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# Solution

It has not been mentioned how many digits the number has.

Let's consider a very large number as a general case.

So, let the number be abc.......z.

When we remove the first digit of this number, we get bc.......z. This number is 57 times smaller than the original number

So, abc.......z = 57 × bc.......z.

This looks clumsy. So, let's write bc.......z as Y.

Therefore, aY = 57 × Y. Please note that aY is not a × Y; "a" is the first digit of the number aY.

Suppose, the number was 78956. I could have written this as 70000 + 8956. This is the same as 7 × 10^{4} + 8956.

Similarly, we can express aY as a × 10^{k} + Y, where "k" is the number of digits following "a".

So, a × 10^{k} + Y = 57Y.

So, a × 10^{k} = 56Y.

So, a × 10^{k} = 7 × 8 × Y.

So, a × (2 × 5)^{k} = 7 × 8 × Y.

So, a × 2^{k} × 5^{k} = 7 × 8 × Y.

There is a 7 and an 8 on the right hand side. In order for the equality to hold true, there should also be a 7 and an 8 on the left hand side. The 7 and 8 on the right hand side must get cancelled out, otherwise, Y will not be a positive integer; it will be a fraction.

For example, suppose we are solving the equation where k = 5, then we have:

a × 2^{5} × 5^{5} = 7 × 2^{3} × Y ..... because 8 = 2^{3}

So, a × 2^{2} × 5^{5} = 7 × Y.

So, a × 4 × 3125 = 7 × Y.

So, a × 12500 = 7 × Y.

So, a × 1785.71428 = Y.

This will mean Y is a fraction and not a positive integer. But Y is a positive integer. So, both 7 and 8 on the right hand side must get cancelled out.

The left hand side does have 2^{k}, which contains 8 (2^{3}). So, 8 gets cancelled out. But what about 7? Neither 2^{k} nor 5^{k} can contain 7.
So, the only possibility is that "a", which is a single digit, should be 7. Therefore a = 7.

Thus, we have 7 × 10^{k} = 7 × 8 × Y.

So, 10^{k} = 8 × Y.

So, 10^{k} should be divisible by 8.

10^{1} = 10; it is not divisible by 8.

10^{2} = 100; it is not divisible by 8.

10^{3} = 1000. It is divisible by 8; 1000 / 8 = 125.

So, Y = 125.

The number is aY, so the number is 7125.

If I delete the first digit, I get 125. Now, 57 × 125 = 7125. Thus, the required number is 7125.

But this is only one solution. There can be other numbers as well. We can take higher powers of 10. For example, 10^{4} = 10000;
10000 / 8 = 1250. So, we get the number as 71250. If I delete the first digit, I get 1250, and 1250 × 57 = 71250. The condition is satisfied.

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