Grains of rice on the last chessboard square

Question

Place 1 grain of rice on the first square of a chessboard, 2 on the second, 4 on the third, doubling each square. Solve in your head. How many grains sit on the 64th square, and roughly how many orders of magnitude is that? Aim: under 30 seconds.

Accepted Answer

The kth square holds 2^{k-1} grains, so the 64th square holds 2^{63}. Convert with 2^{10} \approx 10^3: 2^{63} = 2^{60} \cdot 2^{3} = (2^{10})^{6} \cdot 8 \approx (10^{3})^{6} \cdot 8 \cdot 1.15 \approx 9.2 	imes 10^{18}. (The extra 1.15 is the 2.4\%-per-block drift of 1024 over 1000, compounded six times.) So the last square alone holds about 9 	imes 10^{18} grains, roughly 19 orders of magnitude. The whole board is 2^{64} - 1 \approx 1.8 	imes 10^{19} grains, since doubling sums to just under twice the final term. For scale, the world grows on the order of 5 	imes 10^8 tonnes of rice a …

Grains of rice on the last chessboard square

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