Difference between revisions of "2019 AMC 12B Problems/Problem 3"

(Solution)
(Solution 2)
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==Solution 2==
 
==Solution 2==
  
Divide both sides by n!:
+
Divide both sides by <math>n!</math>:
  
  
(n+1)+(n+1)(n+2)=440
+
<math>(n+1)+(n+1)(n+2)=440</math>
  
factor out (n+1):
+
factor out <math>(n+1)</math>:
  
(n+1)*(n+3)=440
+
<math>(n+1)*(n+3)=440</math>
  
  
prime factorization of 440 and a bit of experimentation gives us n+1=20 and n+3=22, so \boxed{n=19}.
+
prime factorization of <math>440</math> and a bit of experimentation gives us <math>n+1=20</math> and <math>n+3=22</math>, so <math>\boxed{n=19}</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2019|ab=B|num-b=2|num-a=4}}
 
{{AMC12 box|year=2019|ab=B|num-b=2|num-a=4}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 13:34, 14 February 2019

Problem

If (n+1)!+(n+2)! = n!*440, what is the sum of the digits of n?

Solution

n=19 sum is 10 (SuperWill)


Solution 2

Divide both sides by $n!$:


$(n+1)+(n+1)(n+2)=440$

factor out $(n+1)$:

$(n+1)*(n+3)=440$


prime factorization of $440$ and a bit of experimentation gives us $n+1=20$ and $n+3=22$, so $\boxed{n=19}$.

See Also

2019 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

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