Mathematical_Recreations-Kraitchik-2e

54 Mathematical Recreations
which a set of powers of x may be selected from the factors on the
left in such a way that in each case the sum of the exponents is
k andjust one term is selected from each factor.
The solution of the first problem is given by expressing
x .. + 1 - 1
1 + x + ... + X" = ----:--
x-I
as a product of factors of the form
l+X"+···+x k .. =
x(k+l)a - 1
•
x-I
Each such factorization yields a solution consisting of k
weights of a pounds each, corresponding to each factor. Thus
the solutions are seen to depend upon the possible factorizations
of n + 1.
For example, take n = 5.
that is,
xG-I x2-I xG-I x 3 -I xG-I
x-I = x-I' X2 - 1 = x-I' x 3 - l'
1 + x + ... + ;t;5 = (I + x) (1 + X2 + X4)
= (1 + x+ x2)(1 + X3).
The corresponding solutions are: 5 I-pound weights, 1 1-
pound and 2 2-pound weights, 2 I-pound and 1 3-pound
weights.
In the second case the solution is found by expressing
xn + x .. - 1 + ... + x + 1 + X-I + . . . + X-(n-I) + X-n
X 2n+ 1 - 1
xn(x - 1)
as a product of factors of the form
x kG + X(k-I) .. + . . . + X" + 1 + X-" + . . . + X-(k-I) .. + x-kG
X(2k+I)4 - 1
= Xk ..(X" - I)'
Again the solution corresponding to any such factorization