§6 一次不定方程1.问题:有一批货物,共1120件,分装两种规格的木箱。甲种木箱每箱恰好装货12件,乙种恰好装25件,在两种木箱用量不宜悬殊过大的条件下,问甲、乙两种木箱各用多少最为合理?解:设共用甲种木箱x只及乙种木箱y只,恰好把货物装完,则得方程12x+25y=1120因为(12,25)=1,则必有二整数m,n使12m+25n=1易知12·(-2)+25·1=1乘以1120,得 § 6 Once indefinite Equation 1. Problem: There are a total of 1,120 pieces of cargo, which are divided into two sizes of wooden boxes. A wooden box is loaded with exactly 12 pieces of carton per box, and B pieces are loaded with exactly 25 pieces. Under the condition that the amount of two types of wooden boxes should not be disproportionately large, it is reasonable to ask whether the wooden boxes of both types A and B are used differently. A common wooden box x and only wooden box y, just load the goods, you get the equation 12x+25y=1120 because (12,25)=1, there must be two integers m,n to make 12m+25n =1 Easy to know 12·(-2)+25·1=1=1 times 1120