Knowing that $a$ is a natural number, and that $4 a^{2}$ and $\frac{4}{3} \times a^{3}$ are natural numbers with 4 digits, determine $a$.

## List 4
We have $1000 \leq 4 a^{2}<10000$ and also $1000 \leq \frac{4}{3} \times a^{3}<10000$.

From $1000 \leq 4 a^{2}<10000$ it follows that $250 \leq a^{2}<2500$. Since $a$ is an integer and $15^{2}=225,16^{2}=256$ and $50^{2}=2500$, we have that $15<a<50$.

From $1000 \leq \frac{4}{3} a^{3}<10000$ we get $750 \leq a^{3}<7500$. But, $10^{3}=1000,9^{3}=729$, $20^{3}=8000$ and $19^{3}=6859$. Thus, $9<a<20$.

Therefore, we have $a=16,17,18$ or 19.

On the other hand, since $\frac{4}{3} \times a^{3}$ is an integer, we conclude that $a^{3}$ is a multiple of 3 and consequently, $a$ is a multiple of 3. Therefore, $a=18$.

Another way to finish the solution is to substitute the 4 possible values for $a$ and verify that the only value is $a=18$.

## List 4