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# If n is a +ve integer, is (b-a) at least twice the value of

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If n is a +ve integer, is (b-a) at least twice the value of [#permalink]

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03 Aug 2007, 09:23
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If n is a +ve integer, is (b-a) at least twice the value of 3^n - 2^n?

(1) a = 2^(n+1) and b = 3^(n+1)

(2) n = 3
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03 Aug 2007, 09:35
St1:
a = 2^n(2)
2^n = a/2

b = 3^n(3)
3^n = b/3

n is a positive integer, so a/2 and b/3 must be positive and therefore a and b must be positive.

So 3^n - 2^n = a/2 - b/3 = (3a-2b)/6
2(3^n - 2^n) = (3a-2b)/3

If a = 2, b = 1, then LHS = 1, and RHS 4/3 (not at least twice)
If a = 5, b = 3, then LHS = 2, and RHS = 3 (not at least twice)
If a = 50, b = 3, then LHS = 47, and RHS = 144/3 (not at least twice)

Sufficient.

St2:

I go with A
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Re: DS - Number Properties - OG [#permalink]

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03 Aug 2007, 10:53
Andr359 wrote:
If n is a +ve integer, is (b-a) at least twice the value of 3^n - 2^n?

(1) a = 2^(n+1) and b = 3^(n+1)

(2) n = 3

From 1:

b-a = 3*3^n - 2.2^n
=(2+1)*3^n -2*2^n
=2(3^n - 2^n) + 3^n

Since n is a +ve integer 3^n must be positive,

hence, b-a > 2(3^n-2^n)

So A.
Re: DS - Number Properties - OG   [#permalink] 03 Aug 2007, 10:53
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