Let x 1, x2, ..., xn (1, 2, ..., n is the foot code) be an integer and satisfy (see below), and find x 1 3+x2 3+...+xn 3.

a.- 1 & lt； = Xi & lt； = 2(& lt； = less than or equal to), i= 1, 2, ..., n

b.x 1+x2+...+xn= 19

c.x 1^2+x2^2+...+xn^2=99

By. - 1

-m+n+2p= 19，m+n+4p=99 == >m=40-p，n=59-3p

If there are no other conditions, p can be 0, 1, 2, …, 19(n cannot be negative), so.

X 1 3+x2 3+...+xn 3 =-m+n+8p has 20 possible results, so it is easy to substitute them one by one and write the results later.