What's in a number?
Find the least number whose last digit is and which becomes times larger when the last digit is carried to the beginning of the number.
The answer is 142857.
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1 solution
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.