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

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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

\(\displaystyle 4 \times 999\) \(=\) \(\displaystyle 3996\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle 3 + 996\) \(=\) \(\displaystyle 999\)
\(\displaystyle 15 \times 999\) \(=\) \(\displaystyle 14 \, 985\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle 14 + 985\) \(=\) \(\displaystyle 999\)
\(\displaystyle 47 \times 999\) \(=\) \(\displaystyle 46 \, 953\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle 46 + 953\) \(=\) \(\displaystyle 999\)
\(\displaystyle 57 \times 999\) \(=\) \(\displaystyle 56 \, 943\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle 56 + 943\) \(=\) \(\displaystyle 999\)


Proof


Sources