$n$-th term of the series 1 27 125 1000

Question

What will be the nth term of the series 1 27 125 1000

for $n = 1$, it is 1

for $n = 2$, it is 27

for $n = 3$, it is 125

for $n = 4$, it is 1000

Answer 1

For all residue classes modulo $3$, the smallest cube with $n$ digits is the smallest perfect cube $\geq 10^{n - 1}$, so its cube root is the smallest integer $\geq 10^{(n - 1) / 3}$, namely $\lceil 10^{(n - 1) / 3} \rceil$. So, the smallest cube with $n$ digits is $$\boxed{\lceil 10^{(n - 1) / 3} \rceil^3} .$$

If particular, if $n \equiv 1 \pmod 3$, then $10^{n - 1} = 1 \underbrace{0 \cdots 0}_n$, which is the smallest number with $3 n + 1$ digits.

For $n \equiv 2 \pmod 3$, so, for $n = 3 m + 2$, we have $10^{(n - 1) / 3} = \lceil 10^m \cdot 10^{1 / 3} \rceil,$ so to produce the smallest integer whose cube has $n$ digits, we take $10^{1 / 3} = 2.15443\ldots$, move the decimal place rightward $m$ times, and round up, giving $3, 22, 216, \ldots$; their respective cubes are $27, 10648, 10077696, \ldots$.

A similar argument shows that for $n \equiv 0 \pmod 3$ we consider $10^{2 / 3} = 4.64158\ldots$, so the smallest integers are $5, 47, 465, \ldots$, and the corresponding cubes are $125, 103823, 100544625, \ldots$.

Answer 2

Suppose $n\ge 1$

If $10^{n-1}$ is not already a cube , take $$m=\lceil (10^{n-1})^{\frac{1}{3}}\rceil $$ Then, $m^3$ is the smallest cube with $n$ digits.