Question

class ev
{

int c=0;

long lcm(long d,int dd)
{
long div=d;
for(int i=1;i<=dd;i++) // finds the smallest divisible number between 1 to 10
{
if(div%i==0)
{
c=c+1; //counts the number of times the d variable gets divisible and sum them
}
}

if(c==dd) // if sum comes out to be 10 then we know the d variable is completely divisible by every number from 1 to 10
{
return div;// return the number

}
else
{
++div; // increment the number
c=0; // reset

Answer 1

Hi! What's your point?

Answer 2

sir i have attached a java file and i want to calculate the least number divisile by 1 to 20 each , I am able to calculate for 1 to10 but when i do it for 1 to 20 it gave me stackoverflow error

Answer 3

Perhaps it's related to the allowed memory for java.
Try with the -Xms and -Xmx from the java command.
If it doesn't change, then it's related to your program itself.

I will have a look at your file as soon as I get my coffee break :)

Answer 4

You've decided to implement this algorithm recursively. That's fine, but if it takes more recursive calls to reach a number divisible by all numbers from 1 to 10 than there is allowed stack space on the JVM, you'll get a StackOverflow error.

If your prints show that you pass a number divisible by all numbers from 1-10 and don't stop, then you logic for checking that is wrong. If, however, that doesn't happen, then you'll have to implement the algorithm iteratively (= with a loop) instead of recursively in order to reach the answer.

Answer 5

Find the primes that go into each of the numbers 1-20 and then look for overlap. Remove the overlap and multiply the remaining numbers out to get the final number.