Shift that !!

the number is divisible by 5 so it must end with (5 or 0) but 0 is rejected . It will be (y....x57) & Y must be 1 , So easily multiply by 5 and stop when the last digit is 1
Finally The number is 142857

Let me be clear about this, so in this case: - Last digit of the original number is 7 - The resultant number is divisible by 5, or in this case must end in 5 - The resultant number must be 5 times original number - The resultant number must have the same amount of digits as the last

In this case the original number, must start with 1. Because if the head digit of the number is greater than 1. When multiplied by 5, it will have a header which is always greater than 10, but this means an extra digit is added and this doesn't satisfy the criteria. Therefore, the head for the original number is one.

So far it looks like this

However, we don't know how many digits there are. So we test by making a simple multiplication style format.

You require to carry two to transition into the head digit. However, the ending number on the second head digit of the product is 1. Therefore, out of 4 and 5, you must choose 4 as the head digit of the original number. And also for the second read digit of the product is 8 because 5*5 +3 =28

Since the question is asking to place a seven from the end to the front, it shifts the entire digits away from the head, so it will look more like this.

This will repeat over time so that the digit next to 8 will be 2 (8 5+2=42) and so on, until it creates a chain. So then you repeat (2 5+4=14), Oh wait! All we need is the digit 4 as the resultant and need to carry the one for the next multiplication.

The final result will look like this.

Very good Mohamed .!

142857 is a cyclic number.

we have a number with n digits, were the last is 7:

d1 d2 d3 ... d(n-1) dn = d1 d2 d3 ... d(n-1) 7

When the last digit is carried of to the beginning, we have

7 d1 d2 d3 ... d(n-1)

But, when it holds the number becomes 5 times larger. So, we have:

7 d1 d2 d3 ... d(n-1) = 5 x (d1 d2 d3 ... d(n-1) 7)

Well, note first that 5 x 7 = 35

So, we must have 5 x (d1 d2 ... d(n-1) 7 ) ending in 5, which means that d(n-1) = 5 and:

7 d1 d2 d3 ... d(n-2) 5 = 5 x (d1 d2 d3 ... d(n-2) 57)

Note now that 5 x 57 = 285

So, 5 x (d1 d2 d3 ... d(n-2) 57) ends in 85, which means that d(n-2) = 8 and:

7 d1 d2 d3 ... d(n-3) 85 = 5 x (d1 d2 d3 ... d(n-3) 857)

Now, 5 x 857 = 4285, which means that d(n-3) = 2

Continuing on this way, we eventually obtain the number 142857 which, multiplied by 5 becomes 714285.

i used google calculator to arrive at the answer

since 7*5 will always end in 5.

the last and the second last digits are 57.

No matter what digit you put before 57, the result will always contain 85 at the end

So the last 3 digits are 857.

No matter what digit you put before 857, the result will always contain 285. so the last 4 digits are 2857.

and you build the number till 142857.