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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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73% (01:28) correct 27% (01:28) wrong based on 49 sessions
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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, 01:43
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
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 substitutionbased, 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 = (n1)! 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. (C) is our answer.
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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)! = (n1)!*n*(n+1)! ...( we can expand using the definition of n!) n*(n+2)* (n+1)! = (n1)!* n* (n+1)! ...( Cancelling the common terms in LHS & RHS) (n+2) = (n1)! When we put values starting n=1,2,3,4... We find this equation holds true for n=4. (4 + 2) = 6 = ( 41 )! Hence C. Best, Gladi



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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 * (n1)! * (n+1)! => (n+2) = (n1)! ........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) = (n1)! Plug in the value, only 4 fit: (4+2) = (41)! = 6 => Answer (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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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 reexpress (n + 2)! as (n + 2)*(n + 1)! Similarly, we can reexpress n! as (n)*(n  1)! Using those reexpressions, 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 Answer: 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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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 = (n1)! lets start checking values (A) 2 4 = 1....... Incorrect (B) 3 5 = 2!............Incorrect (C) 4 6 = 3! ............... Correct Answer C
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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*(n1)!, 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. Answer: C
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If n*(n + 2)! = n!*(n + 1)!, then n could be which of the following
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