Challenge Level

$$\begin{align}166666&\text{ remainder 4}\\

6\overline{)1000000}&\end{align}$$ So $4$ less than $1000000$ is a multiple of $6$

In order to be a multiple of 6, a number must be both even and a multiple of 3. Of the numbers given, only (b) 999 998 and (d) 999 996 are even.

For a number to be a multiple of 3, the sum of its digits must also be a multiple of 3. From this, we see that, of these two, only (d) 999 996 is a multiple of 3. Hence (d) is the only multiple of 6 here.

This problem is taken from the UKMT Mathematical Challenges.

You can find more short problems, arranged by curriculum topic, in our short problems collection.