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!' 'Godemitè!'"
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 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 $$60 \times 42 + 1 = 312,$$ and $$62 \times 82 + 1 = 632.$$ 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 $61.$
Solutions: 1
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