The three-digit numbers have been represented by ABC, wherein A, B, and C are non-zero digits.

Using 3 distinct digits we can make 3 × 2 × 1 = 6 three-digit numbers.
So, x will be the sum of these 6 three-digit numbers.

We need to find the two values of x closest to 1000, one just below it (which will be the greatest 3-digit value of x), and the other just above it (which will be the lowest 4-digit value of x).

Now, we have to do a bit of hit and try, so that the value of x reaches close to 1000.

Let the three digits be the minimum possible ones, i.e. 1, 2, and 3.
So, we get x = 123 + 132 + 213 + 231 + 312 + 321 = 1332

This is the least possible value of x. So, statement 1 is correct, but statement 2 is incorrect.