2道数学题,
问题描述：
2道数学题,
1、一串珠子,按黑白黑黑白黑黑黑白……的顺序排列,问200颗内有几颗黑子?
2、A玩具25元,B玩具比A玩具便宜一些（整数元）,且B比A少买了2个,买两种玩具共花了280元,问卖了多少个A玩具?
答
(1) 2+3+4+..n = (n-1)(n+2)/2≥200
(n-1)(n+2)≥400
n^2 + n ≥402
n≥20
n=19时一起摆了189颗子（此处摆黑子18颗）,后面继续摆11颗黑子便是200颗珠子
所以黑子有1+2+..+18 + 11 = 182
(2) 设A玩具x个,B玩具价格是b元(b＜25,为整数)
25x + (x - 2)b = 280
若x=2,则花钱为50,矛盾
b = (280-25x)/(x-2)
280＞25x
x＜11.2
3≤x≤11
又b≤24
(280-25x)/(x-2)≤24
280 - 25x≤24x - 48
x≥6.69
x≥7
7≤x≤11
x = 7
b = (280-25x)/(x-2) = 21
x = 8
b = (280-25x)/(x-2) = 13.3
x = 9
b = (280-25x)/(x-2) = 7.8
x = 10
b = (280-25x)/(x-2) = 3.75
x = 11
b = (280-25x)/(x-2) = 0.555
由于b是整数,所以b = 21,x = 7