大师作文网美国数学竞赛题1.Find the area of the region that lies under the gra
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美国数学竞赛题
1.Find the area of the region that lies under the graph of f (x) =|||| 6 − x | −x | −x | −x | and
above the x axis,between x = 0 and x = 12.
2.Alex is playing a card game and tabulating the results.He calculates that he has played
1800 games and won exactly 1542 of them.Rounded to the nearest percent,this is 86%.
What is the smallest number of consecutive games he would have to win in order for his
winning percentage (rounded to the nearest percent) to be equal to 87%?
3.Find the smallest positive integer which is evenly divisible by 225 and whose digits are all zeros
or ones.
1.Find the area of the region that lies under the graph of f (x) =|||| 6 − x | −x | −x | −x | and
above the x axis,between x = 0 and x = 12.
2.Alex is playing a card game and tabulating the results.He calculates that he has played
1800 games and won exactly 1542 of them.Rounded to the nearest percent,this is 86%.
What is the smallest number of consecutive games he would have to win in order for his
winning percentage (rounded to the nearest percent) to be equal to 87%?
3.Find the smallest positive integer which is evenly divisible by 225 and whose digits are all zeros
or ones.
第一题
思路是划分函数区间,单独讨论
Case1
0≤x≤6:f(x)=|||(6-x)-x|-x|-x|=|||6-2x|-x|-x|
Subcase1
0≤x≤3:f(x)=||6-2x-x|-x|=||6-3x|-x|
Subcase 1a
0≤x≤2:f(x)=|6-3x-x|=|6-4x|
Subcase 1a,a
0≤x≤1.5:f(x)=6-4x
Subcase 1a,b
1.5＜x≤2:f(x)=4x-6
Subcase 1b
2＜x≤3:f(x)=|3x-6-x|=|2x-6|=6-2x
Subcase b
3＜x≤6:f(x)=||2x-6-x|-x|=||x-6|-x|=|6-x-x|=2x-6
Case 2
6＜x≤12:y=|||x-6-x|-x|-x|=||6-x|-x|=|x-6-x|=6
Therefore:
f(x)=6-4x,0≤x≤1.5
f(x)=4x-6,1.5＜x≤2
f(x)=6-2x,2＜x≤3
f(x)=2x-6,3＜x≤6
f(x)=6 ,6＜x≤12
面积按5个矩形拼起来算就行了
A=4.5+1.5+1+9+36=51
第二题
87%是从十分位 四舍五入取到的近似值,所以我们只需要把胜率提高到86.5%
Let x be the smallest number of consecutive games he have to win.
We obtain:
(1542+x)/(1800+x)=86.5%
x=111.11
但是这里的四舍五入必须进一位,因为如果只赢111场,胜率是86.499%,无法取到87%
相反,进一位到112后,胜率将会达到86.51%≈87%
Therefore,x=112
第三题
分解225,得到225=25*9
想让25达到0和1的组合,乘上4是最小的.
想让9达到0和1的组合,各位数加起来必须是9的倍数.9最小的倍数是9,所以前面加9个1
最后的数就是11111111100
思路是划分函数区间,单独讨论
Case1
0≤x≤6:f(x)=|||(6-x)-x|-x|-x|=|||6-2x|-x|-x|
Subcase1
0≤x≤3:f(x)=||6-2x-x|-x|=||6-3x|-x|
Subcase 1a
0≤x≤2:f(x)=|6-3x-x|=|6-4x|
Subcase 1a,a
0≤x≤1.5:f(x)=6-4x
Subcase 1a,b
1.5＜x≤2:f(x)=4x-6
Subcase 1b
2＜x≤3:f(x)=|3x-6-x|=|2x-6|=6-2x
Subcase b
3＜x≤6:f(x)=||2x-6-x|-x|=||x-6|-x|=|6-x-x|=2x-6
Case 2
6＜x≤12:y=|||x-6-x|-x|-x|=||6-x|-x|=|x-6-x|=6
Therefore:
f(x)=6-4x,0≤x≤1.5
f(x)=4x-6,1.5＜x≤2
f(x)=6-2x,2＜x≤3
f(x)=2x-6,3＜x≤6
f(x)=6 ,6＜x≤12
面积按5个矩形拼起来算就行了
A=4.5+1.5+1+9+36=51
第二题
87%是从十分位 四舍五入取到的近似值,所以我们只需要把胜率提高到86.5%
Let x be the smallest number of consecutive games he have to win.
We obtain:
(1542+x)/(1800+x)=86.5%
x=111.11
但是这里的四舍五入必须进一位,因为如果只赢111场,胜率是86.499%,无法取到87%
相反,进一位到112后,胜率将会达到86.51%≈87%
Therefore,x=112
第三题
分解225,得到225=25*9
想让25达到0和1的组合,乘上4是最小的.
想让9达到0和1的组合,各位数加起来必须是9的倍数.9最小的倍数是9,所以前面加9个1
最后的数就是11111111100
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