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# For how many pair(s) of positive integers m and n, such that

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Math Expert
Joined: 02 Sep 2009
Posts: 56304
For how many pair(s) of positive integers m and n, such that  [#permalink]

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17 Jun 2019, 00:54
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46% (02:20) correct 54% (02:10) wrong based on 56 sessions

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For how many pair(s) of positive integers $$m$$ and $$n$$, such that $$n > m > 1$$, is $$m^n$$ not greater than $$n^m$$?

A. 0
B. 1
C. 2
D. 3
E. 4

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Re: For how many pair(s) of positive integers m and n, such that  [#permalink]

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17 Jun 2019, 09:04
1
I could get two possiblites (2,4) & ( 2,3) where n>m>1 and $$m^n$$ not greater than $$n^m$$
i.e 16=16 and 8<9
IMO C

Bunuel wrote:
For how many pair(s) of positive integers $$m$$ and $$n$$, such that $$n > m > 1$$, is $$m^n$$ not greater than $$n^m$$?

A. 0
B. 1
C. 2
D. 3
E. 4

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For how many pair(s) of positive integers m and n, such that  [#permalink]

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17 Jun 2019, 18:58
Bunuel chetan2u. Is substitution the only way?. Is there any other efficient method. Also please share similar questions if available.

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Re: For how many pair(s) of positive integers m and n, such that  [#permalink]

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19 Jun 2019, 18:39
1
Bunuel wrote:
For how many pair(s) of positive integers $$m$$ and $$n$$, such that $$n > m > 1$$, is $$m^n$$ not greater than $$n^m$$?

A. 0
B. 1
C. 2
D. 3
E. 4

Let m = 2, so n ≥ 3; we see that:

2^3 = 8 is not greater than 3^2 = 9, and 2^4 = 16 is not greater than 4^2 = 16. However, when n ≥ 5, we will have 2^n greater than n^2.

Now, let m = 3, so n ≥ 4; we see that:

3^4 = 81 is greater than 4^3 = 64. In fact, for any value n ≥ 4, we will have 3^n greater than n^3.

Since m^n is greater than n^m is already true for m = 3 (and n > m), it should be true for any m > 3 (and n > m).

Therefore, we have only two pairs of (m, n) such that m^n is not greater than n^m, namely, (2, 3) and (2, 4).

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Re: For how many pair(s) of positive integers m and n, such that   [#permalink] 19 Jun 2019, 18:39
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# For how many pair(s) of positive integers m and n, such that

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