Integer whose Digits when Grouped in 3s add to Multiple of 999 is Divisible by 999/Mistake

Source Work

1986: David Wells: Curious and Interesting Numbers:

The Dictionary

$142,857$

1997: David Wells: Curious and Interesting Numbers (2nd ed.):

The Dictionary

$142,857$

Mistake

Now any number whose digits when grouped in $3$s from the units end add up to $999$ is a multiple of $999$, and conversely, so $142857$ must be a multiple of $999$.

Correction

The and conversely is incorrect.

The simplest counterexample is:

\(\ds 1001 \times 999\)

\(=\)

\(\ds 999 \, 999\)

The digits of $999 \, 999$, when grouped in $3$s from the units end, do not add up to $999$:

$999 + 999 = 1998$

However, they do add to a multiple of $999$.

So perhaps the result could be worded:

Now any number whose digits when grouped in $3$s from the units end add up to a multiple of $999$ is a multiple of $999$, and conversely, so $142857$ must be a multiple of $999$.

(emphasis added)

This mistake is repeated in a slightly different form on the page $999$.

Also see

• Integer whose Digits when Grouped in 3s add to Multiple of 999 is Divisible by 999

Sources

• 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $142,857$
• 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $142,857$