About how to find prime factors for n
Boston numbers 371
hi @kumarakash121005
refer this code–>
#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
int main() {
int n, sum=0, sum_n=0;
cin>>n;
int temp=n;
while(temp>0)
{
sum_n=sum_n+(temp%10);
temp/=10;
}
while (n%2 == 0)
{
sum+=2;
n = n/2;
}
for (int i = 3; i <= sqrt(n); i = i+2)
{
while (n%i == 0)
{
int temp1=i;
while(temp1>0)
{
sum=sum+(temp1%10);
temp1/=10;
}
n = n/i;
}
}
if (n > 2)
{ while(n>0)
{
sum=sum+(n%10);
n/=10;
}}
if(sum_n==sum)
cout<<"1";
else
cout<<"0";
return 0;
}
Please tell me the logic in brief
logic simply is to check if sum of all digits of number is equal to sum of all its prime factors, then it’s a Boston number else not…
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