You are not logged in.
Pages: 1
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 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 × 4² + 1 = 31², and 62 × 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.
I have tried to work out an answer to other numbers of original squares like having 11 squares.
does anyone have any answers to this problem
Offline
It took me a while to figure out your post, so I tweaked it a bit so that there are ²'s in there. Hopefully you won't mind.
As for the question, I don't think it has an answer. I couldn't see a way to do it analytically, so I just brute-forced it in Excel. I tried every size of small square from 1 to 60000, and none of them worked. Therefore, if there is a solution, then it will involve more than 219 thousand million people, which is considerably more than the current world population. So Harold couldn't possibly have had a formation like that which worked.
For your other question, 11x3² = 10². I don't think there's a formula that will generate answers though, each case would have to be looked at individually.
Why did the vector cross the road?
It wanted to be normal.
Offline
i was looking for a website with answers on i have the answer for 61
the formula is y²=(61x²)+1
i have worked out a lot but i would like any answers you can find
the answer to 61 is aproximately x=226million and y=1.76billion this is the highest until at least 109 according to my calulations
how do u do ur squares i had to copy urs
Offline
Ah, fair enough. So there was an answer for 61, but I just wasn't quite looking high enough.
And you can get the square symbol by writing ": sup2", without the space after the colon.
Why did the vector cross the road?
It wanted to be normal.
Offline
thanks but can anyone find any solutions online
could u post if u r looking
Offline
halley's comet!
henryzz is correct, I have found identical formulae.
I shall be on leave until I say so...
Offline
i want solutions
are people basicly saying there r no solutions on the web
Offline
i want solutions
are people basicly saying there r no solutions on the web
Barring the use of billions of men, I'd put forward the solution that the word "square" in this case is misused, and he merely had approximately square rectangles at his use. So even if there were an even number of men in the initial squares (12x12 = 144 per square, for example) the giant "square" had 8785 people in it, who would've been standing in a form approximates 93x93 men. But since I don't think they had helicopters, or the capacity for counting that many people at once, it would've merely seemed to be one monstrous square of people, even if it wasn't perfectly geometrically perfect. ;-)
Offline
it was a perfect square
the army bit is just to make it a more fun puzzle
when will i get solutions
Offline
Offline
that answers 61 and 97
i have now worked out up to 300 except for things i cant find
what i want is a list showing what the anwer to each number is
this was a puzzle written by Arcamedis and later written about by Dudeney it is not a literal battle
Offline
Try this link: http://www.bioinfo.rpi.edu/~zukerm/cgi-bin/dq.html
Nifty program somebody wrote up to solve whatever problems you ask it.
Offline
pell's equation is X² - d × Y² = ± 1 not y²=(dx²)+1
i have tried transposing terms and i get X² =dY² ± 1 which is just x and y swapped except that it is ±1 instead of +1
this puzzle's equation is
Last edited by henryzz (2007-06-22 06:19:22)
Offline
Pages: 1