+0

Factorial equation

0
249
8

Find the value of n that satisfies n-(n+1)!+6n!=n!+4, where n!=n*(n-1)*(n-2)***2*1 .

May 31, 2021

#1
0

n ==1  and

n == 4

Both of them satisfy the equation.

May 31, 2021
#2
+117766
0

I'd be interested in seeing a formal solution for this.

Jun 1, 2021
#3
+1

CPhill presents a clear solution here:

https://web2.0calc.com/questions/pls-help-asap_40

GA

Jun 1, 2021
#4
+117766
0

Thanks Ginger, but the questions are not the same.

Melody  Jun 1, 2021
edited by Melody  Jun 1, 2021
#5
+117766
+3

$$n-(n+1)!+6n!=n!+4\\ n-n!(n+1)+5n!=4\\ n+n!(-n-1)+5n!=4\\ n+n!(-n-1+5)=4\\ n!(4-n)=4-n\\ n!(4-n)-(4-n)=0\\ (n!-1)(4-n)=0\\ n!=1\qquad or \qquad n=4\\ n=0,\;\;1\;\;or\;\;4$$

.
Jun 1, 2021
#6
+2403
+2

Thank you for the solution.

I tried this problem yesterday and could not get it.

n - (n+1)n! + 5n! = 4

n + (4 - n)n! = 4

I did not think of factoring it from here, that's a really nice solution.

=^._.^=

catmg  Jun 1, 2021
#7
+124524
+1

Nice solution , Melody   !!!!!

Jun 1, 2021
#8
+117766
0

Thanks guys

Jun 2, 2021