2004 AMC 10A Problem 24

Attempt Problem 24 of the 2004 AMC 10A below, then check your answer against the professionally curated solution 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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24.

Let a1,a_1, a2,a_2, …\ldots be a sequence with the following properties: a1=1,a_1 = 1, and a2n=n⋅ana_{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}
Answer: D
Concepts:recursiontriangular numberpattern recognition
Small hint:
Compute a21,a_{2^1}, a22,a_{2^2}, a23,a_{2^3}, a24a_{2^4} and track the exponent of 22
Big hint:
The exponents are 0,0, 1,1, 1+2,1 + 2, 1+2+3,1 + 2 + 3, …\ldots
Solution:
Applying the rule repeatedly, a21=20,a22=21,a23=21+2,a24=21+2+3,… \begin{aligned} a_{2^1} &= 2^0, \\ a_{2^2} &= 2^1, \\ a_{2^3} &= 2^{1+2}, \\ a_{2^4} &= 2^{1+2+3}, \ldots \end{aligned} so in general a2n=21+2+⋯+(n−1)=2n(n−1)2.a_{2^n} = 2^{1 + 2 + \cdots + (n - 1)} = 2^{\frac{n(n-1)}{2}}.
For n=100,n = 100, the exponent is 100⋅992=4950,\dfrac{100 \cdot 99}{2} = 4950, so a2100=24950.a_{2^{100}} = 2^{4950}.
Thus, the correct answer is D.
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