AMC10 2004 A

AMC10 2004 A · Q24

AMC10 2004 A · Q24. It mainly tests Sequences & recursion (algebra), Counting divisors.

Let $a_1, a_2, \cdots $, be a sequence with the following properties. (i) $a_1 = 1$, and (ii) $a_{2n} = n \cdot a_n$ for any positive integer $n$. What is the value of $a_{2100}$?

设序列$a_1, a_2, \cdots$满足：(i) $a_1 = 1$，(ii) 对任意正整数$n$，$a_{2n} = n \cdot a_n$。求$a_{2100}$的值。

(A) 1 1

(B) 2^{99} 2^{99}

(D) 24950 24950

(E) 2^{9999} 2^{9999}

Answer

Correct choice: (D)

正确答案：（D）

Solution

(D) Note that $a_{2^1}=a_2=a_{2\cdot 1}=1\cdot a_1=2^0\cdot 2^0=2^0,$ $a_{2^2}=a_4=a_{2\cdot 2}=2\cdot a_2=2^1\cdot 2^0=2^1,$ $a_{2^3}=a_8=a_{2\cdot 4}=4\cdot a_4=2^2\cdot 2^1=2^{1+2},$ $a_{2^4}=a_{16}=a_{2\cdot 8}=8\cdot a_8=2^3\cdot 2^{1+2}=2^{1+2+3},$ and, in general, $a_{2^n}=2^{1+2+\cdots+(n-1)}$. Because $1+2+3+\cdots+(n-1)=\frac{1}{2}n(n-1),$ we have $a_{2^{100}}=2^{(100)(99)/2}=2^{4950}.$

（D）注意到 $a_{2^1}=a_2=a_{2\cdot 1}=1\cdot a_1=2^0\cdot 2^0=2^0,$ $a_{2^2}=a_4=a_{2\cdot 2}=2\cdot a_2=2^1\cdot 2^0=2^1,$ $a_{2^3}=a_8=a_{2\cdot 4}=4\cdot a_4=2^2\cdot 2^1=2^{1+2},$ $a_{2^4}=a_{16}=a_{2\cdot 8}=8\cdot a_8=2^3\cdot 2^{1+2}=2^{1+2+3},$ 并且一般地，$a_{2^n}=2^{1+2+\cdots+(n-1)}$。因为 $1+2+3+\cdots+(n-1)=\frac{1}{2}n(n-1),$ 所以有 $a_{2^{100}}=2^{(100)(99)/2}=2^{4950}.$

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