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

Source Work

 * The Dictionary
 * $142,857$
 * $142,857$


 * The Dictionary
 * $142,857$
 * $142,857$

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:

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 999 is Divisible by 999