A number theory problem by mridul jain

If we take any four consecutive even numbers and divide them by 8, we get remainders 0,2,4,6 in some order. Thus there is only one number of the form 8k+4 among them which is divisible by 4 but not by 8. Hence if we take four even consecutive numbers 1000a + 100b + 10c + 2, 1000a + 100b + 10c + 4, 1000a + 100b + 10c + 6, 1000a + 100b + 10c + 8, there is exactly one among these four which is divisible by 4 but not by 8. Now we can divide the set of all 4-digit even numbers with non-zero digits into groups of 4 such consecutive even numbers with a, b, c nonzero. And in each group, there is exactly one number which is divisible by 4 but not by 8. The number of such groups is precisely equal to 9 × 9 × 9 = 729, since we can vary a, b.c in the set {1, 2, 3, 4, 5, 6, 7, 8, 9}.