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# If n*(n + 2)! = n!*(n + 1)!, then n could be which of the following

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If n*(n + 2)! = n!*(n + 1)!, then n could be which of the following  [#permalink]

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20 Mar 2018, 01:04
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If $$n*(n + 2)! = n!*(n + 1)!$$, then n could be which of the following values?

(A) 2
(B) 3
(C) 4
(D) 5
(E) 6

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Posts: 586
Re: If n*(n + 2)! = n!*(n + 1)!, then n could be which of the following  [#permalink]

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20 Mar 2018, 01:43
Bunuel wrote:
If n*(n + 2)! = n!*(n + 1)!, then n could be which of the following values?

(A) 2
(B) 3
(C) 4
(D) 5
(E) 6

By substituting the values, we can get C

4 * 6! = 4! * 5!

4 * 6 * 5! = 4! * 5!

Eliminate 5! on both sides.

24 = 24.

Hence C.
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Re: If n*(n + 2)! = n!*(n + 1)!, then n could be which of the following  [#permalink]

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20 Mar 2018, 01:52
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Bunuel wrote:
If n*(n + 2)! = n!*(n + 1)!, then n could be which of the following values?

(A) 2
(B) 3
(C) 4
(D) 5
(E) 6

In addition to the substitution-based, Alternative approach shown above, we'll also show the complete, Precise calculation.

Since (n+2)! = (n+2)(n+1)!, we'll cancel out (n+1)! on both sides to get n(n+2) = n!
Canceling out n then gives n+2 = (n-1)!
Then for n = 1 this is 3 = 1 which is false.
For n = 2 this is 4 = 1 which is false.
For n = 3 this is 5 = 2 which is false.
For n = 4 this is 6 = 6 which is true.

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Re: If n*(n + 2)! = n!*(n + 1)!, then n could be which of the following  [#permalink]

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20 Mar 2018, 02:05
Bunuel wrote:
If n*(n + 2)! = n!*(n + 1)!, then n could be which of the following values?

(A) 2
(B) 3
(C) 4
(D) 5
(E) 6

Instead of putting in values into the complicated equation provided, we can first simplify a bit.

n*(n + 2)! = n!*(n + 1)!
n*(n+2)*(n+1)! = (n-1)!*n*(n+1)! ...( we can expand using the definition of n!)
n*(n+2)*(n+1)! = (n-1)!*n*(n+1)! ...( Cancelling the common terms in LHS & RHS)
(n+2) = (n-1)!

When we put values starting n=1,2,3,4...

We find this equation holds true for n=4.

(4 + 2) = 6 = ( 4-1 )!

Hence C.

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Re: If n*(n + 2)! = n!*(n + 1)!, then n could be which of the following  [#permalink]

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20 Mar 2018, 06:30
Bunuel wrote:
If n*(n + 2)! = n!*(n + 1)!, then n could be which of the following values?

(A) 2
(B) 3
(C) 4
(D) 5
(E) 6

n * (n+2) * (n+1)! = n * (n-1)! * (n+1)!

=> (n+2) = (n-1)! ........Eq (A)

Eq (A) satisfied by Option (C): 4
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Re: If n*(n + 2)! = n!*(n + 1)!, then n could be which of the following  [#permalink]

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20 Mar 2018, 10:15
Bunuel wrote:
If n*(n + 2)! = n!*(n + 1)!, then n could be which of the following values?

(A) 2
(B) 3
(C) 4
(D) 5
(E) 6

n*(n + 2)! = n!*(n+1)!, or (n+2) = (n-1)!

Plug in the value, only 4 fit: (4+2) = (4-1)! = 6

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Re: If n*(n + 2)! = n!*(n + 1)!, then n could be which of the following  [#permalink]

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21 Mar 2018, 16:15
Bunuel wrote:
If n*(n + 2)! = n!*(n + 1)!, then n could be which of the following values?

(A) 2
(B) 3
(C) 4
(D) 5
(E) 6

We can re-express (n + 2)! as (n + 2)*(n + 1)! Similarly, we can re-express n! as (n)*(n - 1)!

Using those re-expressions, we can simplify the equation as:
n*(n + 2)! = n!*(n + 1)!

n*(n + 2)(n + 1)! = n*(n - 1)!*(n + 1)!

Dividing both sides by n*(n + 1)!, we have:

n + 2 = (n - 1)!

We can see that, in general, (n - 1)! will be much larger than (n + 2). Therefore, the equation can only be true for a small value of n. Let’s consider the answer choices now.

(A) 2

2 + 2 = (2 - 1)! ?

4 = 1! ?

4 = 1 ? → No

(B) 3

3 + 2 = (3 - 1)! ?

5 = 2! ?

5 = 2 ? → No

(C) 4

4 + 2 = (4 - 1)! ?

6 = 3! ?

6 = 6 ? → Yes

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Re: If n*(n + 2)! = n!*(n + 1)!, then n could be which of the following  [#permalink]

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09 Apr 2018, 21:57
Bunuel wrote:
If n*(n + 2)! = n!*(n + 1)!, then n could be which of the following values?

(A) 2
(B) 3
(C) 4
(D) 5
(E) 6

n*(n + 2)! = n!*(n + 1)!
(n + 2)! /(n + 1)! = n!/n
n+2 = (n-1)!

lets start checking values
(A) 2
4 = 1....... Incorrect
(B) 3
5 = 2!............Incorrect
(C) 4
6 = 3! ............... Correct

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If n*(n + 2)! = n!*(n + 1)!, then n could be which of the following  [#permalink]

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10 Apr 2018, 00:01

Solution

Given:
• We are given that n*(n+2)! = n! * (n+1)!

To find:
• We need to find the option that satisfies n*(n+2)! = n! * (n+1)!

Approach and Working:

By applying n! = n*(n-1)!, n*(n+2)! = n! * (n+1)! can be written as n*(n+2) *(n+1)! = n! * (n+1)!
• By cancelling (n+1)! from both the sides, we get n*(n+2) = n!

• For n=2, 2*4≠2!
• For n=3, 3*5 ≠3!
• For n=4, 4*6= 4!=24, hence, the value of n is 4.

Hence, the correct answer is option C.
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If n*(n + 2)! = n!*(n + 1)!, then n could be which of the following   [#permalink] 10 Apr 2018, 00:01
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