2004 AMC 10A Problem 24

Below is the professionally curated solution for Problem 24 of the 2004 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2004 AMC 10A solutions, or check the answer key.

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Concepts:recursiontriangular numberpattern recognition

Difficulty rating: 2010

24.

Let a1,a_1, a2,a_2, \ldots be a sequence with the following properties: a1=1,a_1 = 1, and a2n=nana_{2n} = n \cdot a_n for any positive integer n.n. What is the value of a2100?a_{2^{100}}?

11

2992^{99}

21002^{100}

249502^{4950}

299992^{9999}

Solution:

Applying the rule repeatedly, a21=20,a22=21,a23=21+2,a24=21+2+3, a_{2^1} = 2^0, \quad a_{2^2} = 2^1, \quad a_{2^3} = 2^{1+2}, \quad a_{2^4} = 2^{1+2+3}, \ldots so in general a2n=21+2++(n1)=2n(n1)/2.a_{2^n} = 2^{1 + 2 + \cdots + (n - 1)} = 2^{n(n-1)/2}.

For n=100,n = 100, the exponent is 100992=4950,\dfrac{100 \cdot 99}{2} = 4950, so a2100=24950.a_{2^{100}} = 2^{4950}.

Thus, the correct answer is D.

Problem 24 in Other Years