Factorials and Figurate Numbers. 413 
nent and common difference of the other to its exponent, may be 
reduced to one. 
For let m"'© and [m-+-nc]"'l* be the two factorials, the base 
of the latter being formed as announced in the proposition : then 
because’c is the common difference, and m is the first factor of 
the factorial m”'’, the second factor will be m-+c, the third 
m+ 2c, the fourth m+3c, andso on. Therefore in m+ 1 factors, 
the (7+ 1)th factor from the first will be the first factor, together 
with 7 times the common difference c; therefore if the factorial] 
(m-+inc)*' be annexed to the Stel m'|° as two factors, the 
product will be the factorial m"*“°, 
PROBLEM. 
To resolve a given factorial into two factorial factors, in which 
the factors of each shall have the same common difference as the 
factors of the given factorial, and the one a given exponent less 
than that of the given factorial. 
Rule.—1. Take the less from the greater of the two given ex- 
ponents, and the remainder will be the exponent of the factorial 
= which is not given. i 
. To the base of the given factorial apply isthe of the ex- 
tears of the two factorial factors, and the common difference, 
and the quantity thus formed will be one of the factorial factors. 
3. To-the same base add the product of the exponent and 
common difference of the factorial factor thus completed, and to 
the sum as a base apply the remaining exponent of the two fac- 
torial factors, and the common difference, then the quantity thus 
formed will be the other factorial factor. 
Examples.—1. Resolve m”'° into two factorial factors, so that 
one of them may have the given exponent r. 
By rule, n—r will be the exponent of the other. Now if m'!* 
be the one factorial, (m-+rc) n—r\¢ will be the other. 
Or, if m"—"'° be the one factorial [m+ (a—r)c]"!® will be the 
other. 
2. Resolve m"!* into two factorial factors, so that one of them 
may have the given exponent 1. 
By rule, »—1 will be the exponent of the other. Now there- 
—i1 
fore, if. m'I° ao 
will be the other: or if m"—'l° be the one factorial, then will 
[m+ (n— 1c]! =m-+ (n—1)c be the other. 
3. Resolve m”!' into two factorial factors, so that one of them 
may have the given exponent 1. 
By rule, 2—1 will be the exponent of the other. If therefore 
m 
= m be the one factorial factor, then (m-+c) 
