Indija mo 2011

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 $9\times 9\times 9=729$, since we can vary $a,b,c$ in the set $\{1,2,3,4,5,6,7,8,9\}$.