Actually, it was a pretty simple problem.1. Divide 1901 by 6 (which is equal to 316)2. Since there is a remainder, 317(316 1) times 6 gives you the first number divisible by 6 within the range.3. Next, divide 2000 by 6 (which is 333), similarly, 333 times 6 gives you the last number divisible by 6 within this range.The answer is 17 not 16 because there are 17 numbers which are divisible by 6: -317, 318, 319, 320, 321, 322, 323, 324, 325, 326, 327, 328, 329, 330, 331, 332, 333Hope this is helpful.Thanks