int fun(int n,int m) { int sum=n+m; If(countdigits(sum)==count | OFF CAMPUS UPDATES

int fun(int n,int m)
{
int sum=n+m;
If(countdigits(sum)==countdigits(n))
return sum;
else
return n;
}
int countDigit(int n)
{

int count = 0;

while (n != 0)

{

n = n / 10;

++count;

}

return count;
}

Same number of digits code




lower = int(input("Enter lower range: "))
upper = int(input("Enter upper range: "))

for num in range(lower,upper + 1):
if num > 1:
for i in range(2,num):
if (num % i) == 0:
break
else:
print(num)

This is for circular prime u can modify this based on input and output format




def ret_s(n):
result=0
while(n>0):
result+=n%10
n=int(n/10)
return result
n=int(input())
r=int(input())
re=r*ret_s(n)
print(ret_s(re))

For an intelligent agency has recived




n = int(input());

v = int(input());
temp = 0
for i in range(1,n+1):
for j in range(1,n+1):
if i*j == v:
temp = temp + 1
if temp == 0:
print('NO POWER')
else:
print(temp)



No of n houses,
in phython




def ret_s(n):
result=0
while(n>0):
result+=n%10
n=int(n/10)
return result
n=int(input())
r=int(input())
re=r*ret_s(n)
print(ret_s(re))

For an intelligent agency has recived




Same number of digits code

int fun(int n,int m)
{
int sum=n+m;
If(countdigits(sum)==countdigits(n))
return sum;
else
return n;
}
int countDigit(int n)
{
int count = 0;
while (n != 0)
{
n = n / 10;
++count;
}
return count;
}


Intelligent agency code



def ret_s(n):
result=0
while(n>0):
result+=n%10
n=int(n/10)
return result
n=int(input())
r=int(input())
re=r*ret_s(n)
print(ret_s(re))



POWER CODE



#include

using namespace std;

int main()
{

int n,v;
cin>>n;
cin>>v;
int count=0;
for(int i=1;i<=n;i++)
{
for(int j=1;j<=n;j++)
{
if(i*j==v)
count++;
}

}
if(count!=0)
cout<else
cout<<"NO POWER";

return 0;
}




from itertools import combinations
n = int(input());
m = int(input());
p = int(input());
e = int(input());
comb1 = combinations(range(0,n), p);
comb2 = combinations(range(0,m), e);
temp = len(list(comb1)) * len(list(comb2));
print(temp);


Sationary code
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