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If (n+1)! + (n+2)! = n!*440. What is the sum of the digits of n?
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15 May 2019, 03:23
15 May 2019, 03:23
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If (n+1)! + (n+2)! = n!*440. What is the sum of the digits of n?
A. 3
B. 8
C. 10
D. 11
E. 12
A. 3
B. 8
C. 10
D. 11
E. 12
GMAT Club Legend
Joined: 12 Sep 2015
Posts: 6803
Given Kudos: 799
Location: Canada
Re: If (n+1)! + (n+2)! = n!*440. What is the sum of the digits of n?
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15 May 2019, 05:50
15 May 2019, 05:50
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Bunuel wrote:
If (n+1)! + (n+2)! = (n!)(440). What is the sum of the digits of n?
A. 3
B. 8
C. 10
D. 11
E. 12
A. 3
B. 8
C. 10
D. 11
E. 12
Key concepts:
(n+1)! = (n+1)(n!)
(n+2)! = (n+2)(n+1)(n!)
So, we can rewrite the original equation as: (n+1)(n!)! + (n+2)(n+1)(n!) = (n!)(440)
Divide both sides by n! to get: (n+1) + (n+2)(n+1) = 440
Expand left side to get: (n+1) + n² + 3n + 2 = 440
Simplify left side to get: n² + 4n + 3 = 440
Subtract 440 from both sides to get: n² + 4n - 437 = 0
Factor: (n + 23)(n - 19) = 0
So, EITHER n = -23 OR n = 19
Since we n cannot be negative in a factorial, n must equal 19
Sum of digits = 1 + 9 = 10
Answer: C
Cheers,
Brebnt
General Discussion
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WE:Education (Education)
Re: If (n+1)! + (n+2)! = n!*440. What is the sum of the digits of n?
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15 May 2019, 07:09
15 May 2019, 07:09
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Bunuel wrote:
If (n+1)! + (n+2)! = n!*440. What is the sum of the digits of n?
A. 3
B. 8
C. 10
D. 11
E. 12
A. 3
B. 8
C. 10
D. 11
E. 12
Need to know that:
(n+1)! = (n+1)(n!)
(n+2)! = (n+2)(n+1)(n!)
Given, (n+1)! + (n+2)! = n!*440
i.e. (n+1)! [1 + (n+2)] = n!*440
i.e. n!*(n+1) (n+3) = n!*440
i.e. (n+1) (n+3) = 440
i.e. Product of two consecutive Even number = 440
But 20*22 = 440
therefore n+1 = 20
i.e. n = 19
sum of the digits of n = 1+9=10
Answer: Option C
