Factorial

Number Theory Level 2

Find the integer x x satisfying 4 ( x + 1 ) ! = ( 2 x − 6 ) ! x ! 4(x+1)! = (2x-6)! \, x! .

Notation: ! ! is the factorial notation. For example, 8 ! = 1 × 2 × 3 × ⋯ × 8 8! = 1\times2\times3\times\cdots\times8 .


The answer is 5.

Ahmed Mohamed by Ahmed Mohamed , 25, Egypt

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2 solutions

Rolf TheRolf
Jan 11, 2017

The best I have so far:

4 ( x + 1 ) ! = ( 2 x − 6 ) ! x ! 4 (x+1)! = (2x-6)! x!

4 ( x + 1 ) x ! = ( 2 x − 6 ) ! x ! 4 (x+1) x! = (2x-6)! x!

4 x + 4 = ( 2 x − 6 ) ! 4x + 4 = (2x-6)!

Let y = 2 x − 6 y=2x-6 , then x = ( y + 6 ) / 2 x=(y+6)/2 .

2 y + 16 = y ! 2y + 16 = y!

Since y ≥ 0 y \geq 0 , y ! ≥ 16 y! \geq 16 and from that y > 3 y>3 follows (suppose you have a table of values of n!). You try 4, and that's correct, y = 4 , x = 5 y=4, x=5 .

0 Helpful 0 Interesting 0 Brilliant 0 Confused

it's 4(x+1)! on LHS in the first two passages. It's just a written mistake the idea was good

Raseq Ninunine - 4 years, 5 months ago

x x must be equal to 5 5

4 ( 5 + 1 ) ! = [ 2 ( 5 ) − 6 ] ! 5 ! 4(5+1)! = [2(5) - 6]!5!

2880 = 2880 2880=2880

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You used trial and error. Can you prove that x = 5 x = 5 from the start? :)

Michael Huang - 4 years, 5 months ago

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