1. **Problem statement:** Majed has two magic stones. When rubbed once, each stone produces one new stone, effectively doubling the number of stones. We need to find how many times the stones must be rubbed to reach 100 stones starting from 2 stones. 2. **Understanding the problem:** Each rub doubles the number of stones. Starting with 2 stones, after 1 rub, there will be $2 \times 2 = 4$ stones, after 2 rubs, $4 \times 2 = 8$ stones, and so on. 3. **Formula:** The number of stones after $n$ rubs is given by: $$\text{stones} = 2 \times 2^n = 2^{n+1}$$ 4. **Goal:** Find $n$ such that: $$2^{n+1} = 100$$ 5. **Solving for $n$:** Take the logarithm base 2 of both sides: $$n+1 = \log_2(100)$$ Calculate $\log_2(100)$: $$\log_2(100) = \frac{\log_{10}(100)}{\log_{10}(2)} = \frac{2}{0.3010} \approx 6.644$$ So: $$n + 1 = 6.644 \implies n = 5.644$$ 6. **Interpretation:** Since $n$ must be an integer number of rubs, and $n=5.644$ is not an integer, we round up to the next whole number: $$n = 6$$ 7. **Verification:** After 6 rubs: $$2^{6+1} = 2^7 = 128$$ stones, which is more than 100. After 5 rubs: $$2^{5+1} = 2^6 = 64$$ stones, which is less than 100. Therefore, Majed must rub the stones **6 times** to have at least 100 stones. **Final answer:** $$\boxed{6}$$