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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
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0

n ==1  and

n == 4

Both of them satisfy the equation.

 May 31, 2021
 #2
avatar+114462 
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I'd be interested in seeing a formal solution for this.   frown

 Jun 1, 2021
 #3
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+1

CPhill presents a clear solution here:

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

 

 

GA

 Jun 1, 2021
 #4
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Thanks Ginger, but the questions are not the same.

Melody  Jun 1, 2021
edited by Melody  Jun 1, 2021
 #5
avatar+114462 
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\(n-(n+1)!+6n!=n!+4\\ n-n!(n+1)+5n!=4\\ n+n!(-n-1)+5n!=4\\ n+n!(-n-1+5)=4\\ n!(4-n)=4-n\\ n!(4-n)-(4-n)=0\\ (n!-1)(4-n)=0\\ n!=1\qquad or \qquad n=4\\ n=0,\;\;1\;\;or\;\;4\)

.
 Jun 1, 2021
 #6
avatar+2193 
+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
avatar+121054 
+1

Nice solution , Melody   !!!!!

 

 

cool cool cool

 Jun 1, 2021
 #8
avatar+114462 
0

Thanks guys  laugh

 Jun 2, 2021

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