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### Misbehaving Prime Numbers

April21

Two academics have shocked themselves and the world of mathematics by discovering a pattern in prime numbers. Primes – numbers greater than 1 that are divisible only by themselves and 1 – are considered the ‘building blocks’ of mathematics, because every number is either a prime or can be built by multiplying primes together – (84, for example, is 2 x 2 x 3 x 7). Their properties have baffled number theorists for centuries, but mathematicians have usually felt safe working on the assumption they could treat primes as if they occur randomly.

Now, however, Kannan Soundararajan and Robert Lemke Oliver of Stanford University in the US have discovered that when it comes to the last digit of prime numbers, there is a kind of pattern. Apart from 2 and 5, all prime numbers have to end in 1, 3, 7 or 9 so that they can’t be divided by 2 or 5. So if the numbers occurred randomly as expected, it wouldn’t matter what the last digit of the previous prime was. Each of the four possibilities – 1, 3, 7, or 9 – should have an equal 25 per cent (one in four) chance of appearing at the end of the next prime number.

But after devising a computer programme to search for the first 400 billion primes, the two mathematicians found prime numbers tend to avoid having the same last digit as their immediate predecessor – as if, in the words of Dr Lemke Oliver they “really hate to repeat themselves.”

A prime ending in 1 was followed by a prime ending in 1 only 18.5 per cent of the time, significantly less often than the expected 25 per cent. And, the pair found, primes ending in 3 tended be followed by primes ending in 9 more often than in 1 or 7.

The pattern – already being referred to as ‘the conspiracy among primes’ – has left mathematicians amazed that it could have remained undiscovered for so long. “I was floored,” Ken Ono, a number theorist at Emory University in Atlanta, told Quanta Magazine. “I have to rethink how I teach my class in analytic number theory now.” (source: The Independent)

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#### Post Support

10 x 9 x 8 + (7 + 6) x 5 x 4 x (3 + 2) x 1 = 2020

NCEA Level 2 Algebra Problem. Using the information given, the shaded area = 9, that is:
y(y-8) = 9 –> y.y – 8y – 9 =0
–> (y-9)(y+1) = 0, therefore y = 9 (can’t have a distance of – 1 for the other solution for y)
Using the top and bottom of the rectangle,
x = (y-8)(y+2) = (9-8)(9+2) = 11
but, the left side = (x-4) = 11-4 = 7, but rhs = y+? = 9+?, which is greater than the value of the opp. side??
[I think that the left had side was a mistake and should have read (x+4)?]