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.


130.--THE SCULPTOR'S PROBLEM.

An ancient sculptor was commissioned to supply two statues, each on a
cubical pedestal. It is with these pedestals that we are concerned. They
were of unequal sizes, as will be seen in the illustration, and when the
time arrived for payment a dispute arose as to whether the agreement was
based on lineal or cubical measurement. But as soon as they came to
measure the two pedestals the matter was at once settled, because,
curiously enough, the number of lineal feet was exactly the same as the