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

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): $999$