First digits of a cube of a natural number

Question

Can a cube of a number be of form: $2016a_1a_2a_3\dots a_n$? I have no direction, and would love to get a certain direction/proof. Thanks in advance

Answer 1

We have

$$\sqrt[3]{2016} < 12.633 < \sqrt[3]{2017}$$

So $12633^3=2016134440137$ satisfies your conditions.

This works for any integer $n$ in place of $2016$: find a decimal number between $\sqrt[3]n$ and $\sqrt[3]{n+1}$, and remove the decimal point.

Answer 2

Yes, it can. $2722^3 = 20168071048$ Apparently there are a lot of such numbers.

Answer 3

There're three possibilities due to the first few significant figures of

$\sqrt[3]{2016} \approx 12.6327\ldots$,

$\sqrt[3]{20160} \approx 272.216\ldots$

$\sqrt[3]{201600} \approx 5863.58\ldots$

So,

$12633^{3}=2016134440137$

$2722^{3}=20168071048$

$5864^{3}=201642412544$