In a situation in which data are known to three significant digits, we write $6.379\mathrm{m}=6.38\mathrm{m}$ and $6.374\mathrm{m}=$ $6.37\mathrm{m}.$ When a number ends in $5,$ we arbitrarily choose to write $6.375\mathrm{m}=6.38\mathrm{m}.$ We could equally well write $6.375\mathrm{m}=$ $6.37\mathrm{m},$ "rounding down" instead of "rounding up," because we would change the number 6.375 by equal increments in both cases. Now consider an order-of-magnitude estimate, in which factors of change rather than increments are important. We write $500\mathrm{m}\sim 10^3\mathrm{m}$ because 500 differs from 100 by a factor of 5 while it differs from 1000 by only a factor of $2.$ We write $437\mathrm{m}\sim 10^3\mathrm{m}$ and $305\mathrm{m}\sim 10^2\mathrm{m}.$ What distance differs from $100\mathrm{m}$ and from $1000\mathrm{m}$ by equal factors so that we could equally well choose to represent its order of magnitude as $\sim 10^2\mathrm{m}$ or as $\sim 10^3\mathrm{m}?$

Key Concepts: Significant Digits, Order Of Magnitude, Factors Of Change Explanation: The given problem deals with significant digits, approximations, and order of magnitude. We are required to find a distance that differs from 100m and 1000m by equal factors. Step by Step Solution: Step 1. Let's consider the factor by which 100 differs from this distance. Let the distance be 'x', then we can write - Step 2. $\frac{x}{100}=k$ Where, k is a constant factor. Simplifying the above equation we get - Step 3. $x=100k$ Similarly, we can write another equation for the factor by which 1000 differs from the distance 'x' - Step 4. $\frac{x}{1000}=l$ where l is a constant factor. Simplifying the equation, we get - Step 5. $x=1000l$ Now, substituting value of 'x' from equation (3) in equation (1), we get - Step 6. $10l=k$ Equation (4) and (5) gives us - Step 7. $10l=k\Rightarrow \frac{k}{10}=l$$x=1000l\Rightarrow x=10000/k$ Step 8. From the above equations, it is quite clear that distance 'x' varies inversely with the factor by which 100 differs from it. And, it varies directly with the factor by which 1000 differs from it ($k/10$). Therefore, the factor by which 100 differs from the distance 'x' must equal the square root of the factor by which 1000 differs from it. Step 9. Let's try to find such a number by taking an example. Suppose, the distance is 316.2 m, then factor by which 100 differs from it is - Step 10. $\frac{316.2}{100}=3.162$ Similarly, factor by which 1000 differs from it is - Step 11. $\frac{316.2}{1000}=0.3162$ Step 12. Now, we need to check if these factors differ by equal factors. For that, we divide the larger factor by the smaller factor. Step 13. $\frac{3.162}{0.3162}=10$ The factors differ by 10, which means the estimation of distance can be $\sim 10^2\mathrm{m}$ or $\sim 10^3\mathrm{m}$. Step 14. Therefore, distance 316.2 m differs from 100m and 1000m by equal factors. Final Answer: 316.2 m