac must be composite because it is equal to 2*5*a*c.
yes b/c 1 is neither prime or composite