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AccipiterQ
GMAT 2: 730 Q49 V41
Posts: 144
Updated on: 07 Dec 2019, 04:21
Originally posted by AccipiterQ on 29 Oct 2013, 12:26.
Last edited by Bunuel on 07 Dec 2019, 04:21, edited 2 times in total.
Last edited by Bunuel on 07 Dec 2019, 04:21, edited 2 times in total.
Edited the question.
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
73% (02:23) correct
27%
(02:29)
wrong
based on 1300
sessions
History
Date
Time
Result
Not Attempted Yet
If ¡n! = (n!)^2, then ¡17! - ¡16! =
A. ¡1!
B. (¡16!)(¡4!)(2)
C. (¡16!)(12)(2)
D. 172
E. (¡16!)(12^2)(2)
A. ¡1!
B. (¡16!)(¡4!)(2)
C. (¡16!)(12)(2)
D. 172
E. (¡16!)(12^2)(2)
The problem defines a made-up 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 upside-down-exclamation 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).
¡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 upside-down-exclamation 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).
AccipiterQ
GMAT 2: 730 Q49 V41
Posts: 144
29 Oct 2013, 12:33
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AccipiterQ
So you have basically \(17!^2-16!^2\), you can rewrite that as \(16!^2\)*(\(17^2\)-1)
= \(16!^2\)*(289-1)
= \(16!^2\)*(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
General Discussion
Blax0r
Joined: 04 Sep 2012
Last visit: 28 Apr 2018
Posts: 138
Given Kudos: 17
Location: United States
Schools: Booth '16 (D) LBS '16 (D) Tuck '16 (D) CBS '16 (D) Kellogg '16 (D) Stern '16 (WD) Sloan '16 (D) Haas '16 (D)
GMAT 1: 760 Q50 V42
GPA: 3.3
Schools: Booth '16 (D) LBS '16 (D) Tuck '16 (D) CBS '16 (D) Kellogg '16 (D) Stern '16 (WD) Sloan '16 (D) Haas '16 (D)
GMAT 1: 760 Q50 V42
Posts: 138
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.
