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# For how many integers n is 2^n = n^2 ?

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For how many integers n is 2^n = n^2 ?  [#permalink]

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29 Sep 2010, 06:13
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65% (01:23) correct 35% (01:10) wrong based on 671 sessions

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For how many integers n is 2^n = n^2 ?

A. None
B. One
C. Two
D. Three
E. More than Three
Math Expert
Joined: 02 Sep 2009
Posts: 60484

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29 Sep 2010, 06:21
8
12
For how many integers n is 2^n = n^2 ?
A. None
B. One
C. Two
D. Three
E. More than Three

$$2^n= n^2$$ is true for 2 integers:
$$n=2$$ --> $$2^2=2^2=4$$;
$$n=4$$ --> $$4^2=2^4=16$$.

Well, $$2^2=2^2=4$$ is obvious choices, then after trial and error you'll get $$4^2=2^4=16$$ as well. But how do we know that there are no more such numbers? You can notice that when $$n$$ is more than 4 then $$2^n$$ is always more than $$n^2$$ so $$n$$ cannot be more than 4. $$n$$ cannot be negative either as in this case $$2^n$$ won't be an integer whereas $$n^2$$ will be.

NOTE: I think it's worth remembering that $$4^2=16=2^4$$, I've seen several GMAT questions on number properties using this (another useful property $$8^2=4^3=2^6=64$$).

Hope it helps.
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Re: For how many integers n is 2^n = n^2?  [#permalink]

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08 Apr 2011, 20:14
1
1
=> 2^n = n^2
Taking nth root on both sides
=> 2 = (n^2)^1/n
=> 2 = n ^ 2/n
Lets consider positive even multiples of 2 for n (since LHS = 2)
For n = 2
=> 2 = 2 ^ 2/2 - First value that satisfier

For n = 4
=> 2 = 4 ^ 2/4 - Second value that satisfier

For n = 8
=> 2 = 8 ^ 2/8 - Doesnt satisfy

For n = 16
=> 2 = 16 ^ 2/16 - Doesnt satisfy

Two values. Ans = C
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29 Sep 2010, 08:03
vanidhar wrote:
for how many integers n is 2^n= n^2 ?
0
1
2
3
>3

It helps to know that the function 2^x is more expansive than x^2 for large positive x and converges quickly to 0 for negative x.

So we know we only have to check small values of x. For positive x, it is easy to see this is true for x=2,4 and then the function 2^x explodes

For negative x, 2^-1 is less than -1^2 already so no negative integers can satisfy the equality

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Re: For how many integers n is 2^n = n^2 ?  [#permalink]

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13 Nov 2016, 22:14
This question might very well baffle us under the exam stress . So what is the methodology.
We can assume that the number wont be very large as - the bigger the numbers will get the difference between the two algebraic expression will increase.
we realise 2^0 is not equal to 0^2.
The continue with 1, 2, 3 4, 5, 6, 7, 8 by then you will get n=2, 4 suits the criterion, others don't and any bigger number will go super off limit.

trust me it took my 59 sec to do it using this long method.
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Re: For how many integers n is 2^n = n^2 ?  [#permalink]

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13 Nov 2016, 23:17
vanidhar wrote:
For how many integers n is 2^n = n^2 ?

A. None
B. One
C. Two
D. Three
E. More than Three

$$n^2 = 2^n$$

So, $$n$$ = $$2^\frac{n}{2}$$

Now, n must be Even and a multiple of 2

Plug in the even values only 2 and 4 remains...

$$2$$ = $$2^\frac{2}{2}$$

$$4$$ = $$2^\frac{4}{2}$$

Hence, answer will be (C) 2

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Re: For how many integers n is 2^n = n^2 ?  [#permalink]

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12 Jun 2018, 15:11
why is the answer 2 and not 3? shouldn't zero be counted too?
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Re: For how many integers n is 2^n = n^2 ?  [#permalink]

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12 Jun 2018, 21:03
spatel2 wrote:
why is the answer 2 and not 3? shouldn't zero be counted too?

If n = 0:

2^n = 2^0 = 1 (recall that any nonzero number to the power of 0, is 1).
n^2 = 0^2 = 0.
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Re: For how many integers n is 2^n = n^2 ?  [#permalink]

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25 Nov 2019, 19:18
Hello chetan2u

Is there another approach to solving this question? I do not want to rely on hit and trial method.
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For how many integers n is 2^n = n^2 ?  [#permalink]

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25 Nov 2019, 19:56
vanidhar wrote:
For how many integers n is 2^n = n^2 ?

A. None
B. One
C. Two
D. Three
E. More than Three

akash7gupta11

$$2^n = n^2$$
What all does this tell you
(a) n cannot be negative as LHS 2^n will become fraction, while n^2 will remain integer.
(b) Since we have LHS as power of 2, the RHS or n will also be in terms of 2

So, let $$n=2^x$$....
$$2^{2^x}=(2^x)^2=2^{2x}.......2^x=2x.....2^{x-1}=x$$
Now moment x>2, the equality fails, so x=1 and 2

C
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For how many integers n is 2^n = n^2 ?   [#permalink] 25 Nov 2019, 19:56
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