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# 16!^100 – 9!^100 or (4!^100 – 3!^100)(4!^100 + 3!^10

Author Message
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Senior Manager
Joined: 20 May 2014
Posts: 282
Followers: 18

Kudos [?]: 50 [0], given: 220

16!^100 – 9!^100 or (4!^100 – 3!^100)(4!^100 + 3!^10 [#permalink]  29 Oct 2017, 04:10
00:00

Question Stats:

12% (01:12) correct 87% (01:11) wrong based on 8 sessions
 Quantity A Quantity B 16!^100 – 9!^100 (4!^100 – 3!^100)(4!^100 + 3!^100)

A. The quantity in Column A is greater
B. The quantity in Column B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given

Kudos for correct solution.
[Reveal] Spoiler: OA
Director
Joined: 03 Sep 2017
Posts: 521
Followers: 1

Kudos [?]: 343 [1] , given: 66

Re: 16!^100 – 9!^100 or (4!^100 – 3!^100)(4!^100 + 3!^10 [#permalink]  30 Oct 2017, 08:14
1
KUDOS
Using the formula of difference of squares, quantity B becomes $$(4!^{100})^2-(3!^{100})^2 = 4!^{200}-3!^{200}$$. Then, comparing the two quantities, it is easy to notice that A is larger.

Re: 16!^100 – 9!^100 or (4!^100 – 3!^100)(4!^100 + 3!^10   [#permalink] 30 Oct 2017, 08:14
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