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If a and b are positive integers and a^3b^2 = 72, then a + b =

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Joined: 02 Sep 2009
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If a and b are positive integers and a^3b^2 = 72, then a + b =  [#permalink]

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New post 21 Feb 2018, 22:09
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A
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E

Difficulty:

  5% (low)

Question Stats:

92% (00:50) correct 8% (01:14) wrong based on 81 sessions

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Re: If a and b are positive integers and a^3b^2 = 72, then a + b =  [#permalink]

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New post 21 Feb 2018, 22:14
Factors of 72 are: \(2^3\)*\(3^2\)

So \(a^3\)\(b^2\)= \(2^3\)*\(3^2\)

So a= 2 and b=3

a + b= 5

Answer: E.
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Re: If a and b are positive integers and a^3b^2 = 72, then a + b =  [#permalink]

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New post 22 Feb 2018, 02:07
Bunuel wrote:
If a and b are positive integers and \(a^3b^2 = 72\), then a + b =

(A) 36

(B) 17

(C) 8

(D) 6

(E) 5



Prime factorization

\(a^3b^2 = 72\) = \(2^3 3^2\)

a +b = 2+3 =5

Answer: E
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Re: If a and b are positive integers and a^3b^2 = 72, then a + b =  [#permalink]

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New post 22 Feb 2018, 23:30
Bunuel wrote:
If a and b are positive integers and \(a^3b^2 = 72\), then a + b =

(A) 36

(B) 17

(C) 8

(D) 6

(E) 5

IMO E 72=2^3*3^2 so a=2 and b=3 and 3+2=5
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Re: If a and b are positive integers and a^3b^2 = 72, then a + b =  [#permalink]

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New post 26 Feb 2018, 09:41
Bunuel wrote:
If a and b are positive integers and \(a^3b^2 = 72\), then a + b =

(A) 36

(B) 17

(C) 8

(D) 6

(E) 5


Breaking 72 into prime factors we have:

72 = 3^2 x 2^3

Since a and b are positive integers, (a^3)(b^2) = (2^3)(3^2) is satisfied only when a = 2 and b = 3. Thus, a + b = 5.

Answer: E
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Re: If a and b are positive integers and a^3b^2 = 72, then a + b = &nbs [#permalink] 26 Feb 2018, 09:41
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If a and b are positive integers and a^3b^2 = 72, then a + b =

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