Problem 5 was relatively easier than the ones we saw so far. Here’s the description:

2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder.

What is the smallest positive number that is evenly divisible by all of the numbers from 1 to 20?

As usual, brute force for the win:

```
#include <stdio.h>
int main(){
int i,j,counter;
for (i=10;i<1000000000;i++){
counter=0;
for (j=11;j<21;j++){
if (i%j==0)
counter++;
}
if (counter==10){
printf("%dn",i);
break;
}
}
return 0;
}
```

Notice that if the number is evenly divisible by all numbers from 11 to 20 it’s also divisible by all numbers from 2 to 10.

SubhaprasadI don’t have much experience in programming. I just had a doubt whether i can write i<1000000000 as i is an integer. Thank you.