# Multiple of 999 can be Split into Groups of 3 Digits which Add to 999

## Contents

## Theorem

Let $n$ be a (strictly positive) integer multiple of $999$.

Then it is possible to split $n$ up into groups of $3$ digits, counting from the unit position, to form $3$-digit integers which add up to $999$.

### Examples

\(\displaystyle 4 \times 999\) | \(=\) | \(\displaystyle 3996\) | $\quad$ | $\quad$ | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle 3 + 996\) | \(=\) | \(\displaystyle 999\) | $\quad$ | $\quad$ |

\(\displaystyle 15 \times 999\) | \(=\) | \(\displaystyle 14 \, 985\) | $\quad$ | $\quad$ | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle 14 + 985\) | \(=\) | \(\displaystyle 999\) | $\quad$ | $\quad$ |

\(\displaystyle 47 \times 999\) | \(=\) | \(\displaystyle 57 \, 942\) | $\quad$ | $\quad$ | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle 57 + 942\) | \(=\) | \(\displaystyle 999\) | $\quad$ | $\quad$ |

## Proof

## Sources

- 1997: David Wells:
*Curious and Interesting Numbers*(2nd ed.) ... (previous) ... (next): $999$