What will be left?

Let the number be $(10^k \times x) +y$

Here x is a one-digit natural number and y is any natural number.

if we delete x, resulting number is y

=> $y=\frac{1}{57}\times [(10^k \times x) +y]$

=> $\frac{56}{57}\times y =\frac{10^k}{57}\times x$

=> $56\times y=10^k\times x$

Now, see that x is a 1-digit number. If x is not equal to 7, then RHS will not be divisible by 7.

But LHS will still be divisible by 7 as $56\times k$ is divisible by 7

This is a contradiction, thus $x=7$ .

Thus $8\times y = 10^k$

thus 8 divides $10 ^k$ completely

thus minimum value of k is 3

At $k=3, y=125$

Hence $n=7125$ (n was the original number)

This is the least value of n.

For k=4, 5, 6 and so on, n=71250, 712500 and so on.

A simple Java code will do the trick (I'm aware this is not how the question was intended to be solved).

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class test {
    public static void main (String[] args) throws java.lang.Exception {
        int x = 10;
        boolean found = false;
        while (!found) {
            String temp = Integer.toString(x);
            temp = temp.substring(1);
            int y = Integer.parseInt(temp);
            if (y*57 == x && !found) {
                found = true;
            } else {
                x++;
            }
        }
        System.out.println(x);
    }
}

The output of this code when run gives us 7125.