Bifunctional Problem

Number Theory Level 2

Given that $(x+1)! = \dfrac{(3!)!}{3!}$ . Find the value of $\dfrac{(x+1)!}{x}$ .

Notation : $!$ denotes the factorial notation. For example, $8! = 1\times2\times3\times\cdots\times8$ .

2 solutions

Akash Patalwanshi
May 13, 2016

$(x+1)! = \frac{(3!)!}{3!} = \frac{6!}{3!} = 120$ $→ (x+1)! = 120 → x = 4$ $→ \frac{(x+1)!}{x} = \frac{120}{4} =\boxed{30}$

David Hontz
May 13, 2016

Part 1

First: $(x+1)! = \frac{(3!)!}{3!} = \frac{(3\times2\times1)!}{3\times2\times1}= \frac{6\times5\times4\times3\times2\times1}{6} = 120$

Second: $(x+1)! = 120 = 5\times4\times3\times2\times1 = 5!$

Third: $(x+1)! = 5!$

Fourth: $x+1=5$

Fifth: $x=4$

Part 2

$\frac{(x+1)!}{x} = \frac{(4+1)!}{4}= \frac{5\times4\times3\times2\times1}{4} = 5\times3\times2\times1 = 30$

The answer is therefore $\boxed{30}$

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Samara Simha Reddy - 5 years, 1 month ago

Nice comment (+1)

Ashish Menon - 5 years ago