n = int(input())
l=[]
for i in range(2,n+1):
flag =0
for j in range(2,i):
if(i%j ==0):
flag =1
break
if(flag == 0):
l.append(i)
print(l)
tot=0
for i in range(len(l)):
tot = tot+l[i]
print(tot)
l=int(input())
r=int(input())
l1=[]
for i in range(l,r+1):
a1=i
c=0
if a1>1:
for i in range(2,a1):
if a1%i==0:
c=1
break
if c==0:
l1.append(a1)
print(sum(l1))
int prime(int num){
if(num==1)
return 0;
int flag = 1;
for(int i =2; i>start>>end;
if(start>=end)
cout<<"Invalid input";
else{
int sum=0, average=0, count =0;
cout<<"\nprime number in given range are \n";
for(int i = start; i<=end; i++){
int Yesprime = 0;
Yesprime = prime(i);
if(Yesprime){
cout<<" "<<i;
sum = sum + i;
count++;
}
}
#include
int main()
{
int i,j,k,l,m,n,sum=0;
printf(“enter values within which we need to find out sum of all prime number\n”);
scanf(“%d%d”,&k,&l);
for(i=k;i<=l;++i)
{
int count=0;
for(j=2;j<=(i/2);j++)
{
if(i%j==0)
{
count++;
break;
}
}
if(count==0)
sum=sum+i;
#Sum of odd number in given range
a,b=map(int, input().split(” “))
sum=0
for i in range(a,b+1):
if (i%2!=0):
sum =sum+ i
print(“sum of odd number is:”,sum)
n = int(input())
l=[]
for i in range(2,n+1):
flag =0
for j in range(2,i):
if(i%j ==0):
flag =1
break
if(flag == 0):
l.append(i)
print(l)
tot=0
for i in range(len(l)):
tot = tot+l[i]
print(tot)
l=int(input())
r=int(input())
l1=[]
for i in range(l,r+1):
a1=i
c=0
if a1>1:
for i in range(2,a1):
if a1%i==0:
c=1
break
if c==0:
l1.append(a1)
print(sum(l1))
#include
int main()
{
int r_1,r_2,k,sum=0;;
int flag=1;
scanf(“%d%d”,&r_1,&r_2);
k=r_1;
while(k>=r_1 && k<=r_2)
{
for(int i=2;i<k;i++)
{
if(k%i==0){
flag=0;
break;}
}
if(flag==1){
sum=sum+k;}
k++;
flag=1;
}
if(r_1==1)
printf("%d",sum-1);
else
printf("%d",sum);
return 0;
}
//sum of prime no in a given range
#include
int main(){
int l,u,i,j;
scanf(“%d%d”,&l,&u);
int sum= 0,f=0;
if(l==1) l=2;
for(i=l;i<=u;i++){
for(j=2;j<=i/2;j++){
if(i%j==0){
f=1;
break;
}
}
if(f==0){
sum += i;
printf("%d ",i);
}
f=0;
}
printf("\n%d",sum);
return 0;
}
#include
using namespace std;
int prime (int);
int prime(int num){
if(num==1)
return 0;
int flag = 1;
for(int i =2; i>start>>end;
if(start>=end)
cout<<"Invalid input";
else{
int sum=0, average=0, count =0;
cout<<"\nprime number in given range are \n";
for(int i = start; i<=end; i++){
int Yesprime = 0;
Yesprime = prime(i);
if(Yesprime){
cout<<" "<<i;
sum = sum + i;
count++;
}
}
cout<<"\ncount "<<count;
cout<<"\nsum "<<sum;
cout<<"\naverage "<<(float)sum/count;
}
return 0;
}
n = int(input())
sum1 = 2
for i in range(3,n):
for j in range(2,i):
if i%j!=0:
a = “prime”
else:
a = “not prime”
break
if a == ‘prime’:
sum1 += i
print(sum1)
Enjoy
// Online C compiler to run C program online
#include
int main() {
int l,r,i,j;
int sum=0,temp;
scanf(“%d”,&l);
scanf(“%d”,&r);
for(i=l;i<=r;i++){
temp=0;
for(j=2;j<l;j++){
if(i%j==0 && i!=2){
temp=1;
}
}
if(temp==0){
sum=sum+i;
}
}
printf("%d",sum);
return 0;
}
#include
int main()
{
int i,j,k,l,m,n,sum=0;
printf(“enter values within which we need to find out sum of all prime number\n”);
scanf(“%d%d”,&k,&l);
for(i=k;i<=l;++i)
{
int count=0;
for(j=2;j<=(i/2);j++)
{
if(i%j==0)
{
count++;
break;
}
}
if(count==0)
sum=sum+i;
}
if(k==1)
sum=sum-1;
printf("%d",sum);
return 0;
}
#include
int prime(int n)
{
int i,c=0;
for(i=1;i<=n;i++)
{
if(n%i==0)
c++;
}
if(c==2)
return 1;
else
return 0;
}
int main()
{
int x,i,q,a[100],j,sum=0,l;
scanf("%d",&l);
scanf("%d",&x);
q=0;
for(i=l;i<=x;i++)
{
if(prime(i))
{
a[q]=i;
q++;
}
}
for(i=0;i<q;i++)
{
sum=sum+a[i];
}
printf("%d",sum);
return 0;
}
Thanks for contributing your code Sadique
#Sum of odd number in given range
a,b=map(int, input().split(” “))
sum=0
for i in range(a,b+1):
if (i%2!=0):
sum =sum+ i
print(“sum of odd number is:”,sum)
#python
n1,n2=map(int,input().split(” “))
x=0
for i in range(n1,n2+1):
if(i%2!=0):
x=x+i
print(f”sum of odd number between {n1} and {n2} is:{x}”)
#include
int main()
{
int sum=0,i,max;
scanf(“%d”,&max);
for(i=0;i<max;i++)
{
if(i%2!=0)
{
sum=sum+i;
}
}
printf("%d",sum);
return 0;
}