Smallest Number which is Multiplied by 99 by Appending 1 to Each End
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Theorem
The smallest positive integer which is multiplied by $99$ when $1$ is appended to each end is:
- $112 \, 359 \, 550 \, 561 \, 797 \, 752 \, 809$
Proof
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 $N = \dfrac {10^{22} + 1} {89} = 112 \, 359 \, 550 \, 561 \, 797 \, 752 \, 809$.
$\blacksquare$
Sources
- December 1975: J.L. Hunsucker and Carl Pomerance: On an Interesting Property of 112359550561797752809 (The Fibonacci Quarterly Vol. 13, no. 4: pp. 331 – 334)
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $112,359,550,561,797,732,809$
- but note the transcription error in the $5$th digit from the end: it should be $5$ not $3$.