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16!^100 – 9!^100 or (4!^100 – 3!^100)(4!^100 + 3!^10
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29 Oct 2017, 04:10
29 Oct 2017, 04:10
2
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Question Stats:
38% (01:17) correct
61% (01:30) wrong based on 76 sessions
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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.
Re: 16!^100 – 9!^100 or (4!^100 – 3!^100)(4!^100 + 3!^10
[#permalink]
30 Oct 2017, 08:14
30 Oct 2017, 08:14
2
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.
Answer A
Answer A
Re: 16!^100 9!^100 or (4!^100 3!^100)(4!^100 + 3!^10
[#permalink]
28 Jan 2026, 21:22
28 Jan 2026, 21:22
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Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
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