Multiple of 999 can be Split into Groups of 3 Digits which Add to 999/Mistake
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Source Work
1986: David Wells: Curious and Interesting Numbers:
- The Dictionary
- $999$
1997: David Wells: Curious and Interesting Numbers (2nd ed.):
- The Dictionary
- $999$
Mistake
- In fact, any multiple at all of $999$ can be separated into groups of $3$ digits from the unit position, which when added will total $999$. The same principle applies to multiples of $9 \quad 99 \quad 9999$ and so on.
Correction
Not every multiple of $999$, $9$, $99$ and so on.
The simplest counterexamples are:
\(\ds 1001 \times 999\) | \(=\) | \(\ds 999 \, 999\) | ||||||||||||
\(\ds 11 \times 9\) | \(=\) | \(\ds 99\) | ||||||||||||
\(\ds 101 \times 99\) | \(=\) | \(\ds 9999\) | ||||||||||||
\(\ds 10001 \times 9999\) | \(=\) | \(\ds 99 \, 999 \, 999\) |
This mistake is repeated in a slightly different form on the page $142,857$.
Sources
- 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $999$
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $999$