int main()
{
int i,j,n,f=0;
printf("Enter the length\n");
scanf("%d",&n);
for(i=2;i<=n;i++)
{
for(j=2;j<i;j++)
{
if((i%j)==0)
{
f=1;
break;
}
f=0;
}
if(f==0)
printf("%d ",i);
}
}
No 5291 is not a prime using 2 numbers. It is a prime using three numbers.
17The prime numbers between 10 and 20 are 11, 13, 17 and 19. There is only one square number between 10 and 20: 16. Using the criteria given, the number that fits is 17.
Oh, what a lovely request! In FoxPro, you can create a program to print all prime numbers from 1 to 100 by using a loop to check each number for divisibility only by 1 and itself. If it meets this criteria, you can print it out on the screen. Remember, every number is unique and special, just like a happy little tree in a vast forest.
#include<iostream.h> #include<conio.h> void prime(int n) { clrscr(); int num; cout<<"enter the numbers"<<endl; cin>>num; prime(num); getch(); } void prime(int n) { int prime=1,i; for(i=2;i<=n/2;i++) if(n%i==1) prime=0; if(prime==1) cout<<"the number"<<n>>"is prime"; else cout<<"the number"<<n<<"is not prime"; }
47 is a prime number.
11 is a prime number.
41 is a prime number.
None. 887 is a prime number.
NA: 13 is a prime number.
To write a C program to find prime numbers between 1 to 500, you can use a nested loop structure. In the outer loop, iterate from 2 to 500, and in the inner loop, check if the number is divisible by any number from 2 to the square root of the number. If it is not divisible by any number other than 1 and itself, then it is a prime number. Print out all prime numbers found within the specified range. Remember to include necessary header files, such as <stdio.h>, and use appropriate logic to implement the program efficiently.
The question is quite ambiguous. If you're talking about all the even prime numbers between 4 and 50, then the sum is zero. No even number, except '2', is a prime number, so there aren't any between 4 and 50..
No prime power exists since there are no duplicate prime numbers in the prime factorization.