Smallest Number which is Multiplied by 99 by Appending 1 to Each End

Theorem

The smallest positive integer which is multiplied by $99$ when $1$ is appended to each end is:

$112 \, 359 \, 550 \, 561 \, 797 \, 752 \, 809$

We have that:

$112 \, 359 \, 550 \, 561 \, 797 \, 752 \, 809 = 101 \times 1 \, 052 \, 788 \, 969 \times 1 \, 056 \, 689 \, 261$

while:

$\ds 11 \, 123 \, 595 \, 505 \, 617 \, 977 \, 528 \, 091$

$=$

$\ds 3^2 \times 11 \times 101 \times 1 \, 052 \, 788 \, 969 \times 1 \, 056 \, 689 \, 261$

$\ds $

$=$

$\ds 3^2 \times 11 \times 112 \, 359 \, 550 \, 561 \, 797 \, 752 \, 809$

Let $N$ be the smallest integer satisfying $99 N = \sqbrk {1N1}$ when expressed in decimal notation.

Suppose $N$ is $k$ digits long.

Then:

$\sqbrk {1 N 1} = 10^{k + 1} + 10 N + 1$

Subtracting $10 N$ from $99 N$ gives:

$89 N = 10^{k + 1} + 1$

One can show, by trial and error, that the smallest $k$ where $10^{k + 1} + 1$ is divisible by $89$ is $21$.

Then $N = \dfrac {10^{22} + 1} {89} = 112 \, 359 \, 550 \, 561 \, 797 \, 752 \, 809$.

$\blacksquare$

but note the transcription error in the $5$th digit from the end: it should be $5$ not $3$.