AMC10 2011 B

AMC10 2011 B · Q10

AMC10 2011 B · Q10. It mainly tests Rounding & estimation, Arithmetic sequences basics.

Consider the set of numbers $\{1, 10, 10^2, 10^3, \dots, 10^{10}\}$. The ratio of the largest element of the set to the sum of the other ten elements of the set is closest to which integer?

考虑集合 $\{1, 10, 10^2, 10^3, \dots, 10^{10}\}$。该集合中最大元素与其它十个元素之和的比值最接近哪个整数？

(A) 1 1

(B) 9 9

(D) 11 11

(E) 101 101

Answer

Correct choice: (B)

正确答案：（B）

Solution

Answer (B): The sum of the smallest ten elements is $1+10+100+\cdots+1{,}000{,}000{,}000=1{,}111{,}111{,}111.$ Hence the desired ratio is $\dfrac{10{,}000{,}000{,}000}{1{,}111{,}111{,}111}=\dfrac{9{,}999{,}999{,}999+1}{1{,}111{,}111{,}111}=9+\dfrac{1}{1{,}111{,}111{,}111}\approx 9.$ OR The sum of a finite geometric series of the form $a(1+r+r^2+\cdots+r^n)$ is $\dfrac{a}{1-r}(1-r^{n+1})$. The desired denominator $1+10+10^2+\cdots+10^9$ is a finite geometric series with $a=1$, $r=10$, and $n=9$. Therefore the ratio is $\dfrac{10^{10}}{1+10+10^2+\cdots+10^9}=\dfrac{10^{10}}{\dfrac{1}{1-10}(1-10^{10})}=\dfrac{10^{10}}{10^{10}-1}\cdot 9\approx \dfrac{10^{10}}{10^{10}}\cdot 9=9.$

答案（B）：最小的十个元素之和为 $1+10+100+\cdots+1{,}000{,}000{,}000=1{,}111{,}111{,}111.$ 因此所求比值为 $\dfrac{10{,}000{,}000{,}000}{1{,}111{,}111{,}111}=\dfrac{9{,}999{,}999{,}999+1}{1{,}111{,}111{,}111}=9+\dfrac{1}{1{,}111{,}111{,}111}\approx 9.$ 或者形如 $a(1+r+r^2+\cdots+r^n)$ 的有限等比数列之和为 $\dfrac{a}{1-r}(1-r^{n+1})$。所需分母 $1+10+10^2+\cdots+10^9$ 是一个有限等比数列，其中 $a=1$，$r=10$，$n=9$。因此该比值为 $\dfrac{10^{10}}{1+10+10^2+\cdots+10^9}=\dfrac{10^{10}}{\dfrac{1}{1-10}(1-10^{10})}=\dfrac{10^{10}}{10^{10}-1}\cdot 9\approx \dfrac{10^{10}}{10^{10}}\cdot 9=9.$

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