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For how many (not necessarily positive) integer values of n is the val

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For how many (not necessarily positive) integer values of n is the val  [#permalink]

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New post 18 Mar 2019, 06:04
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A
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E

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  45% (medium)

Question Stats:

40% (02:17) correct 60% (01:38) wrong based on 5 sessions

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For how many (not necessarily positive) integer values of n is the value of \(4000.(\frac{2}{5})^n\)an integer?

(A) 3

(B) 4

(C) 6

(D) 8

(E) 9

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Re: For how many (not necessarily positive) integer values of n is the val  [#permalink]

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New post 18 Mar 2019, 06:19
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Noshad wrote:
For how many (not necessarily positive) integer values of n is the value of \(4000.(\frac{2}{5})^n\)an integer?

(A) 3

(B) 4

(C) 6

(D) 8

(E) 9


4000 = 2^5 * 5^3

So the denominator of the fraction \((\frac{2}{5})^n\) could be 1 or 5^1 or 5^2 or 5^3 or 2^1 or 2^2 or 2^3 or 2^4 or 2^5 for the expression to be an integer.

Hence n could be 0 (2/5 = 1/1)
n could be 1 [(2/5)^n = (2/5)]
n could be 2 [(2/5)^n = (2/5)^2]
n could be 3 [(2/5)^n = (2/5)^3]
n could be -1 [(2/5)^n = (2/5)^-1 = (5/2)^1]
and similarly, n could be -2, -3, -4 and -5

In all, we get 9 values of n.

Answer (E)
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Re: For how many (not necessarily positive) integer values of n is the val   [#permalink] 18 Mar 2019, 06:19
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