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# ((n-1)!+n!+(n+1)!)/n!

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Intern
Joined: 06 Nov 2013
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Updated on: 18 Feb 2016, 15:26
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55% (hard)

Question Stats:

64% (01:33) correct 36% (01:54) wrong based on 169 sessions

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$$\frac{(n-1)!+n!+(n+1)!}{n!}$$

Which of the following values of n will result between 16 and 17 for the equation?

A. 14
B. 15
C. 16
D. 17
E. 18

If I factor (n-1) and reduce the equation and then substitute the answer choices then result becomes a. But is there a shortcut to solving this.

Originally posted by jahanafsana on 08 Mar 2015, 18:25.
Last edited by ENGRTOMBA2018 on 18 Feb 2016, 15:26, edited 1 time in total.
Reformatted the question and added the OA.
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Joined: 02 Aug 2009
Posts: 6561

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08 Mar 2015, 18:41
jahanafsana wrote:
(n-1)!+n!+(n+1)!
n!

Which of the following values of n will result between 16 and 17 for the equation?

If I factor (n-1) and reduce the equation and then substitute the answer choices then result becomes a. But is there a shortcut to solving this.

a. 14
b. 15
c. 16
d. 17
e. 18

hi jahanafsana,
it depends how you have reduced the equation...
1) first is you open each term .... find common terms... get an equation and then substitute values..
2) second , which may be slightly less time consuming.....
$$\frac{((n-1)!+n!+(n+1)!)}{n!}$$
=$$\frac{(n-1)!}{n!}$$+$$\frac{n!}{n!}$$+$$\frac{(n+1)!}{n!}$$
=$$\frac{1}{n}$$+1+n+1.... now you can easily see n+2 should be greater than or equal to 16.... 14 is the value...
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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
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3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html

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09 Mar 2015, 23:50
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jahanafsana wrote:
(n-1)!+n!+(n+1)!
n!

Which of the following values of n will result between 16 and 17 for the equation?

If I factor (n-1) and reduce the equation and then substitute the answer choices then result becomes a. But is there a shortcut to solving this.

a. 14
b. 15
c. 16
d. 17
e. 18

$$\frac{(n - 1)! + n! + (n + 1)!}{n!} = \frac{1}{n} + 1 + (n+1)$$

Note that out of the options, whatever value you give n, 1/n will be a very small term, much smaller than 1.
In rest of the expression, you have 1 + (n + 1) which you want to be 16. So n must be 14.

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11 Mar 2015, 23:58
Easiest way is simplify the equation and then check the answer choice.
(n-1)! +n! + (n+1)! /n! = (I/n) +1+ (n+1)
(n+2) + (I/n) is 16 or more than that so only option is 14.
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12 Mar 2015, 09:44
CrackVerbalGMAT wrote:
Easiest way is simplify the equation and then check the answer choice.
(n-1)! +n! + (n+1)! /n! = (I/n) +1+ (n+1)
(n+2) + (I/n) is 16 or more than that so only option is 14.

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12 Mar 2015, 09:46
CrackVerbalGMAT wrote:
Easiest way is simplify the equation and then check the answer choice.
(n-1)! +n! + (n+1)! /n! = (I/n) +1+ (n+1)
(n+2) + (I/n) is 16 or more than that so only option is 14.

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18 Feb 2016, 13:51
VeritasPrepKarishma wrote:
jahanafsana wrote:
(n-1)!+n!+(n+1)!
n!

Which of the following values of n will result between 16 and 17 for the equation?

If I factor (n-1) and reduce the equation and then substitute the answer choices then result becomes a. But is there a shortcut to solving this.

a. 14
b. 15
c. 16
d. 17
e. 18

$$\frac{(n - 1)! + n! + (n + 1)!}{n!} = \frac{1}{n} + 1 + (n+1)$$

Note that out of the options, whatever value you give n, 1/n will be a very small term, much smaller than 1.
In rest of the expression, you have 1 + (n + 1) which you want to be 16. So n must be 14.

Could you show how you simplified this part in steps? I seem to get a different answer.
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18 Feb 2016, 15:20
saiesta wrote:
VeritasPrepKarishma wrote:
jahanafsana wrote:
(n-1)!+n!+(n+1)!
n!

Which of the following values of n will result between 16 and 17 for the equation?

