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If ¡n! = (n!)^2, then ¡17!  ¡16! = [#permalink]
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Updated on: 04 Feb 2014, 05:13
Question Stats:
60% (01:16) correct 40% (01:02) wrong based on 477 sessions
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If ¡n! = (n!)^2, then ¡17!  ¡16! = A. ¡1! B. (¡16!)(¡4!)(2) C. (¡16!)(12)(2) D. 172 E. (¡16!)(12^2)(2) This was actually the most fun I've had solving a gmat problem The problem defines a madeup symbol; this type of problem is essentially asking you to follow directions. The weird symbol means a sort of “double factorial”: take the factorial and then take it again! For example, ¡5! would be (5)(4)(3)(2)(1)(5)(4)(3)(2)(1).
¡17! and ¡16! would work the same way as ¡5!, but nobody would want to multiply those out without a calculator. Is there a shortcut? Glance at the form of the answers—most still use the weird upsidedownexclamation symbol. There must be a way, then, to rewrite ¡17! and ¡16! in another form.
There is! The two numbers are close to identical, but ¡17! contains two additional factors: 17 and 17. Factor out a ¡16! from the two terms: ¡17!  ¡16! = ¡16! (17×17 – 1)
Glance at the answers. ¡16! is in three of them. Do any match? First, simplify the (17×17 – 1) part. 172 = 289, and 289 – 1 = 288. Find something that matches ¡16! (288) (B) (¡16!)(¡4!)(2) (C) (¡16!)(12)(2) (E) (¡16!)(122)(2) Answer (C) is too small. Answer (E) is easier to process than (B); start with (E). 122 = 144. (144)(2) = 288. Bingo!
The correct answer is (E).
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Originally posted by AccipiterQ on 29 Oct 2013, 12:26.
Last edited by Bunuel on 04 Feb 2014, 05:13, edited 1 time in total.
Edited the question.



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Re: If ¡n! = (n!)2, then ¡17!  ¡16! = [#permalink]
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Updated on: 29 Oct 2013, 12:35
I think 99% of people will interpret ¡n! = (n!)2 as (n!)*2.
Just write it as n!! or (n!)!.
Would give negative kudos if it were possible.
*edit* OP fixed mistake previous mistake, problem is written correctly now.
removing myself from this question; clearly something weird's going on from the OP.
Originally posted by Blax0r on 29 Oct 2013, 12:33.
Last edited by Blax0r on 29 Oct 2013, 12:35, edited 2 times in total.



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Re: If ¡n! = (n!)2, then ¡17!  ¡16! = [#permalink]
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29 Oct 2013, 12:33
AccipiterQ wrote: If ¡n! =\((n!)^2\), then ¡17!  ¡16! = A) ¡1! B) (¡16!)(¡4!)(2) C) (¡16!)(12)(2) D) 172 E) (¡16!)(12^2)(2) This was actually the most fun I've had solving a gmat problem The problem defines a madeup symbol; this type of problem is essentially asking you to follow directions. The weird symbol means a sort of “double factorial”: take the factorial and then take it again! For example, ¡5! would be (5)(4)(3)(2)(1)(5)(4)(3)(2)(1).
¡17! and ¡16! would work the same way as ¡5!, but nobody would want to multiply those out without a calculator. Is there a shortcut? Glance at the form of the answers—most still use the weird upsidedownexclamation symbol. There must be a way, then, to rewrite ¡17! and ¡16! in another form.
There is! The two numbers are close to identical, but ¡17! contains two additional factors: 17 and 17. Factor out a ¡16! from the two terms: ¡17!  ¡16! = ¡16! (17×17 – 1)
Glance at the answers. ¡16! is in three of them. Do any match? First, simplify the (17×17 – 1) part. 172 = 289, and 289 – 1 = 288. Find something that matches ¡16! (288) (B) (¡16!)(¡4!)(2) (C) (¡16!)(12)(2) (E) (¡16!)(12^2)(2) Answer (C) is too small. Answer (E) is easier to process than (B); start with (E). 122 = 144. (144)(2) = 288. Bingo!
The correct answer is (E). So you have basically \(17!^216!^2\), you can rewrite that as \(16!^2\)*(\(17^2\)1) = \(16!^2\)*(2891) = \(16!^2\)*(288) now look at the answers, and what the stem tells you; you're told that ¡n!=n!^2, so \(16!^2\) = ¡16!, and 288 is just 12^2 [144]*2, so \(16!^2\)*(288) = ¡16!(12^2)(2) E



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Re: If ¡n! = (n!)^2, then ¡17!  ¡16! = [#permalink]
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06 Mar 2014, 19:20
Bunuel, can you please tell if factorial questions are asked in GMAT? Thanks!!
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Re: If ¡n! = (n!)^2, then ¡17!  ¡16! = [#permalink]
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07 Mar 2014, 00:52



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Re: If ¡n! = (n!)^2, then ¡17!  ¡16! = [#permalink]
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30 Dec 2015, 20:44
J17! = 17! * 17! THIS can be rewritten as: 16! * 17 * 16! * 17.
now, 16! * 16! is j16!.
we can factor out j16! we are left with j16! (17*17  1) = j16! * 288 288=144 *2 or 12^2 * 2. answer E.



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Re: If ¡n! = (n!)^2, then ¡17!  ¡16! = [#permalink]
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10 Nov 2016, 20:14
Can someone tell me why you subtract the 1 when you're factoring out 16!



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Re: If ¡n! = (n!)^2, then ¡17!  ¡16! = [#permalink]
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12 Nov 2017, 13:37
Can someone explain me why D is not correct, I can't get it: the only difference in (17!)^2 and (16!)^2 isn't just the 17^2??



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Re: If ¡n! = (n!)^2, then ¡17!  ¡16! = [#permalink]
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12 Nov 2017, 21:21




Re: If ¡n! = (n!)^2, then ¡17!  ¡16! =
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12 Nov 2017, 21:21






