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# If n is a positive integer, is the value of b - a at least

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If n is a positive integer, is the value of b - a at least [#permalink]

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18 Dec 2012, 09:00
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If n is a positive integer, is the value of 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
[Reveal] Spoiler: OA
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Re: If n is a positive integer, is the value of b - a at least [#permalink]

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18 Dec 2012, 09:03
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If n is a positive integer, is the value of b - a at least twice the value of 3^n - 2^n?

Question: is $$b-a\geq{2(3^n - 2^n)}$$?

(1) a= 2^(n+1) and b= 3^(n+1). The question becomes: is $$3^{n+1}-2^{n+1}\geq{2*(3^n - 2^n)}$$? --> is $$3*3^{n}-2*2^{n}\geq{2*3^n-2*2^{n}$$? --> is $$3^{n}\geq{0}$$? 3^n is always more than zero, so this statement is sufficient.

(2) n = 3. Clearly insufficient.

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Re: If n is a positive integer, is the value of b - a at least [#permalink]

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29 Sep 2013, 02:59
Bunuel wrote:
If n is a positive integer, is the value of b - a at least twice the value of 3^n - 2^n?

Question: is $$b-a\geq{2(3^n - 2^n)}$$?

(1) a= 2^(n+1) and b= 3^(n+1). The question becomes: is $$3^{n+1}-2^{n+1}\geq{2*(3^n - 2^n)}$$? --> is $$3*3^{n}-2*2^{n}\geq{2*3^n-2*2^{n}$$? --> is $$3^{n}\geq{0}$$? 3^n is always more than zero, so this statement is sufficient.

(2) n = 3. Clearly insufficient.

For statement (1), I understand how you arrived at $$3*3^{n}-2*2^{n}\geq{2*3^n-2*2^{n}$$ but how does this equation tell you whether b-a is at least twice the value of 3^n-2^n?

For example, if i sub in n=1 into the equation, it is true that b-a is at least twice the value (b-a= 5 and 3^n-2^n = 1 then 1*2). However, if i sub in n=2, the equation no longer is true as b-a then becomes 19 and the right side of the equation then becomes 10, which is not at least twice the value.

What am i missing here?
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Re: If n is a positive integer, is the value of b - a at least [#permalink]

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29 Sep 2013, 13:01
bulletpoint wrote:
Bunuel wrote:
If n is a positive integer, is the value of b - a at least twice the value of 3^n - 2^n?

Question: is $$b-a\geq{2(3^n - 2^n)}$$?

(1) a= 2^(n+1) and b= 3^(n+1). The question becomes: is $$3^{n+1}-2^{n+1}\geq{2*(3^n - 2^n)}$$? --> is $$3*3^{n}-2*2^{n}\geq{2*3^n-2*2^{n}$$? --> is $$3^{n}\geq{0}$$? 3^n is always more than zero, so this statement is sufficient.

(2) n = 3. Clearly insufficient.

For statement (1), I understand how you arrived at $$3*3^{n}-2*2^{n}\geq{2*3^n-2*2^{n}$$ but how does this equation tell you whether b-a is at least twice the value of 3^n-2^n?

For example, if i sub in n=1 into the equation, it is true that b-a is at least twice the value (b-a= 5 and 3^n-2^n = 1 then 1*2). However, if i sub in n=2, the equation no longer is true as b-a then becomes 19 and the right side of the equation then becomes 10, which is not at least twice the value.

What am i missing here?

For the first statement after some manipulation the question becomes: is $$3^{n}\geq{0}$$? Irrespective of the actual value of n, 3^n will always be greater than zero, thus the answer to the question is YES.
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Re: If n is a positive integer, is the value of b - a at least [#permalink]

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30 Sep 2013, 18:49
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If n is a positive integer, is the value of 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

The Question asks: b - a > 2*(3^n - 2^n)

The answer should be a definitive yes/no

Statement 1:

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

Lets take the smallest possible value of a Positive Integer i.e 1 and put it in for "n"

3^(1+1) - 2^ (1+1) > 2*(3^1 - 2^1)
3^2 - 2^2 > 2* (3 - 2)
9 - 4 > 2 *1

Lets test another number (just to be on the Safe Side), lets test n=2

3^(2+1) - 2^ (2+1) > 2*(3^2 - 2^2)
3^3 - 2^3 > 2* (9 - 4)
27 - 8 > 2* 5

Thus, Sufficient.

Statement 2:

n=3

Let's Put it in the in-equality b - a > 2*(3^n - 2^n)

b - a> 2*(3^3 - 2^3)
b - a> 2*(27 - 8)
b - a> 2*(19)
b - a> 38

If, b=100 and a=10, than definitive answer
If, b= 2 and a= 10, than definitive answer
But since it doesn't provide any value for either "a" or "b"

Thus, Not Sufficient

[Reveal] Spoiler:
A
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Re: If n is a positive integer, is the value of b - a at least [#permalink]

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20 Jul 2014, 00:12
Do you think that it's better to approach this problem by plugging in numbers for n (as long as you're sure you can do it fast for cases n=1 and n=2), or to do the problem algebraically (which you know will take ~45-90seconds to crunch)?

Would be particularly interested in getting Bunuel's take...Thanks!
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Re: If n is a positive integer, is the value of b - a at least [#permalink]

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20 Jul 2014, 06:06
warriorsquared wrote:
Do you think that it's better to approach this problem by plugging in numbers for n (as long as you're sure you can do it fast for cases n=1 and n=2), or to do the problem algebraically (which you know will take ~45-90seconds to crunch)?

