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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
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Question Stats:
35% (01:28) correct
64% (01:18) wrong based on 31 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
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22 Jun 2023, 03:54
22 Jun 2023, 03:54
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