Woah! 7 is so cool!
Find the least number whose last digit is 7 and which becomes 5 times larger when this last digit is carried to the beginning of the number.
Details and assumptions:
You're not adding any 0's or any other digit when you shift the last digit to the beginning. So for example, say the number was 321157, then after the shift, the number will be 732115.
The answer is 142857.
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1 solution
Let's say that the number is . . . a b c 7 . We are given that ( . . . a b c 7 ) 5 = 7 . . . a b c . Through elementary multiplication, we can see that the last digit, i.e, c = 5. From there, we can see that by multiplying c = 5 by 5 and adding the carry of 3, we get the units digit of 8, which is b. And we carry out this process until we get a 7 in the resultant product form. That number happens to be 142857.
Cool fact: 7 1 = 0 . 1 4 2 8 5 7 1 4 2 8 5 7 . . . . I'm pretty sure that this problem is related to this, but I don't know how just yet.