Question

某商场文具专柜以每支a元的价格购进一批英雄牌钢笔

某商场文具专柜以每只a（a为整数）元的价格购进一批钢笔，决定每只加价2元销售。商品很快卖没了，结账时，售货员发现这批钢笔总额为339a+805元。求文具专柜共购进多少支钢笔及每只钢笔的进价a？ 给出解题过程，O(∩_∩)O谢谢

Answer 1

设文具专柜共购进X支钢笔:
(a+2)X=339a+805
X=(339a+805)/(a+2)
=(339a+678+127)/(a+2)
=[339(a+2)+127]/(a+2)
=339+127/(a+2)
X为整数,127/(a+2)最小值=1,a+2=127,a=125(元)
X=339+1=340(支)

Answer 2

设文具专柜共购进X支钢笔:
(a+2)X=339a+805
X=(339a+805)/(a+2)
=(339a+678+127)/(a+2)
=[339(a+2)+127]/(a+2)
=339+127/(a+2)
X为整数,127/(a+2)最小值=1,a+2=127,a=125(元)
X=339+1=340(支)

Answer 3

设文具专柜共购进钢笔y支，则有

y= = = =399+

∵ a>0且为整数，y为正整数

∴ a十2=7或a十2=1

∴ a=5或 a=－1(不合题意)

当a=5时 y=400

答：文具专柜共购进400支，每支进价5元

Answer 4

设文具专柜共购进X支钢笔:
(a+2)X=339a+805
X=(339a+805)/(a+2)
=(339a+678+127)/(a+2)
=[339(a+2)+127]/(a+2)【678=2X339】
=339+127/(a+2)
X为整数,127/(a+2)最小值=1,a+2=127,a=125(元)
X=339+1=340(支)

Answer 5

399a+805
=399*(a+2)+7
故售价为7元。400支。进价5元。

Answer 6

000