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> 三位男子A、B、C带着他们的妻子a、b、c到超市购物,至于谁是谁的妻子就不知道了,只能从下列条件来推测:他们6人,每人花在买商品的钱数(单位:元)正好等于商品数量的平方,而-数学
三位男子A、B、C带着他们的妻子a、b、c到超市购物,至于谁是谁的妻子就不知道了,只能从下列条件来推测:他们6人,每人花在买商品的钱数(单位:元)正好等于商品数量的平方,而-数学
题目简介
三位男子A、B、C带着他们的妻子a、b、c到超市购物,至于谁是谁的妻子就不知道了,只能从下列条件来推测:他们6人,每人花在买商品的钱数(单位:元)正好等于商品数量的平方,而-数学
题目详情
三位男子A、B、C带着他们的妻子a、b、c到超市购物,至于谁是谁的妻子就不知道了,只能从下列条件来推测:他们6人,每人花在买商品的钱数(单位:元)正好等于商品数量的平方,而且每位丈夫都比自己的妻子多花48元钱,又知A比b多买9件商品,B比a多买7件商品.试问:究竟谁是谁的妻子?
题型:解答题
难度:中档
来源:不详
答案
设一对夫妻,丈夫买了x件商品,妻子买了y件商品.
则有x2-y2=48,即(x十y)(x-y)=48.(4分)
∵x、y都是正整数,且x+y与x-y有相同的奇偶性,
又∵x+y>x-y,48=24×2=12×4=8×6,
∴
x+y=24
x-y=2
或
x+y=12
x-y=4
或
x+y=8
x-y=6
.(7分)
解得x=13,y=11或x=8,y=4或x=7,y=1.(9分)
符合x-y=9的只有一种,可见A买了13件商品,b买了4件.
同时符合x-y=7的也只有一种,可知B买了8件,a买了1件.
∴C买了7件,c买了11件.(12分)
由此可知三对夫妻的组合是:A、c;B、b;C、a.(14分)
故答案为:A、c;B、b;C、a.
上一篇 :
______ wants to stay in a ho
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题目简介
三位男子A、B、C带着他们的妻子a、b、c到超市购物,至于谁是谁的妻子就不知道了,只能从下列条件来推测:他们6人,每人花在买商品的钱数(单位:元)正好等于商品数量的平方,而-数学
题目详情
答案
则有x2-y2=48,即(x十y)(x-y)=48.(4分)
∵x、y都是正整数,且x+y与x-y有相同的奇偶性,
又∵x+y>x-y,48=24×2=12×4=8×6,
∴
解得x=13,y=11或x=8,y=4或x=7,y=1.(9分)
符合x-y=9的只有一种,可见A买了13件商品,b买了4件.
同时符合x-y=7的也只有一种,可知B买了8件,a买了1件.
∴C买了7件,c买了11件.(12分)
由此可知三对夫妻的组合是:A、c;B、b;C、a.(14分)
故答案为:A、c;B、b;C、a.