is p = 2^37 ?
if yes,
D = 2^32 and 2^P = D^D
D^D = (2^32)^ 2^32
D^D = {(2^(2^5)}^ (2^32)
D^D = (2)^{(2^5)*(2^32)}
D^D = {2^(2^37)} = 2^P
Therefore, P = 2^37
Exponents of exponents
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Source: Beat The GMAT — Problem Solving |
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akshatsingh
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Magellan
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I don't think it is the answer.
IMO, this transition is not correct:
D^D = {(2^(2^5)}^ (2^32)
D^D = (2)^{(2^5)*(2^32)}
I think the answer is the following:
D^D
= 2^32^2^32
= 2^2^5^2^32
= 2^2^320
P = 2^320
IMO, this transition is not correct:
D^D = {(2^(2^5)}^ (2^32)
D^D = (2)^{(2^5)*(2^32)}
I think the answer is the following:
D^D
= 2^32^2^32
= 2^2^5^2^32
= 2^2^320
P = 2^320
-
akshatsingh
- Senior | Next Rank: 100 Posts
- Posts: 77
- Joined: Thu Apr 10, 2008 10:13 pm
- Thanked: 4 times
