485] 
PROBLEMS AND SOLUTIONS. 
599 
if x be the number of times of carrying for the sum m + n, or of borrowing for the 
difference (m + n) — in or (m + n) — n ; that is, JS T = x, the required theorem. I remark 
that although the foregoing expression of the number N is a very elegant and 
ingenious one, yet the original form of N, as given at the end of (3), is the natural 
and proper expression of the number of times that the factor p occurs in the 
binomial coefficient ^, . 
LI (in) 11 (n) 
[Vol. x. p. 98.] 
2756. (Proposed by J. Griffiths, M.A.)—Show that an infinite number of triangles 
can be described such that each has the same circumscribing, nine-point, and self 
conjugate circles as a given triangle. 
Solution by Professor Cayley. 
It is a known theorem that if two triads of points, say A, B, C and A', B', C', 
are self-conjugate in regard to a conic S, they lie in a conic 2. Take the conic S 
and the points A, B, G as given; then 2 will be a conic passing through A, B, G\ 
and if on this conic we take any point A', and then take B' to be either of the 
intersections of the conic 2 by the polar of A in regard to S, and finally construct 
C' as the pole of A'B' in regard to S, then, by what precedes, G' will be on a conic 
through A, B, G, A', B', that is, on the conic 2. Or given the conic S, the triangle 
ABG, and the conic 2 through A, B, C, we obtain an infinity of triangles A'B'G', 
self-conjugate in regard to S and inscribed in 2. That is, if S, 2 are circles, we 
have an infinity of triangles self-conjugate in regard to the circle S and inscribed in 
the circle 2; and inasmuch as the nine-points circle can be constructed by means of 
the two circles S, 2 alone, the triangles have all of them the same nine-points circle. 
[Vol. x. p. 108.] 
2737. (Proposed by Professor Cayley.)—Find in solido the locus of a point P, 
such that from it two given points A, G, and two given points B, D, subtend equal