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THE BATTLE OF HASTINGS.





(Money Puzzles)
All historians know that there is a great deal of mystery and
uncertainty concerning the details of the ever-memorable battle on that
fatal day, October 14, 1066. My puzzle deals with a curious passage in
an ancient monkish chronicle that may never receive the attention that
it deserves, and if I am unable to vouch for the authenticity of the
document it will none the less serve to furnish us with a problem that
can hardly fail to interest those of my readers who have arithmetical
predilections. Here is the passage in question.
"The men of Harold stood well together, as their wont was, and formed
sixty and one squares, with a like number of men in every square
thereof, and woe to the hardy Norman who ventured to enter their
redoubts; for a single blow of a Saxon war-hatchet would break his lance
and cut through his coat of mail.... When Harold threw himself into the
fray the Saxons were one mighty square of men, shouting the
battle-cries, 'Ut!' 'Olicrosse!' 'Godemite!'"
Now, I find that all the contemporary authorities agree that the Saxons
did actually fight in this solid order. For example, in the "Carmen de
Bello Hastingensi," a poem attributed to Guy, Bishop of Amiens, living
at the time of the battle, we are told that "the Saxons stood fixed in a
dense mass," and Henry of Huntingdon records that "they were like unto a
castle, impenetrable to the Normans;" while Robert Wace, a century
after, tells us the same thing. So in this respect my newly-discovered
chronicle may not be greatly in error. But I have reason to believe that
there is something wrong with the actual figures. Let the reader see
what he can make of them.
The number of men would be sixty-one times a square number; but when
Harold himself joined in the fray they were then able to form one large
square. What is the smallest possible number of men there could have
been?
In order to make clear to the reader the simplicity of the question, I
will give the lowest solutions in the case of 60 and 62, the numbers
immediately preceding and following 61. They are 60 x 4 squared + 1 = 31 squared,
and 62 x 8 squared + 1 = 63 squared. That is, 60 squares of 16 men each would be 960
men, and when Harold joined them they would be 961 in number, and so
form a square with 31 men on every side. Similarly in the case of the
figures I have given for 62. Now, find the lowest answer for 61.


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Next: THE SCULPTOR'S PROBLEM.

Previous: A PROBLEM IN SQUARES.



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