## January 2016

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In my January diary I left this hanging:

2016 is the number of ways of placing two white knights on a chessboard. What is the probability, if this is done at random, that they defend each other?

I also posed the following question:

This month a new largest known prime was announced. Its value is 2^74,207,281 minus one. That's a number of 22,338,618 digits: pretty big, though of course nothing like as big as A(4,3).

Call that numberPand imagine that, defying the actuary, I live to bePdays old. What, on average, will be the proportion ofprimzahlfreimonths at that stage in my life?

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**• Solutions** (courtesy of two readers)

Reader A:

I won't bother with the prime-free-month problem but the two knights problem has the answer 1/12.

Knights on the chessboard have 2 moves from 4 squares (i.e. the corner squares), 3 moves from 8 squares (side squares adjacent to corners), 4 moves from 20 squares (the 16 other side squares plus the 4 that are one in diagonally from a corner), 6 moves from 16 squares (the others except the inmost four by four), and 8 moves from 16 squares (the inmost four by four). This averages 5.25 moves, so the probability that the other knight is on one of the squares a knight's move away is 5.25/63 = 1/12.

Reader B:

Let's see. The prime number theorem (which I learned about from you inPrime Obsession!) says that the probability of a number being prime in the neighborhood of a numberNis roughly 1 / ln(N), right? And since ln(x) = ln(2) * log_{2}(x), and log_{2}(P) = 74,207,281 to a high approximation, the odds of a given number being prime in that region is pretty small: 1 / 51,436,568, more or less.

There go an average 30.437 days to a month; so if only one in 51,436,568 days is a prime, only an average one in 1,689,936 months will have one. The other 1,689,935 months will beprimzahlfrei.

Put it another way, wait time for a month that is notprimzahlfreiis around 141,000 years.