For positive integer nn, the factorial notation n!n! represents the product of the integers from nn to 1. (For example, 6!=6⋅5⋅4⋅3⋅2⋅16! = 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1.) What value of NN satisfies the following equation?
5!⋅9!=12⋅N!5! \cdot 9! = 12 \cdot N!
A. 10
B. 11
C. 12
D. 13
E. 14

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Worked solution

1. 2020 AMC 8 官方解析

The given equation implies that
N!=5!⋅9!12=(5⋅4⋅3⋅2⋅1)⋅(9!)12=10⋅9!=10!N! = \frac{5! \cdot 9!}{12} = \frac{(5 \cdot 4 \cdot 3 \cdot 2 \cdot 1) \cdot (9!)}{12} = 10 \cdot 9! = 10!
so N=10N = 10.

由所给的方程可以推断出:
N!=5!⋅9!12=(5⋅4⋅3⋅2⋅1)⋅(9!)12=10⋅9!=10!N! = \frac{5! \cdot 9!}{12} = \frac{(5 \cdot 4 \cdot 3 \cdot 2 \cdot 1) \cdot (9!)}{12} = 10 \cdot 9! = 10!
故N=10N = 10 .

Answer

A. 10

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