Hi.
I'm currently learning C++ programming and I have problems coding a program.
I need to write a program that finds the integer from 1 to 1000 with the most divisors that produce no remainder.
For example, the integer 60 has 12 divisors that produce no remainder.
They are 1,2,3,4,5,6,10,12,15,20,30,60.

Can anyone help code this program? Thanks in advance.

Lotsa loops :D

loop from 1 to 1000, duh. Then for each one, start at the number you are at, and count down from it. At each number counting down, and keep subtracting the number(lets call it "test") from the original. If you get to a point where the test is bigger than the remainder, but its not 0, then it doesnt work. If it does work, however, increase a counter for that number that counts the number of divisors. Once test reaches 0, then compare the number of divisors to a set of 2 variables you have: One for the current record number, and another for the number of divisors that number has. If your test is greater, and set it as the current record. And once the parent loop reachers 1000, your done, and output the result.

Hope it helps, lol

Yea, sure... lots of loops too :P

#include [iostream.h]

int main(void)
{
long count, input, j, i, values[100] = { 0 };

cout [[ "Enter a number: ";
cin ]] input;

j = 0;
count = 0;

for(i=1; i [= input; i++)
{
if((input % i) == 0)
{
values[j++] = i;
count++;
}
}

cout [[ endl [[ "The number " [[ input
[[ ", has " [[ count [[ " factors:" [[ endl [[ endl ;

for(j = 0; values[j] != '\0'; j++)
cout [[ values[j] << ' ';

return 0;

}

or something along those lines. :)

#include

int main()
{
for(int m = 1; m < 1001; m++) {

cout 0)

if(m % n-- == 0)

cout << n + 1 << " ";

cout << "\n\n";
}
return 0;
}

#include
int main()
{
for(int m = 1; m 0)
if(m % n-- == 0)
cout << n + 1 << " ";
cout << "\n\n";
}
return 0;
}

:-)

Thank you guys so much!!!!!!!!!

ya considering we wrote the code for u..now go out and learn it c++ haha :-p