Would be particularly interested in getting Bunuel's take...Thanks!

You should take the approach which fits you the best.
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Re: If n is a positive integer, is the value of b - a at least [#permalink]

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20 Jul 2014, 09:17
Can someone please show me the simplification as to how the equation becomes whether 3^n > 0?
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Re: If n is a positive integer, is the value of b - a at least [#permalink]

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20 Jul 2014, 09:43
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suhaschan wrote:
Can someone please show me the simplification as to how the equation becomes whether 3^n > 0?

$$3^{n+1}-2^{n+1}\geq{2*(3^n - 2^n)}$$;

$$3*3^{n}-2*2^{n}\geq{2*3^n-2*2^{n}$$;

Cancel $$-2*2^{n}$$: $$3*3^{n}\geq{2*3^n}$$;

Subtract 2*3^n from both sides: $$3*3^{n}-2*3^n\geq{0}$$;

$$3^{n}\geq{0}$$

Hope it's clear.
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Re: If n is a positive integer, is the value of b - a at least [#permalink]

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21 Jul 2014, 05:50
Thanks a lot Bunuel. The explanation couldn't have been more lucid.
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If n is a positive integer, is the value of b - a at least [#permalink]

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03 Aug 2014, 10:09
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Bunuel wrote:
suhaschan wrote:
Can someone please show me the simplification as to how the equation becomes whether 3^n > 0?

$$3^{n+1}-2^{n+1}\geq{2*(3^n - 2^n)}$$;

$$3*3^{n}-2*2^{n}\geq{2*3^n-2*2^{n}$$;

Cancel $$-2*2^{n}$$: $$3*3^{n}\geq{2*3^n}$$;

Subtract 2*3^n from both sides: $$3*3^{n}-2*3^n\geq{0}$$;

$$3^{n}\geq{0}$$

Hope it's clear.

Hi Bunuel,
In the end instead of subtracting $$2*3^n$$ from both sides, can we cancel $$3^n$$ from both sides (as it's always positive) and reach $$3>2$$ making the statement sufficient? Is that also a correct approach?
Thanks.
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Re: If n is a positive integer, is the value of b - a at least [#permalink]

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12 Aug 2014, 02:55
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Dienekes wrote:
Bunuel wrote:
suhaschan wrote:
Can someone please show me the simplification as to how the equation becomes whether 3^n > 0?

$$3^{n+1}-2^{n+1}\geq{2*(3^n - 2^n)}$$;

$$3*3^{n}-2*2^{n}\geq{2*3^n-2*2^{n}$$;

Cancel $$-2*2^{n}$$: $$3*3^{n}\geq{2*3^n}$$;

Subtract 2*3^n from both sides: $$3*3^{n}-2*3^n\geq{0}$$;

$$3^{n}\geq{0}$$

Hope it's clear.

Hi Bunuel,
In the end instead of subtracting $$2*3^n$$ from both sides, can we cancel $$3^n$$ from both sides (as it's always positive) and reach $$3>2$$ making the statement sufficient? Is that also a correct approach?
Thanks.

Yes, this also would be a correct way of solving.
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If n is a positive integer, is the value of b - a at least [#permalink]

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04 May 2015, 06:45
Bunuel wrote:
suhaschan wrote:
Can someone please show me the simplification as to how the equation becomes whether 3^n > 0?

$$3^{n+1}-2^{n+1}\geq{2*(3^n - 2^n)}$$;

$$3*3^{n}-2*2^{n}\geq{2*3^n-2*2^{n}$$;

Cancel $$-2*2^{n}$$: $$3*3^{n}\geq{2*3^n}$$;

Subtract 2*3^n from both sides: $$3*3^{n}-2*3^n\geq{0}$$;

$$3^{n}\geq{0}$$

Hope it's clear.

You could stop at the part I have boldfaced to realize that the statement is sufficient, no? I mean $$3*3^{n}-2*2^{n}\geq{2*3^n-2*2^{n}$$; clearly shows that it it IS bigger. (Given that n is a positive integer)
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Re: If n is a positive integer, is the value of b - a at least [#permalink]

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22 Oct 2015, 14:20
bulletpoint wrote:
Bunuel wrote:
If n is a positive integer, is the value of b - a at least twice the value of 3^n - 2^n?

Question: is $$b-a\geq{2(3^n - 2^n)}$$?

(1) a= 2^(n+1) and b= 3^(n+1). The question becomes: is $$3^{n+1}-2^{n+1}\geq{2*(3^n - 2^n)}$$? --> is $$3*3^{n}-2*2^{n}\geq{2*3^n-2*2^{n}$$? --> is $$3^{n}\geq{0}$$? 3^n is always more than zero, so this statement is sufficient.

(2) n = 3. Clearly insufficient.

For statement (1), I understand how you arrived at $$3*3^{n}-2*2^{n}\geq{2*3^n-2*2^{n}$$ but how does this equation tell you whether b-a is at least twice the value of 3^n-2^n?

For example, if i sub in n=1 into the equation, it is true that b-a is at least twice the value (b-a= 5 and 3^n-2^n = 1 then 1*2). However, if i sub in n=2, the equation no longer is true as b-a then becomes 19 and the right side of the equation then becomes 10, which is not at least twice the value.

What am i missing here?

How did you get that the right side of the equation is 10?? 3^2 = 9 and 2^2 = 4 which means 5. 5 X 2 = 10. 19 > 10.
Re: If n is a positive integer, is the value of b - a at least   [#permalink] 22 Oct 2015, 14:20
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