数字 50

因为 $$100000 = 10^5 = 2^5\cdot 5^5,$$ 而质数数字只有 $2,3,5,7$。

如果某一位出现了 3 或 7，那么整个乘积里就会出现因子 3 或 7，这与 $100000=2^5\cdot5^5$ 不符。所以每一位只能是 2 或 5。

设数字中有 $x$ 个 2、$y$ 个 5，则必须满足 $$x+y=10,$$ 且 $$2^x5^y=2^55^5.$$

于是得到 $$x=5,\qquad y=5.$$

也就是说，合法的 10 位数恰好由 5 个 2 和 5 个 5 组成。只需从 10 个位置中选出 5 个位置放 2，其余放 5：

$$\binom{10}{5}=\boxed{252}.$$

---

Original Explanation

We need 10-digit numbers whose digits are prime digits and whose digit product equals $100000 = 10^5 = 2^5 * 5^5.$

• Any appearance of 3 or 7 would introduce a factor 3 or 7 into the product, which cannot occur, so no digit can be 3 or 7.
• Thus every digit must be either 2 or 5. If there are $x$ digits equal to $2$ and $y$ digits equal to 5, then

$$x + y = 10 \\ 2^x5^y = 2^55^5 \\ \text{Hence} \ x = 5, y = 5$$

So each valid number is a length - 10 sequence with exactly five 2's and five 5's. The leading digit may be 2 or 5 (nonzero), so no further restriction.

The number of such sequences is the number of ways to choose positions for the five 2's among 10 slots:

$$\begin{pmatrix} 10 \\ 5 \end{pmatrix} = \boxed{252}$$