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# Find the least wose last digit is 7 and which becomes 5 times larger when this last digit is carried to the beginning of the no. 9 years ago

As I understand, 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.

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