If I factor (n-1) and reduce the equation and then substitute the answer choices then result becomes a. But is there a shortcut to solving this.

a. 14
b. 15
c. 16
d. 17
e. 18

$$\frac{(n - 1)! + n! + (n + 1)!}{n!} = \frac{1}{n} + 1 + (n+1)$$

Note that out of the options, whatever value you give n, 1/n will be a very small term, much smaller than 1.
In rest of the expression, you have 1 + (n + 1) which you want to be 16. So n must be 14.

Could you show how you simplified this part in steps? I seem to get a different answer.

You can adopt 2 methods to verify: either substitute n=3 and cross check. For algebraic method look below:

Realize that (n+1)! = (n+1)*n*(n-1)(n-2)... = (n+1)*n!

Similarly, n!=n*(n-1)! ---> (n-1)!= n!/n

Thus, $$\frac{(n - 1)! + n! + (n + 1)!}{n!} =\frac{(n!/n)+n!+(n+1)n!}{n!}$$ = (cancelling n! from both the numerator and denominator you get) $$\frac{1}{n}+1+(n+1)$$

Hope this helps.
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18 Feb 2016, 20:46
saiesta wrote:
VeritasPrepKarishma wrote:
jahanafsana wrote:
(n-1)!+n!+(n+1)!
n!

Which of the following values of n will result between 16 and 17 for the equation?

If I factor (n-1) and reduce the equation and then substitute the answer choices then result becomes a. But is there a shortcut to solving this.

a. 14
b. 15
c. 16
d. 17
e. 18

$$\frac{(n - 1)! + n! + (n + 1)!}{n!} = \frac{1}{n} + 1 + (n+1)$$

Note that out of the options, whatever value you give n, 1/n will be a very small term, much smaller than 1.
In rest of the expression, you have 1 + (n + 1) which you want to be 16. So n must be 14.

Could you show how you simplified this part in steps? I seem to get a different answer.

$$\frac{((n-1)!+n!+(n+1)!)}{n!}$$
=$$\frac{(n-1)!}{n!}$$+$$\frac{n!}{n!}$$+$$\frac{(n+1)!}{n!}$$
=$$\frac{1}{n}$$+1+n+1....
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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html

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18 Feb 2016, 20:59
2
saiesta wrote:
VeritasPrepKarishma wrote:
jahanafsana wrote:
(n-1)!+n!+(n+1)!
n!

Which of the following values of n will result between 16 and 17 for the equation?

If I factor (n-1) and reduce the equation and then substitute the answer choices then result becomes a. But is there a shortcut to solving this.

a. 14
b. 15
c. 16
d. 17
e. 18

$$\frac{(n - 1)! + n! + (n + 1)!}{n!} = \frac{1}{n} + 1 + (n+1)$$

Note that out of the options, whatever value you give n, 1/n will be a very small term, much smaller than 1.
In rest of the expression, you have 1 + (n + 1) which you want to be 16. So n must be 14.

Could you show how you simplified this part in steps? I seem to get a different answer.

Note that
$$(n-1)! = 1*2*3*...*(n-1)$$
$$n! = 1*2*3*...*(n-1)*n$$

So n! just has an extra n in it.

When you divide (n-1)! by n!, you are just left with an n in the denominator.

Similarly,
$$n! = 1*2*3*...*(n-1)*n$$
$$(n+1)! = 1*2*3*...*(n-1)*n*(n+1)$$
So (n+1)! only has an extra (n+1).

When you divide (n+1)! by n!, you are left with (n+1) in the denominator.

$$\frac{(n - 1)! + n! + (n + 1)!}{n!}$$

$$= \frac{(n - 1)!}{n!} + \frac{n!}{n!} + \frac{(n+1)!}{n!}$$

$$= \frac{1}{n} + 1 + (n+1)$$
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19 Feb 2016, 13:01
VeritasPrepKarishma Thank you. Your explanation has been quite useful.
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11 May 2017, 02:49
jahanafsana wrote:
$$\frac{(n-1)!+n!+(n+1)!}{n!}$$

Which of the following values of n will result between 16 and 17 for the equation?

A. 14
B. 15
C. 16
D. 17
E. 18

If I factor (n-1) and reduce the equation and then substitute the answer choices then result becomes a. But is there a shortcut to solving this.

$$\frac{(n-1)!+n!+(n+1)!}{n!}$$
=1/n+1+n+1
=n+2+(1/n)
16<n+2+1/n<17
->14<n+1/n<15
n=14

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