Mathematical_Recreations-Kraitchik-2e

Numerical Pastimes 53
If the standard weights may be placed in one scale pan
only, the first set alone gives the answer. For every number
from 1 to 40 can be represented as a sum of multiples by 0 or
1 of the successive powers of 2 from the Oth (= 1) to the 5th
(= 32). In fact, all numbers from 1 to 63 can be so represented.
If the standard weights may be placed in either pan, then
the second set gives a more economical solution, both in the
number of the standard weights and in their total weight.
For every number from 1 to 40 may be represented as a sum
of multiples by 0, 1, or -1 of the successive powers of 3 from
the Oth to the 3rd.
Bachet's solution is correct but not complete. It was completed
by Major MacMahon, who generalized the medieval
problem and posed it as follows:
Find all sets of standard weights, not necessarily unequal,
with which one can weigh every whole number of pounds
from 1 to n, in each of the following cases: (1) If the standard
weights may be placed in only one scale pan. (2) If the
standard weights may be placed in either scale pan.
MacMahon added two other conditions: (a) It is not permissible
to measure additional weights with the set of standard
weights.
(b) Each weight can be measured in just one
way with these standard weights.
The solution of MacMahon's problem is based on the following
theorem:
Let a product
(1+ x+ x2 + ... + x .. ) (1 + x+ x2 + .,. + xn)
(1+ x+ x 2 + ... + x p )
be written as a polynomial,
1+ Ax+ Bx 2 + Cx 3 +
Then the coefficient oj any particular power oj x, say oj x", in
the expanded product is the total number oj different ways in