Question

in a recent year, a total of 6.3 × 10⁸ metric tons of wheat were produced in the world. at this level, how many years of wheat production would be required to fill the request of circe the gather? (note: one metric ton equals 1000 kg. a grain of wheat weighs 0.000008 kg. round your answer to the nearest year.)

Explanation:

Step1: Convert total wheat production to kg

First, we know that \(1\) metric ton \( = 1000\) kg, and the total wheat production is \(6.3\times10^{8}\) metric tons. So we convert this to kg by multiplying by \(1000\):
\(6.3\times 10^{8}\text{ metric tons} \times 1000\ \frac{\text{kg}}{\text{metric ton}} = 6.3\times 10^{11}\text{ kg}\)

Step2: Find total mass needed for Sissa Ben Dahir's wheat

A grain of wheat weighs \(0.00008\) kg. Sissa Ben Dahir's wheat request (from the story) is \(2^{64}- 1\) grains (since he asked for 1 grain on the first square, 2 on the second, ..., \(2^{63}\) on the 64th square; the sum is \(2^{64}-1\)). Let's calculate the total mass:
First, \(2^{64}-1\approx2^{64}\) (since \(2^{64}\) is much larger than 1).
\(2^{64}=1.8446744073709552\times 10^{19}\)
Total mass \(m=(2^{64}-1)\times0.00008\ \text{kg}\approx1.8446744073709552\times 10^{19}\times8\times 10^{-5}\ \text{kg}\)
\(m\approx1.4757395258967642\times 10^{15}\text{ kg}\)

Step3: Calculate number of years

Let \(N\) be the number of years. The annual production is \(6.3\times 10^{11}\text{ kg/year}\). So:
\(N=\frac{\text{Total mass needed}}{\text{Annual production}}=\frac{1.4757395258967642\times 10^{15}\text{ kg}}{6.3\times 10^{11}\text{ kg/year}}\)
\(N=\frac{1.4757395258967642}{6.3}\times 10^{4}\approx0.2342443692\times 10^{4}=2342.443692\)

Wait, no—wait, I think I messed up the Sissa Ben Dahir story. Wait, actually, the standard problem: he asked for 1 grain on the first square, 2 on the second, ..., \(2^{63}\) on the 64th. The total number of grains is \(2^{64}-1\approx1.844674407\times 10^{19}\) grains.

Wait, let's recalculate the total mass:
Each grain is \(0.00008\) kg, so total mass \(M=(2^{64}-1)\times0.00008\) kg.
\(2^{64}-1 = 18446744073709551615\)
\(M = 18446744073709551615\times0.00008\) kg
\(M = 1.4757395258967641292\times 10^{15}\) kg

Annual production: \(6.3\times 10^{8}\) metric tons \(= 6.3\times 10^{8}\times 1000 = 6.3\times 10^{11}\) kg/year

Now, number of years \(n=\frac{M}{\text{Annual production}}=\frac{1.4757395258967641292\times 10^{15}}{6.3\times 10^{11}}\)
\(n=\frac{1.4757395258967641292}{6.3}\times 10^{4}\)
\(1.4757395258967641292\div6.3\approx0.234244\)
\(0.234244\times 10^{4}=2342.44\), so approximately 2342 years? Wait, no—wait, maybe I misread the annual production. Wait, the problem says "a total of \(6.3\times 10^{8}\) metric tons"—wait, \(6.3\times 10^{8}\) metric tons is \(6.3\times 10^{11}\) kg, correct. But maybe the Sissa Ben Dahir's total is different? Wait, no, let's check the problem again. Wait, the problem says "how many years of wheat production would be required to fill the request of Sissa Ben Dahir's wheat". Let's confirm the total grains: the chessboard problem: 64 squares, each with \(2^{k - 1}\) grains (k from 1 to 64), so total grains \(S=\sum_{k = 0}^{63}2^{k}=2^{64}-1\approx1.844674407\times 10^{19}\) grains.

Each grain is \(0.00008\) kg, so total mass \(=1.844674407\times 10^{19}\times0.00008 = 1.475739526\times 10^{15}\) kg.

Annual production: \(6.3\times 10^{8}\) metric tons \(= 6.3\times 10^{11}\) kg/year.

So \(n=\frac{1.475739526\times 10^{15}}{6.3\times 10^{11}}=\frac{1.475739526}{6.3}\times 10^{4}\approx0.2342\times 10^{4}=2342\) years. Wait, but maybe I made a mistake in the exponent. Wait, \(6.3\times 10^{8}\) metric tons: \(10^{8}\) metric tons is 100 million metric tons, times 6.3 is 630 million metric tons. 630 million metric tons is \(6.3\times 10^{8}\times 1000 = 6.3\times 10^{11}\) kg. The total mass needed is ~\(1.47\times 10^{15}\) kg. So dividing, \(1.47\times 10^{15}/6.3\times 10^{11}= (1.47/6.3)\times 10^{4}=0.2333\times 10^{4}=2333\) (approx). Wait, maybe my calculation of the total grains is wrong? Wait, no—the standard chessboard problem is \(2^{64}-1\) grains. Let's recalculate the total mass:

\(2^{64}-1 = 18446744073709551615\) grains.

Each grain: 0.00008 kg = \(8\times 10^{-5}\) kg.

Total mass: \(18446744073709551615\times8\times 10^{-5}\) kg.

Calculate that: \(18446744073709551615\times8 = 1.4757395258967641292\times 10^{19}\)

Multiply by \(10^{-5}\): \(1.4757395258967641292\times 10^{14}\)? Wait, no! Wait, \(18446744073709551615\) is \(1.8446744073709552\times 10^{19}\) (since \(10^{19}\) is 1 followed by 19 zeros; \(1.8446744073709552\times 10^{19}\) is correct for \(2^{64}\)). Then \(1.8446744073709552\times 10^{19}\times8\times 10^{-5}=1.47573952589676416\times 10^{15}\) kg (because \(10^{19}\times10^{-5}=10^{14}\)? Wait, no: \(10^{19}\times10^{-5}=10^{14}\)? Wait, \(19 - 5 = 14\)? No, \(19 + (-5)=14\)? Wait, no: \(a^m\times a^n=a^{m + n}\). So \(10^{19}\times10^{-5}=10^{14}\). Wait, so \(1.8446744073709552\times 10^{19}\times8\times 10^{-5}= (1.8446744073709552\times8)\times 10^{19 - 5}=14.7573952589676416\times 10^{14}=1.47573952589676416\times 10^{15}\) kg. Yes, that's correct.

Annual production: \(6.3\times 10^{8}\) metric tons = \(6.3\times 10^{8}\times 1000\) kg = \(6.3\times 10^{11}\) kg/year.

Now, \(n=\frac{1.47573952589676416\times 10^{15}}{6.3\times 10^{11}}=\frac{1.47573952589676416}{6.3}\times 10^{4}\)

\(1.47573952589676416\div6.3\approx0.234244\)

\(0.234244\times 10^{4}=2342.44\), so approximately 2342 years. Rounding to the nearest year, 2342.

Wait, but maybe the problem is using a different total for Sissa Ben Dahir? Wait, the problem says "fill the request of Sissa Ben Dahir's wheat"—maybe it's the sum \(1 + 2 + 4 + ... + 2^{63}=2^{64}-1\) grains, which is what we used. So the calculation seems correct.

Answer:

\(\boxed{2342}\)