The inexhaustible board | THE GAME OF SCIENCE | EUROtoday

The estimation of the peak of a mock of the Eiffel Tower, proposed final week, is a type of instances wherein the OBC (good Cubero eye) normally fails loudly. Most of the folks requested estimate that an iron mannequin of 1 kilo would have kind of the scale of a standard liter bottle, when the reality is that their top could be a couple of meter and a half. And, seen from our tiny human perspective, the Eiffel Tower is a big mass of iron, however in actuality it’s an especially swish construction.

Rounding, the Eiffel tower measures about 300 meters excessive and weighs about 8000 tons, that’s, it’s 8 million occasions higher than an iron mannequin of 1 kilo and, subsequently, 200 occasions greater (200³ = 8000000). Then the peak of the mannequin could be 300/200 = roughly 1.5 m.

Where the OBC normally fails is in estimates associated to exponential progress processes, that’s, with geometric progressions. The finest recognized instance is that of the legendary inventor of chess that asks for a wheat grain by the primary field of the board, two for the second, 4 for the third and so forth. It will not be straightforward to think about that with the grains comparable to the 64 bins of the board, the Iberian Peninsula might be lined with a layer of wheat of a number of meters thick. Or that bending a folio 43 occasions in a row would acquire a thickness just like the gap of the earth to the moon.

By the best way, are you aware why geometric progressions are referred to as that? And the arithmetic? And if you do not know (what’s more than likely), are you able to consider any affordable rationalization?

THE DAMERO AND ITS VARIANTS

Speaking of the chess board and its variants, inexhaustible helps of riddles of every kind, I keep in mind one which despatched me not lengthy a pal, vaguely associated to the basic downside of the 8 girls (place 8 girls on an empty chess board in order that none of them threatens another). It is about having 18 chips on a 6×6 board in order that in every row, in every column and within the two diagonals, there are 3 and solely 3 chips.

Can the issue be generalized to different boards of a few bins?

On the 2×2 trivial board, it’s evident that we can’t place 2 chips in response to the issue circumstances: both they’re in adjoining or diagonal bins.

On a 4×4 board, can we place 8 chips in order that there are 2 in every row, column and diagonal?

On the standard chess board, 8×8, can we place 32 chips in order that there are 4 in every row, column and diagonal?

Even harder:

Can or not it’s demonstrated that in each 2nx2n board, being n any pure quantity higher than 1, we are able to dispose 2n² chips in order that there are n in every row, column and diagonal?