Find the least positive integer whose last digit is 7 and which becomes 5 times larger when this last digit is carried to front of the number.
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Okay, so this is how I got it.
Since, the last digit is 7 and on shifting it to the starting the number becomes 5 times the original, hence the second last digit must be 5. Now, since the last 2 digits are 57 hence the last 2 digitsof the product will be 85 and therefore, the third last digit is 8.
Continue, using the same logic to arrive at this number i.e. 142857
Also, note that all the results for the following summation where 'n' is any non-negative integer, have the required property.
k = 0 ∑ n 1 4 2 8 5 7 × 1 0 6 k