六年级五个班的同学共植树100棵.已知每个班植树的棵数都不相同,且按数量从多到少的排名恰好是一、二、三
推荐回答(3个)
设1~5班种的数量分别为a、b、c、d、e,根据题意,有:
a+b+c+d+e=100; ----(1)
a>b>c>d>e;
a=b+c;
b=d+e;
设二班比三班多种x,即b=c+x,则x>0;
a=b+c=2c+x; ------------(2)
b=d+e=c+x; -------------(3)
(2)、(3)代入(1),得:
5c+3x=100
c、x都是整数,x不为0,则5c=97, 则c最大为19。
我们假设a b c d+e分别如下
【1】 39 20 19 20 总和98 不足100,由于c不可能往上增加,所以离100尚缺的2只能在a b d+e上面找补,很明显这三个数值是同步增长的,最少一次加3,所以只能靠减少c的数值来满足题意
【2】 40 21 19 21 总和101 超过100
所以三班栽种19棵不肯能更实现
假设3班种18
【3】 38 20 18 20 总和96 不足100
【4】 40 22 18 22 总和102 超过100
【5】 39 21 18 21 总和99 不足100
继续往下假设,3班种17,从上面我们可以看出,每次a b d+e可以同步增长,增幅相等,所以可以构造出离100的差值为3的倍数
【6】 37 20 17 20 总和 94 离100差6,为3倍的2
【7】 39 22 17 22 总和100,可以满足
所以实际上三班最多种17棵
希望回答对你有帮助
分别设五个班植树的是a,b,c,d,e
已知a+b+c+d+e=100
a=b+c
b=d+e
带入
a+b+c+b=100
3b+2c=100
因为b>c
所以3b+2c=100>5c
所以c<20
所以三班最多植树19棵
设1~5班种的数量分别为a、b、c、d、e,根据题意,有:
a+b+c+d+e=100; ----(1)
a>b>c>d>e;
a=b+c;
b=d+e;
设二班比三班多种x,即b=c+x,则x>0;
a=b+c=2c+x; ------------(2)
b=d+e=c+x; -------------(3)
(2)、(3)代入(1),得:
5c+3x=100
c、x都是整数,x不为0,则5c=97, 则c最大为19。
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