Integer which is Multiplied by 9 when moving Last Digit to First

Theorem
Let $N$ be the positive integer:
 * $N = 10 \, 112 \, 359 \, 550 \, 561 \, 797 \, 752 \, 808 \, 988 \, 764 \, 044 \, 943 \, 820 \, 224 \, 719$

$N$ is the smallest positive integer $N$ such that if you move the last digit to the front, the result is the positive integer $9 N$.

Proof
First we demonstrate that this number has this property:

A way to find this positive integer is given below.

Suppose $N = \sqbrk {a_k a_{k - 1} \dots a_1}$ satisfies the property above.

Consider the rational number represented by $q = 0.\dot {a_k} a_{k - 1} \dots \dot {a_1}$.

Since $a_k \ne 0$, we require $q \ge 0.1$.

By the property we have $9q = 0.\dot {a_1} a_k \dots \dot {a_2}$.

Then $90 q = a_1.\dot {a_k} a_{k - 1} \dots \dot {a_1}$.

Notice that the recurring part of $90 q$ is the same as the recurring part of $q$.

Subtracting, we get $89 q = a_1$.

We have $a_1 \ge 89 \times 0.1 = 8.9$, so $a_1 = 9$, which gives $q = \dfrac 9 {89}$.

In decimal form:


 * $q = 0. \dot 1011235955056179775280898876404494382022471 \dot 9$.

which gives us $N = 10 \, 112 \, 359 \, 550 \, 561 \, 797 \, 752 \, 808 \, 988 \, 764 \, 044 \, 943 \, 820 \, 224 \, 719$ as the smallest positive integer satisfying the property.