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

Question

A. 3
B. 8
C. 10
D. 11
E. 12

BrentGMATPrepNow · Accepted Answer

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

GMATinsight · Answer

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)!   = 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

Archit3110 · Answer

(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

ScottTargetTestPrep · Answer

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!*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

Pritishd · Answer

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

anirudhjay · Answer

It took over 5 mins with the hit-and-trial method to get this question right. Factorizing 437 as 23 x 19 would never occur to me in an instant. Is the high scoring GMAT test taker really able to quickly factorize 437 or am I missing something in my preparation? The question stats indicate 69% got it right in 2:38 on average.

remtempora · Answer

I also struggle with factorizing bigger numbers such as 437. What works best for me is to just use the pq-formula. In this case it gives 441 as the root in the formula, which is easy to convert into 21^2.

earumsit · Answer

I always think of it in a more simple way, there is 4n so basically divide 437 by 4 and you see there's a 19 with a remainder of 1, leave the remainder and try dividing 437 by 19 and you see that you will get 23, then just simply see if 23 - 19 = 4, if it does then you're all good.

AndyRugger · Answer

I could never factorize 437 in my head!

However, the answer gives you an easier way out than setting a quadratic expression equal to zero.

Once you collect terms, you get to (n+3)(n+1) = 440.

Break down 440 into factor pairs to find a factor pair with a difference of two between them.

440 = 10*44 = 20*22 - done!

Recognizing that 20= n+1, therefore n=19. Sum of 1+9 = 10.

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

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GMAT Problem Solving (PS) Questions

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

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