When the creator of chess (Sissa Ibn Dahir) showed his invention to the emperor of India, the emperor was so impressed by the new game, that he gave the inventor the right to name his prize.
The man asked the emperor this: give me for the first square of the chess board one grain of rice, two for the second square, four for the third, and so forth, doubling the amount of rice each time.
Calculate how much of rice should the emperor have given to the inventor. Also compare the total weight of the rice to the global production of rice in 2010 (464,000,000 metric tons). Assume that the weight of one grain of rice is 25 mg.

To calculate the total amount of rice the emperor should have given to the inventor, we need to determine the sum of grains of rice placed on all 64 squares of a chessboard, with the amount doubling on each subsequent square. This is a geometric progression.

### Step-by-Step Solution:

1. **Understanding the Geometric Progression**:
   - The number of grains of rice on each square doubles from the previous square:
   - 1 grain on the first square, 2 on the second, 4 on the third, etc.
   - This is a geometric sequence where the first term \(a = 1\) and the common ratio \(r = 2\).

2. **Sum of Grains on All 64 Squares**:
   - The formula for the sum \(S_n\) of the first \(n\) terms of a geometric progression is:
   \[
   S_n = a \frac{r^n - 1}{r - 1}
   \]
   - Here, \(a = 1\), \(r = 2\), and \(n = 64\).

   Substituting these values into the formula gives:
   \[
   S_{64} = 1 \cdot \frac{2^{64} - 1}{2 - 1} = 2^{64} - 1
   \]

3. **Calculating \(2^{64} - 1\)**:
   - \(2^{64} = 18,446,744,073,709,551,616\)
   - Therefore,
   \[
   S_{64} = 18,446,744,073,709,551,616 - 1 = 18,446,744,073,709,551,615 \text{ grains of rice}
   \]

4. **Convert Grains to Weight**:
   - The weight of one grain of rice is 25 mg (milligrams).
   - To find the total weight in milligrams:
   \[
   \text{Total weight in mg} = 18,446,744,073,709,551,615 \times 25
   \]

5. **Convert Milligrams to Metric Tons**:
   - 1 milligram (mg) = \(10^{-6}\) grams (g)
   - 1 gram = \(10^{-3}\) kilograms (kg)
   - 1 kilogram = \(10^{-3}\) metric tons

   Thus, 1 milligram = \(10^{-9}\) metric tons.

   The total weight in metric tons:
   \[
   \text{Total weight in metric tons} = 18,446,744,073,709,551,615 \times 25 \times 10^{-9}
   \]

   Simplifying, we get:
   \[
   \text{Total weight in metric tons} = 4.611686018427387905375 \times 10^{17} \times 10^{-9}
   \]

   \[
   \text{Total weight in metric tons} \approx 4.611686018 \times 10^8 \, \text{metric tons}
   \]

6. **Comparison with 2010 Global Rice Production**:
   - The global rice production in 2010 was 464,000,000 metric tons.
   - The total weight of rice needed is approximately \(4.61 \times 10^8\) metric tons.

   Comparing these values:
   \[
   \text{Ratio} = \frac{4.611686018 \times 10^8}{4.64 \times 10^8} \approx 0.994
   \]

   The rice requested by the inventor is approximately **99.4%** of the total global rice production in 2010.

### Conclusion:
The emperor would have needed to give approximately **461,168,601,842,738 metric tons** of rice to the inventor, which is almost the entire global rice production of 2010.

Would you like more details on this calculation or have any questions?

### Related Questions:
1. What is the sum of a geometric series formula and when is it used?
2. How does the concept of exponential growth apply to real-world scenarios like this?
3. What would be the total amount of rice if the chessboard had 100 squares instead of 64?
4. How does the weight of rice compare to other major crops globally?
5. How much space would be needed to store 461,168,601,842,738 metric tons of rice?

**Tip:** Remember, exponential growth can lead to extremely large numbers very quickly, as seen in this rice-and-chessboard problem!

Explore the chessboard rice problem where grains double on each square. Calculate the total grains, convert to weight, and compare to global rice production in 2010. Understand geometric progression and exponential growth in this advanced high school level math problem.

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