Question: how many digits does $12^{10} \times 6^{-8}$ have
1. **State the problem:** We need to find how many digits the number $12^{10} \times 6^{-8}$ has.
2. **Recall the formula for the number of digits:** The number of digits of a positive integer $N$ is given by $\lfloor \log_{10}(N) \rfloor + 1$.
3. **Rewrite the expression:**
$$12^{10} \times 6^{-8} = \frac{12^{10}}{6^{8}}$$
4. **Express bases in prime factors:**
$$12 = 2^{2} \times 3$$
$$6 = 2 \times 3$$
5. **Rewrite powers:**
$$12^{10} = (2^{2} \times 3)^{10} = 2^{20} \times 3^{10}$$
$$6^{8} = (2 \times 3)^{8} = 2^{8} \times 3^{8}$$
6. **Divide the powers:**
$$\frac{12^{10}}{6^{8}} = \frac{2^{20} \times 3^{10}}{2^{8} \times 3^{8}} = 2^{20-8} \times 3^{10-8} = 2^{12} \times 3^{2}$$
7. **Calculate the value:**
$$2^{12} = 4096$$
$$3^{2} = 9$$
$$2^{12} \times 3^{2} = 4096 \times 9 = 36864$$
8. **Count digits:**
The number $36864$ has $5$ digits.
**Final answer:** The number $12^{10} \times 6^{-8}$ has **5** digits.