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What I'm Reading: 'Port Out, Starboard Home and other language myths'

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Edited by Richard Walker, Monday 22 June 2026 at 00:23

The author, Michael Quinion, is a man after my own heart. (I mean because he is an etymologist, he's not looking for a transplant or anything.) His book is really fascinating and a very enjoyable read. I've only read part way through at the minute but when I've finished I'll post a short review. Watch this space!

I'd not heard of this book before and I simply stumbled across when searching for the origin of the phrase 'All at sixes and sevens'.

The phrase is found in Chaucer and Shakespeare with the same meaning as today (to be flummoxed, or in a state of confusion) and from Quinion's book I learned the most probable explanation is that it comes from a dice game where 'sixes and sevens' were proverbially risky bets.

The reason I was looking up the expression is that in my Quora Digest feed the redoubtable Alon Amit answered the question: to prove that for any n we can find a n -digit number which is divisible by two super n (or some greater power) and entirely made up of the digits six and seven

For example today is the 22nd and here is number satisfying those conditions

7777777777666766667776 equals two super 22 multiplication 11 multiplication 47 multiplication 337 multiplication 10643272511

There are other pairs of digits that could be used instead of six and seven but I think the Original Poster probably recalled the phrase 'at sixes and sevens' and thought those numbers would be a good choice. 

If you are interested here is a Python program that implements Amit's algorithm at the end of this post.

# Generates a number with num digits, all 6s or 7, 
# which is divisble by 2^num or some higher power of 2.

def next(n):
    digits = len(str(n))
    if n % 2**(digits + 1):
        return int('7' + str(n))
    else:
      return int('6' + str(n))

num = 22

n = 6
for step in range(1,num):
    n = next(n)
print(n)
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Richard Walker

A Lovely Geometry Theorem

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Suppose equilateral triangles ABE, ACF and ADG share a common vertex A as shown. Then the midpointss of the segments EC, FD and GB form the vertices of an equilateral triangle.

The special case where points BECFDG lie on a circe centred at A was proved by Alon Amit (April 5 2020) in response to a question posed on Quora:

A hexagon is inscribed in a unit circle such that three alternate sides are of unit length with the other three arbitrary. Connect the midpoints of the three arbitrary length sides. Can you prove those points form an equilateral triangle?

Amit's clever proof, using complex numbers, does not in fact depend on the points lying on the unit circle and holds as long as we have the configuration in the figure above, and this more general case is even more surprising than the original version.


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