Answer already given, but some comments on how to find it.
Let the starting number be N. As Sushma noted,
N % 7 = 0
N-2 % 6 = 0
N-5 % 5 = 0
N-10 % 4 = 0
N-17 % 3 = 0
From N % 7 = 0, we know N is divisible by 7. From N-5 % 5 = 0, we also know it is divisible by 5, which means it's also divisible by 35 (the product of 7*5). This gives us only 21 numbers (35, 70, 105,...) below 750, but we can do better.
From N-2 % 6 = 0, we know N is even, giving us 10 numbers (70, 140, 210,...)
From N-10 % 4 = 0, and knowing all our numbers end in 0, we know the 10's place must be odd, giving us 5 numbers (70, 210, 350, 490, 630)
Since the number we are incrementing by at this point is not divisible by 3 (we are adding 140 each iteration), we can break them up into groups of three, test which one minus 17 is divisible by 3, and take that one for all other groups of 3. In our example, {70, 210, 350} is the first group, meaning only the 3rd number in each set of 3, after subtracting 17, will be divisible by 3 (350), changing our increment to 140 * 3 (420). This gives the equation from tojo where any number that fits in 350 + 420*k will match our criteria.