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1991年11月号问题解答 (解答由问题提供人给出) 11.设a_1,a_2,…,a_n为正整数,N为整数,解方程[a_1x]+[a_2x]+…+[a_nx]=N。解不失普遍性,可设(a_1,a_2,…,a_n)=1,否则如果(a_1,a_2,…,a_n)=d。令x=dx就转化为这种情况,为了方便, 令 M=a_1+a_2+…+a_n, F(x)=[a_1x]+[a_2x]+…+[a_nx]。很明显,F(x)是不减的且F(x+1)=F(x)+M。由此可得出,F(x+s)=F(x)+sM,(s为整数) 如果N=N_1+Ms, 则F(x)=N有解的充分必要条件是F(x)=N_1有解。如果F(x)=N_1的解为a,则F(x)=
November 1991 issue No. (Answer is given by the problem provider) 11. Let a_1, a_2, ..., a_n be positive integers, N be an integer, and solve the equation [a_1x]+[a_2x]+...+[a_nx]= N. The solution does not lose its universality. It can be set (a_1, a_2, ..., a_n) = 1, otherwise (a_1, a_2, ..., a_n) = d. Let x = dx be transformed into this case. For convenience, let M=a_1+a_2+...+a_n, F(x)=[a_1x]+[a_2x]+...+[a_nx]. Obviously, F(x) is undiminished and F(x+1)=F(x)+M. From this, it can be concluded that F(x+s)=F(x)+sM, (s is an integer) If N=N_1+Ms, then F(x)=N has a solution with F(x). =N_1 has a solution. If the solution of F(x)=N_1 is a, then F(x)=