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

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Bunuel
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If r and s are positive integers such that (2^r)(4^s) = 16, then 2r + [#permalink] New post   10 Jan 2016, 07:47
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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
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D
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If r and s are positive integers such that (2^r)(4^s) = 16, then 2r + [#permalink] New post   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] New post   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

Final Answer:
Show Spoiler
D


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

Answer: option D
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Re: If r and s are positive integers such that (2^r)(4^s) = 16, then 2r + [#permalink] New post   15 Mar 2017, 16:34
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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


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.

Answer: D
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Re: If r and s are positive integers such that (2^r)(4^s) = 16, then 2r + [#permalink] New post   15 Jan 2016, 02:01
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(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] New post   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, \(2r + s = 5\)

Answer must be (D) 5
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Re: If r and s are positive integers such that (2^r)(4^s) = 16, then 2r + [#permalink] New post   19 Feb 2018, 10:52
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Nice question. I solved it the following way:

(2^r) (4^s) = 16
(2^r) (4^s) = 2^2 * 4^1

therefore, this means:
r = 2 and s = 1

plug into the target question:
2r + s = ?
2(2) + 1 = 5
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If r and s are positive integers such that (2^r)(4^s) = 16, then 2r + [#permalink] New post   02 Jun 2019, 21:55
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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
As, 4^s = (2)^2s
=> (2^r)(2^2s) = (2^2)(2^2)

So, r = 2 and 2s = 2, so s = 1.

2r + s = 2(2) + 1 = 4 + 1
= 5

Option D.
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If r and s are positive integers such that (2^r)(4^s) = 16, then 2r + [#permalink] New post   15 Sep 2019, 11:16
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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


Given: r and s are positive integers such that (2^r)(4^s) = 16

Asked: 2r + s = ?

\(2^{r+2s} = 16 = 2^4\)
r+2s = 4
r =2, s=1

2r + s = 2*2 + 1 = 5

IMO D
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Re: If r and s are positive integers such that (2^r)(4^s) = 16, then 2r + [#permalink] New post   12 Nov 2020, 07:26
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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



r+2s=4 [as r and s are positive integers, the highest possible values of and s are]
2+2*1=4

So, 2r+s=2*2+1=5

Ans. D
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Re: If r and s are positive integers such that (2^r)(4^s) = 16, then 2r + [#permalink] New post   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] New post   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] New post   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
therefore the answer is D
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Re: If r and s are positive integers such that (2^r)(4^s) = 16, then 2r + [#permalink] New post   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

Posted from my mobile device
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Re: If r and s are positive integers such that (2^r)(4^s) = 16, then 2r + [#permalink] New post   21 Feb 2018, 20:01
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Anantz wrote:
If there was 8 in option then it would be a problem.

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

Posted from my mobile device


Hi Anantz,

If this question were to appear on Test Day, then it's certainly possible that '8' could be among the answer choices. However, we're told that S and R are POSITIVE INTEGERS, so S=0, R=4 is NOT an option here (and '8' is not a correct answer). In that same way, S=2, R=0 is not a correct answer either (even though the answer "2" does appear among the five answer choices).

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Re: If r and s are positive integers such that (2^r)(4^s) = 16, then 2r + [#permalink] New post   16 Jul 2018, 20:00
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


Bunuel, as per Wiley Efficient Learning (app.efficientlearning.com), it is sub-600 level question.
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Re: If r and s are positive integers such that (2^r)(4^s) = 16, then 2r + [#permalink] New post   22 Aug 2020, 09:32
only possible value of r and s are 2 and 1. hence 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] New post   08 Aug 2023, 22:00
2^r*4^s= 16= 2^2 * 4^1
r=2, s=1
Now we can solve easily.
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Re: If r and s are positive integers such that (2^r)(4^s) = 16, then 2r + [#permalink] New post   09 Aug 2023, 00:23
(2^r)(4^s) = 16
=>(2^r)(2^2s) = 2^4
=> r +2s = 4 (Since r & s are positive integers)
=> r = 2 and s = 1

Therefore 2r+s=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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