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

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
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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
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(n+1)! + (n+2)! = n!*440
(n+1)*n!+n!*(n+1)*(n+2) = n!*440
simplify we get
(n+1)*(n+3)=440
n=19
sum of digits ; 1+9 ; 10
IMO C

Recall that (n + 1)! = (n + 1)(n!) and (n + 2)! = (n + 2)(n + 1)(n!), so we can factor out n! from the left-hand side of the equation:

n![(n+1) + (n+2)(n+1)] = n!*440

Canceling out the n! from both sides, we have:

n + 1 + n^2 + 3n + 2 = 440

n^2 + 4n - 437 = 0

(n -19)(n + 23) = 0

n = 19 or n = -23

Since n can’t be negative, n = 19, and thus the sum of the digits of n is 1 + 9 = 10.

Answer: C
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Hi Brent,

I was able to make my way till the highlighted part but then could not figure out the factorization piece. I remember leaving out a question in between because of the same issue during my first attempt. Any shortcuts on the same? Am I expected to memorize all multiplication tables from 1 to 20?

Warm Regards,
Pritishd
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