2^n=n^2

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Oxana: Junior | Next Rank: 30 Posts; Posts: 18; Joined: Sun Jul 27, 2008 8:59 am; Thanked: 1 times

by Oxana » Tue Jul 29, 2008 6:54 pm

For how many integers N is 2^N=N^2

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Source: Beat The GMAT — Problem Solving | Tue Jul 29, 2008 6:54 pm

oldgeezer: Junior | Next Rank: 30 Posts; Posts: 12; Joined: Fri Jun 13, 2008 11:11 am; Thanked: 1 times

by oldgeezer » Tue Jul 29, 2008 7:30 pm

for 2? (2 and 4)

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parallel_chase: Legendary Member; Posts: 1153; Joined: Wed Jun 20, 2007 6:21 amFollowed by: 2 members; Thanked: 146 times

by parallel_chase » Wed Jul 30, 2008 1:27 am

I'd agree with oldgeezer, only 2 and 4 are the possible values of N in such expression.

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acmeheadshot: Newbie | Next Rank: 10 Posts; Posts: 1; Joined: Tue Apr 22, 2008 8:46 am

by acmeheadshot » Thu Jul 31, 2008 6:24 am

n 2^n n^2
0 1 0
1 2 1
2 4 4
3 8 9
4 16 16
5 32 25
6 64 36
.
.
.

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VP_RedSoxFan: GMAT Instructor; Posts: 85; Joined: Thu May 01, 2008 12:56 pm; Location: Salt Lake City, UT; Thanked: 24 times; GMAT Score:750+

by VP_RedSoxFan » Thu Jul 31, 2008 7:15 am

I like the approach by acmeheadshot. If you run a quick chart of possible N and then the solutions to 2^N and N^2 you can

a) see that there are 2 (N=2,4) solutions for N,

b) that as N gets bigger 2^N and N^2 move farther apart so we can be reasonable sure there aren't more as N gets bigger, and

c) (this is not shown in acmeheadshot's post but I'm sure he/she thought about it anyway) if we throw a few negative integers for N, because the problem doesn't preclude them, we can know that no negative integers are solutions for N

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