Integer which is Multiplied by 9 when moving Last Digit to First/Proof 1

Theorem

$N = 10 \, 112 \, 359 \, 550 \, 561 \, 797 \, 752 \, 808 \, 988 \, 764 \, 044 \, 943 \, 820 \, 224 \, 719$

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

First we demonstrate that $N$ has this property:

$\ds $

$\ds 10 \, 112 \, 359 \, 550 \, 561 \, 797 \, 752 \, 808 \, 988 \, 764 \, 044 \, 943 \, 820 \, 224 \, 719 \times 9$

$\ds $

$=$

$\ds 91 \, 011 \, 235 \, 955 \, 056 \, 179 \, 775 \, 280 \, 898 \, 876 \, 404 \, 494 \, 382 \, 022 \, 471$

A way to find $N$ 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:

$9 q = 0 . \dot {a_1} a_k \dots \dot {a_2}$

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:

$q = \dfrac 9 {89}$

$\dfrac 9 {89} = 0 \cdotp \dot 10112 \, 35955 \, 05617 \, 97752 \, 80898 \, 87640 \, 44943 \, 82022 \, 471 \dot 9$

$\Box$

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.

$\blacksquare$