Let [tex]abc[/tex] be the starting number, where the catenated letters represent digits, and not factors like in the product [tex]a\times b\times c[/tex]. Each of [tex]a,b,c[/tex] can be elements of the list [tex]\{0,1,2,3,\ldots,9\}[/tex], with the exception of [tex]a\neq0[/tex].
Since [tex]abc[/tex] is a base-10 numeral, we can also write it as
[tex]10^2a + 10^1b + 10^0c = 100a + 10b + c[/tex]
Moving the last digits to the first slot makes the new number,
[tex]100c + 10a + b[/tex]
so we must also have [tex]c\neq0[/tex].
The digits in the base-10 numeral [tex]cab[/tex] are identical to those of the base-7 numeral [tex]cab_7[/tex]. The latter number we can write as
[tex]7^2c + 7^1a + 7^0b = 49c + 7a + b[/tex]
but since we're working in base 7 now, the digits [tex]a,b,c[/tex] cannot exceed 6.
Now, we want to solve for [tex]a,b,c[/tex] such that
[tex]100c + 10a + b = 49c + 7a + b \implies 51c + 3a = 3 (17c + a) = 0[/tex]
Both [tex]a,c[/tex] must be positive, but this equation has solution [tex]a=-17c[/tex]. So there are no positive integers solutions, and hence no such number.
