At first you would think you can test all of them at once by placing the product of 2 through 9 on the right side of the % operator.
if (i % (2 * 3 * 4 * 5 * 6 * 7 * 8 * 9) == 0)
But because certain numbers include previous numbers in their factorization, you should use a lower number, specifically, the least common multiple. 8 is a multiple of 2 and 4, 9 is a multiple of 3, and if 8 and 9 are in the product, then 6 (2 * 3) is covered too.
if (i % (5 * 7 * 8 * 9) == 0)
That turns out to be 2520, which is the least common multiple. It would much more readable to use 2520 and explain in a comment why this number is used.
/**
* The goal is to test if the number is a multiple of all integers
* from 2 through 9. Mathematically, the least common multiple to is a
* multiple of all its input numbers. Here, the LCM of 2, 3, ..., 9 is 2520.
*/
public static final int LCM_2_THRU_9 = 2520;
I’ve declared a constant and I’ll use it here:
if (i % LCM_2_THRU_9 == 0)