Fatorial #5

Algebra Level 2

If x! = x . (x - 1)! , log 10 = 1 ,

A = an indentity matrix of order 3x3 ,

|A| = the determinant of the matrix A .

Find the value of x.

$\frac{100! - x!} {x!}$ = ( $\log 10^{100}$ ) - |A|

101 100 98 99

1 solution

Victor Porto
Oct 25, 2014

$\frac{100! - x!} {x!}$ = ( $\log 10^{100}$ ) - |A|

|A| = 1 and ( $\log 10^{100}$ ) = 100

$\frac{100! - x!} {x!}$ = 100 - 1

$\frac{100! - x!} {x!}$ = 99

100! - x! = 99 x!

100! = 100 x!

100 99! = 100 x!

99! = x!

x = 99

given x = 1, (what satisfies the first equation), (100!-x!)/x! = 100! - 1, witch is not equal to 99... substituting this value for X in the second equation, we can only find the value of the determinant... I didn't understand your answer... could you be more specific?

Júlio Vinicius Rodrigues Miguel - 6 years, 7 months ago

x! = x . (x - 1)! is how to calculate a factorial. ("first equation")

The value of the determinant of an indentity matrix of order 3x3 is 1.

And log 10^100 = 100

That way, we substitue the values we already get in the "second equation".

(100! - x!)/(x!) = 100 - 1

(100! - x!)/(x!) = 99

100! - x! = 99 . x!

100! = 100 x!

Using the "first equation", we know that (100! = 100 . 99!)

100 . 99! = 100 . x!

99! = x!

x = 99

Okay?

Victor Porto - 6 years, 7 months ago

Fatorial #5

1 solution

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