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# 2^10= 1024. Which is the largest of the following numbers?

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2^10= 1024. Which is the largest of the following numbers?  [#permalink]  12 Jul 2006, 15:42
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2^10= 1024. Which is the largest of the following numbers?

(A) 10^16
(B) 16^14
(C) 40^8
(D) 80^7
(E) 9000^4
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Joined: 04 Jul 2006
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Kudos [?]: 2 [0], given: 0

a) 10^16 = (10^3)^(16/3) <(2^10)^16/3=2^(160/3)
b) 16^14= (2^4)^14=2^56=2^(168/3)
c) 40^8=10^8*4^8=(10^3)^(8/3)*2^16=(2^10)^(8/3)*2^16<2^56
d) 80^7=(2^3)^7*10^7<2^21*(2^10)^(7/3)<2^56
e) 9000^4 = (10^3)^4 * 9^4 < 2^40 * 2^14 [9^4=6561, 2^14=16382) < 2^56

=> b
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Schools: Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008
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B

(A) 10^16 = 10 * 10^15
(B) 16^14 = 2^56 = (2^10)^5 * 2^6 = 64 * 10^15 approx
(C) 40^8 = 10^8 * 4^8 = 10^8 * 2^10 * 2^6 = 64 * 10^11 approx
(D) 80^7 = 10^7 * 8^7 = 10^7 * 2^21 = = 2 * 10^13 approx
(E) 9000^4 = 1000^4 * 9^4 = 6561 * 10^12 = 65 * 10^15

Among B an E , I would choose B because we have considered 2^10 as 1000 for approximation in B. E is accurate.
_________________

SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

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