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

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Math Expert
Joined: 02 Sep 2009
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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, 07:47
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45% (medium)

Question Stats:

68% (01:12) correct 32% (01:05) wrong based on 496 sessions

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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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Math Expert
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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, 08:11
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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, 16:41
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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, 23:33
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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

(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, 02:01
(2^r ) (4^s) =16
=> (2^2)(4^1)=16

r=2
s=1
Hence 2(r)+s =2(2)+1=5
Hence D

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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 Mar 2017, 16:34
1
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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

We can re-express 4 as 2^2:

2^r * (2^2)^s = 2^4

2^r * 2^(2s) = 2^4

When we have an exponential equation in which the bases are the same, the exponents are equal. Thus we have:

2^(r + 2s) = 4

r + 2s = 4

Since r and s must be positive integers, we see that the only possible choice for r and s is r = 2 and s = 1 (notice that if s = 2, then r = 0, and if s > 2, then r < 0). Therefore, 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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18 Mar 2017, 14:58
2^(r+2s)=2^4
since r,s are integers, r=2, s=1
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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22 Mar 2017, 00:10
Tricky and very nice question. I oversighted that s and r are positive.
was trying to solve an algebric equation to get 2r+s.
Pfff.......
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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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22 Mar 2017, 08:43
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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

Least possible value of s here will be 1 , as $$4^2 = 16$$

Now, we have -

$$(2^r)(4^1) = 16$$

Or, $$(2^r)(2^2) = 2^4$$

Or, $$2^{ r + 2} = 2^4$$

So, $$r + 2 = 4$$

Or, $$r = 2$$

Then $$2r + s = 2*2 + 1$$

Or, $$2r + s = 5$$

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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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07 Oct 2017, 02:49
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

1. First step is to simplify what has been given.

(2^r) (4^s) = 16 <---can be simplified into this ---> (2^r)(2^2s) = 2^4

(2^r)(2^2s) = 2^4 is simpd further into ---> (2^r +2s) = 2^3
(2^r +2s) = 2^3 ---> r + 2s = 4

2. Second is to find the value of the variables, by isolating the above simplified expression

r = 4 - 2s (isolating r)
2(4-2s) + s =
8-4s + s =
8 - 3s =
8 = 3s
8/3 = s (value of s)
r = 4 - 2(8/3)
r= 2/3(value of r)

3. finally, plug in the values of r and s

2(2/3) + 8/3
2 2/3 + 2 2/3 = 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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07 Nov 2017, 06:46
If there was 8 in option then it would be a problem.

I mean s=0 and r=4 making sum 8

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Re: If r and s are positive integers such that (2^r)(4^s) = 16, then 2r +   [#permalink] 07 Nov 2017, 06:46
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