1. **State the problem:** You want to reach the number 0.6328125 starting from some initial value (assumed 0) by applying a sequence of allowed functions one by one without order of operations or brackets. 2. **Allowed functions:** - Add 1: $+1$ - Multiply by 2: $\times 2$ - Divide by 3: $\div 3$ - Multiply by 69: $\times 69$ - Multiply by 911: $\times 911$ - Multiply by $10^{10}$: $\times 10^{10}$ 3. **Goal:** Find a sequence of these operations applied one after another to reach exactly $0.6328125$. 4. **Observation:** $0.6328125$ is a decimal fraction. Let's convert it to a fraction: $$0.6328125 = \frac{81}{128}$$ 5. **Strategy:** Since the functions are applied sequentially, and no brackets or order of operations, we can think of starting from 0 and applying operations stepwise. However, starting from 0 and applying multiplication or division will keep it 0, so we must start by adding 1 first. 6. **Try to express $\frac{81}{128}$ using allowed operations:** - Note that $128 = 2^7$ and $81 = 3^4$. 7. **Rewrite target as:** $$\frac{81}{128} = \frac{3^4}{2^7}$$ 8. **Allowed operations include $\times 2$ and $\div 3$, so we can try to build this fraction by applying these operations in reverse:** 9. **Start from 1 (after applying +1):** - Apply $\div 3$ four times: $1 \to \frac{1}{3} \to \frac{1}{9} \to \frac{1}{27} \to \frac{1}{81}$ - Apply $\times 2$ seven times: $\frac{1}{81} \to \frac{2}{81} \to \frac{4}{81} \to \frac{8}{81} \to \frac{16}{81} \to \frac{32}{81} \to \frac{64}{81} \to \frac{128}{81}$ 10. **But this is $\frac{128}{81}$, the reciprocal of our target.** 11. **So instead, start from 1, apply $\times 2$ seven times, then $\div 3$ four times:** - $1 \to 2 \to 4 \to 8 \to 16 \to 32 \to 64 \to 128$ - $128 \to \frac{128}{3} \to \frac{128}{9} \to \frac{128}{27} \to \frac{128}{81}$ 12. **This is $\frac{128}{81}$, again reciprocal.** 13. **Therefore, to get $\frac{81}{128}$, start from 1, apply $\div 3$ four times, then $\times 2$ seven times, then apply $\div 3$ once more and $\times 2$ once less?** 14. **Try applying $\div 3$ four times and $\times 2$ seven times, then multiply by $\frac{1}{2}$ (not allowed) or divide by 2 (not allowed). Since division by 2 is not allowed, only multiplication by 2 is allowed, we cannot adjust further.** 15. **Conclusion:** The only way to get $\frac{81}{128}$ exactly is to start from 1 (after +1), apply $\div 3$ four times, then $\times 2$ seven times. 16. **Sequence of operations:** - Start at 0 - Apply $+1$ to get 1 - Apply $\div 3$ four times - Apply $\times 2$ seven times 17. **Check final value:** $$1 \times \left(\frac{1}{3}\right)^4 \times 2^7 = 1 \times \frac{1}{81} \times 128 = \frac{128}{81} \neq 0.6328125$$ 18. **But $\frac{128}{81} \approx 1.58$, not our target.** 19. **Try reversing order:** $$1 \times 2^7 \times \left(\frac{1}{3}\right)^4 = 128 \times \frac{1}{81} = \frac{128}{81}$$ 20. **Try starting from 0, apply $+1$, then $\times 2$ seven times, then $\div 3$ four times:** - $0 + 1 = 1$ - $1 \times 2^7 = 128$ - $128 \div 3^4 = 128 \div 81 = \frac{128}{81} \approx 1.58$ 21. **Try starting from 0, apply $+1$, then $\div 3$ four times, then $\times 2$ seven times:** - $0 + 1 = 1$ - $1 \div 3^4 = \frac{1}{81}$ - $\frac{1}{81} \times 2^7 = \frac{128}{81} \approx 1.58$ 22. **Try starting from 0, apply $+1$, then $\div 3$ three times, then $\times 2$ six times:** - $1 \div 3^3 = \frac{1}{27}$ - $\frac{1}{27} \times 2^6 = \frac{64}{27} \approx 2.37$ 23. **Try starting from 0, apply $+1$, then $\div 3$ five times, then $\times 2$ eight times:** - $1 \div 3^5 = \frac{1}{243}$ - $\frac{1}{243} \times 2^8 = \frac{256}{243} \approx 1.05$ 24. **Try starting from 0, apply $+1$, then $\div 3$ seven times, then $\times 2$ four times:** - $1 \div 3^7 = \frac{1}{2187}$ - $\frac{1}{2187} \times 2^4 = \frac{16}{2187} \approx 0.0073$ 25. **Try starting from 0, apply $+1$, then $\div 3$ four times, then $\times 2$ six times:** - $1 \div 3^4 = \frac{1}{81}$ - $\frac{1}{81} \times 2^6 = \frac{64}{81} \approx 0.79$ 26. **Try starting from 0, apply $+1$, then $\div 3$ five times, then $\times 2$ seven times:** - $1 \div 3^5 = \frac{1}{243}$ - $\frac{1}{243} \times 2^7 = \frac{128}{243} \approx 0.53$ 27. **Try starting from 0, apply $+1$, then $\div 3$ five times, then $\times 2$ eight times:** - $1 \div 3^5 = \frac{1}{243}$ - $\frac{1}{243} \times 2^8 = \frac{256}{243} \approx 1.05$ 28. **Try starting from 0, apply $+1$, then $\div 3$ five times, then $\times 2$ seven times, then $+1$ (not allowed to add after multiplication?), no, only one function at a time. 29. **Since the target is $\frac{81}{128}$, and $81 = 3^4$, $128 = 2^7$, the exact sequence is:** - Start at 0 - Apply $+1$ to get 1 - Apply $\times 2$ seven times to get $128$ - Apply $\div 3$ four times to get $\frac{128}{81} \approx 1.58$ 30. **But this is not the target. The target is $0.6328125 = \frac{81}{128}$. So try the reverse:** - Start at 0 - Apply $+1$ to get 1 - Apply $\div 3$ four times to get $\frac{1}{81}$ - Apply $\times 2$ seven times to get $\frac{128}{81} \approx 1.58$ 31. **No way to get $\frac{81}{128}$ exactly with these operations starting from 0 and applying only these functions one by one.** 32. **Therefore, the closest you can get is $\frac{128}{81} \approx 1.58$ or $\frac{64}{81} \approx 0.79$ or $\frac{128}{243} \approx 0.53$.** 33. **Final answer:** It is impossible to reach exactly $0.6328125$ using only the allowed functions starting from 0 and applying them one by one without order of operations or brackets.