Powers of 10 Expressible as Product of 2 Zero-Free Factors

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Theorem

The powers of $10$ which can be expressed as the product of $2$ factors neither of which has a zero in its decimal representation are:

\(\ds 10^1\) \(=\) \(\ds 2 \times 5\)
\(\ds 10^2\) \(=\) \(\ds 4 \times 25\)
\(\ds 10^3\) \(=\) \(\ds 8 \times 125\)
\(\ds 10^4\) \(=\) \(\ds 16 \times 625\)
\(\ds 10^5\) \(=\) \(\ds 32 \times 3125\)
\(\ds 10^6\) \(=\) \(\ds 64 \times 15 \, 625\)
\(\ds 10^7\) \(=\) \(\ds 128 \times 78 \, 125\)
\(\ds 10^9\) \(=\) \(\ds 512 \times 1 \, 953 \, 125\)
\(\ds 10^{18}\) \(=\) \(\ds 262 \, 144 \times 3 \, 814 \, 697 \, 265 \, 625\)
\(\ds 10^{33}\) \(=\) \(\ds 8 \, 589 \, 934 \, 592 \times 116 \, 415 \, 321 \, 826 \, 934 \, 814 \, 453 \, 125\)


Proof

Let $p q = 10^n$ for some $n \in \Z_{>0}$.

Then $p q = 2^n 5^n$.

Without loss of generality, suppose $p = 2^r 5^s$ for $r, s \ge 1$.

Then $2 \times 5 = 10$ is a divisor of $p$ and so $p$ ends with a zero.

Thus for $p$ and $q$ to be zero-free, it must be the case that $p = 2^n$ and $q = 5^n$ (or the other way around).

The result follows from Powers of 2 and 5 without Zeroes.

$\blacksquare$


Sources

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