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Number theory Difficulty 3.6 AMC 10/12 Find the answer Canada

If $n$ is a positive integer,
the notation $n!$ (read “ $n$ factorial”) is used to represent the
product of the integers from $1$ to
$n$ . That is, $n! = n \times (n-1) \times (n-2) \times \cdots \times 3 \times 2 \times 1$ . For example, $5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$ . If $n! = 3! \times 5! \times 7!$ , the value of $n$ is

Pick one

Solution

In this solution we make use of the fact that $n\times(n-1)!=n!$ .

For example, $4\times3!=4\times3\times2\times1=4!$ .

Manipulating the factors in the given product, we obtain the following
equivalent equations: $\begin{align*} n! & = 3!\times5!\times7! \\ & = 3\times2\times1\times5\times4\times3\times2\times1\times7!\ \ (\text{expanding } 3! \text{ and } 5!)\\ & = 5\times2\times3\times3\times4\times2\times7!\ \ (\text{rearranging the factors})\\ & = 10\times9\times8\times7!\\ &=10! \ \ (\text{using the fact three times})\end{align*}$
and so $n=10$ .

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Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.