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 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!)(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 \(17!^2-16!^2\), 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