Find the sum of all three-digit natural numbers that are not exactly divisible by 3.

$(100 + 101 + 102 +  \ldots  + 999) - (102 + 105 +  \ldots  + 999)$     (M1)

$= \frac{{900}}{2}(100 + 999) - \frac{{300}}{2}(102 + 999)$     M1A1A1

$= 329\,400$     A1     N5

Note: A variety of other acceptable methods may be seen including for example $\frac{{300}}{2}(201 + 1995)$ or $\frac{{600}}{2}(100 + 998)$.

[5 marks]

There were many good solutions seen by a variety of different methods.