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If r and s are positive integers such that (2^r)(4^s) = 16, then 2r +

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If r and s are positive integers such that (2^r)(4^s) = 16, then 2r + [#permalink]

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10 Jan 2016, 06:47
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If r and s are positive integers such that (2^r)(4^s) = 16, then 2r + s =

(A) 2
(B) 3
(C) 4
(D) 5
(E) 6
[Reveal] Spoiler: OA

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Re: If r and s are positive integers such that (2^r)(4^s) = 16, then 2r + [#permalink]

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10 Jan 2016, 07:11
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Bunuel wrote:
If r and s are positive integers such that (2^r)(4^s) = 16, then 2r + s =

(A) 2
(B) 3
(C) 4
(D) 5
(E) 6

lets get the eq into simplest orm..
(2^r)(4^s) = 16..
(2^r)(2^2s) = 2^4..
or r+2s=4..
since r and s are positive integers, only r as 2 and s as 1 satisfy the Equation..
so 2r+s=2*2+1=5..
D
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Re: If r and s are positive integers such that (2^r)(4^s) = 16, then 2r + [#permalink]

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13 Jan 2016, 15:41
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Hi All,

This question has a great 'brute force' element to it - you don't need to do any advanced math, but you have to be willing to 'play around' with the prompt to figure out what's possible (and what's not).

We're told that R and S are POSITIVE INTEGERS and that (2^R)(4^S) = 16. We're asked for the value of 2R + S....

Since the two variables are positive integers, that significantly restricts the possibilities. Each 'term' (2^R) and (4^S) will end up being a positive integer greater than 1 (remember, the variables are positive integers, so neither R nor S can equal 0 and neither 'term' can equal 1).

IF...
S = 2, then (2^R)(16) = 16 but we know that R CANNOT be 0, so this option is impossible. We now know that S can ONLY be 1...

When...
S = 1
(2^R)(4) = 16
2^R = 4
R = 2

Now we know that S=1 and R=2 is the only possible solution, so the answer to the question is (2)(2) + 1 = 5

[Reveal] Spoiler:
D

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If r and s are positive integers such that (2^r)(4^s) = 16, then 2r + [#permalink]

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13 Jan 2016, 22:33
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Expert's post
Bunuel wrote:
If r and s are positive integers such that (2^r)(4^s) = 16, then 2r + s =

(A) 2
(B) 3
(C) 4
(D) 5
(E) 6

(2^r)(4^s) = 16
2^(r+2s) = (2^4)

i.e. r+2s = 4
i.e. r=2 and s=1

i.e. 2r+s=2*2+1 = 5

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Re: If r and s are positive integers such that (2^r)(4^s) = 16, then 2r + [#permalink]

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15 Jan 2016, 01:01
(2^r ) (4^s) =16
=> (2^2)(4^1)=16

r=2
s=1
Hence 2(r)+s =2(2)+1=5
Hence D
Re: If r and s are positive integers such that (2^r)(4^s) = 16, then 2r +   [#permalink] 15 Jan 2016, 01:01
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