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Difficulty: 555-605 Levelx   Algebrax   Exponents/Powersx                           
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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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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]
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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]
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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]
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]
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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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]
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]
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]
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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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Re: If r and s are positive integers such that (2^r)(4^s) = 16, then 2r + [#permalink]
Expert Reply
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]
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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If r and s are positive integers such that (2^r)(4^s) = 16, then 2r + [#permalink]
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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]
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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]
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]
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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]
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]
(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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