The answer is [C], Two.
Obviously n=2 is a valid choice since then 2^2=2^2.
The other solution is n=4. Then 4^2 = (2^2)^2 = 2^4
Any other way to express n as an exponent of 2 won't equal 2 to the nth power.
Exponent ?s
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VP_RedSoxFan
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Ryan S.
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lunarpower
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yeah, this problem is somewhat unappetizing, as trial and error is the only way to solve it.imadummy wrote:From a Veritas Prep workbook:
For how many integers, n, does 2^n = n^2
A. Zero
B. One
C. Two
D. Four
E. Eight
this problem strikes me as something unlikely to show up on the gmat, because of its combination of the following 2 factors:
1, soluble only by trial and error;
2, technically an infinity of trials to do (although it becomes perfectly clear after trying x = 3, 4, 5 that no bigger numbers will work).
1) this is not to say that the gmat will never feature a trial-and-error problem, because of course it often does. however, if there is a problem based on trial and error, then it will almost certainly feature a finite set of 'trials', usually based on some sort of combinatorial issue. here is an example from another forum (i don't normally like to cross-post, but i don't recall any equally juicy examples i've posted here).
2) and of course there are problems that would feature an infinity of trials were they to be solved by trial and error - but these problems are usually soluble with some sort of theory / number properties / something else that obviates the need for trial-and-error in the first place.
Last edited by lunarpower on Sat May 24, 2008 2:51 am, edited 1 time in total.
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lunarpower
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incidentally, it is possible to use number properties to figure out that 'n' has to be a power of 2.
here's how:
consider prime factorizations.
obviously, the prime factorization of 2^n contains only 2's, and no other primes.
but because 2^n and n^2 are equal, this means that the prime factorization of n^2 also contains nothing but 2's, which means that n itself must be a power of 2.
therefore, the only integers you have to try are 1, 2, 4, 8, ...
which makes your life a lot easier.
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this insight makes the problem seem a bit more gmat-like.
here's how:
consider prime factorizations.
obviously, the prime factorization of 2^n contains only 2's, and no other primes.
but because 2^n and n^2 are equal, this means that the prime factorization of n^2 also contains nothing but 2's, which means that n itself must be a power of 2.
therefore, the only integers you have to try are 1, 2, 4, 8, ...
which makes your life a lot easier.
--
this insight makes the problem seem a bit more gmat-like.
Ron has been teaching various standardized tests for 20 years.
--
Pueden hacerle preguntas a Ron en castellano
Potete chiedere domande a Ron in italiano
On peut poser des questions à Ron en français
Voit esittää kysymyksiä Ron:lle myös suomeksi
--
Quand on se sent bien dans un vêtement, tout peut arriver. Un bon vêtement, c'est un passeport pour le bonheur.
Yves Saint-Laurent
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Learn more about ron
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Pueden hacerle preguntas a Ron en castellano
Potete chiedere domande a Ron in italiano
On peut poser des questions à Ron en français
Voit esittää kysymyksiä Ron:lle myös suomeksi
--
Quand on se sent bien dans un vêtement, tout peut arriver. Un bon vêtement, c'est un passeport pour le bonheur.
Yves Saint-Laurent
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Learn more about ron
