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2001 AMC 12 Problem 5

Below is the professionally curated solution for Problem 5 of the 2001 AMC 12, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2001 AMC 12 solutions, or check the answer key.

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Concepts:factorialalgebraic manipulation

Difficulty rating: 1370

5.

What is the product of all positive odd integers less than 10,000?10{,}000?

10000!(5000!)2\dfrac{10000!}{(5000!)^2}

10000!25000\dfrac{10000!}{2^{5000}}

9999!25000\dfrac{9999!}{2^{5000}}

10000!25000⋅5000!\dfrac{10000!}{2^{5000} \cdot 5000!}

5000!25000\dfrac{5000!}{2^{5000}}

Solution:

The product of every integer from 11 to 1000010000 is 10000!,10000!, so the product of the odd ones is 10000!10000! divided by the product of the even ones.

The even numbers factor as 2⋅4⋯10000=25000(1⋅2⋯5000)=25000⋅5000!. 2 \cdot 4 \cdots 10000 = 2^{5000}(1 \cdot 2 \cdots 5000) = 2^{5000} \cdot 5000!.

Therefore the product of the odd integers is 10000!25000⋅5000!. \dfrac{10000!}{2^{5000} \cdot 5000!}.

Thus, the correct answer is D.

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Problem 5 in Other Years

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