What is the number of digits of the smallest number $n$, such that $20\mid n$ and $n^2$ is a perfect cube, $n^3$ is a perfect square?

Question

Let $n$ be the smallest positive integer such that n is divisible by $20$, $n^2$ is a perfect cube, and $n^3$ is a perfect square. What is the number of digits of $n$ ?

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How to express in algebra $n=2^2 \cdot 5\cdot k$ , $k \in \mathbb{Z}$, $ k= 2^i\cdot 5^j$, $n^2$ is a perfect cube, $n^3$ is a perfect square?

Make the prime factorization of $n,n^2,n^3$ more explicit and tied to the two constraints.

Answer 1

Let $(a,b,c)\in\mathbb Z_{+}^3$ then basically, you have the following system of equation :

$$\begin{cases}n=20a\\ n^2=b^3 \\ n^3=c^2\end{cases}$$

Since

$$n=b\sqrt b=\frac {c}{\sqrt [3]c}\tag 1$$

then we have $b=b_1^2$ and $c=c_1^3$, where $(b_1,c_1)\in\mathbb Z_{+}^2$ . This implies,

$$n=b_1^3=c_1^2\tag 2$$

Thus, we find that $n=2^2\cdot 5\cdot a$ must be both a perfect square and a perfect cube . Therefore,

$$a=5p^2=2\cdot 5^2\cdot q^3\tag 3$$

where $(p,q)\in\mathbb Z_{+}^2$ .

This leads to the following :

$$\begin{align}&p^2=2\cdot 5\cdot q^3\\ \implies &p=q\sqrt {2\cdot 5\cdot q}\\ \implies &q=2\cdot 5\cdot q_1^2,\thinspace q_1\in\mathbb Z_{≥1}\end{align}\tag 4$$

Finally, we conclude that :

$$\begin{align}\min \{q\}&=10\\ \min\{p\}&=100\end{align}\tag 5$$

which implies,

$$\begin{align}&a=5p^2=5\cdot 10^4\\ \implies &n=20a=10^6\thinspace.\end{align}\tag 6$$

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Thus, the number of digits of $n$ is $6+1=\color{#c00}{7}$ .

Answer 2

You have $n = 2^{2+i} \cdot 5^{1+j}$ for $i,j≥0$ .

$n^2$ has the exponents $4+2i$ and $2+2j$. This is a cube, so $4+2i$ and $2+2j$ are multiples of $3$, making $i = 1, 4, 7$ etc and $j = 2, 5, 8$ etc.

$n^3$ has exponents $6+3i$ and $3+3j$. This is a square, so $6+3i$ and $3+3j$ are even, making $i$ even and $j$ odd.

So $i = 4, 10, 16$ etc and $j = 5, 11, 17$ etc. $i = 4 + 6i’$ and $j = 5 + 6j’$ . n itself is $2^{6a} \cdot 5^{6b}$ with $a, b ≥ 1$. The smallest $n$ is $n = 10^6$.