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# Kiana has two older twin brothers. The product of their three ages is

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Math Expert
Joined: 02 Sep 2009
Posts: 64947
Kiana has two older twin brothers. The product of their three ages is  [#permalink]

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28 Mar 2019, 22:35
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Difficulty:

35% (medium)

Question Stats:

74% (02:04) correct 26% (02:31) wrong based on 31 sessions

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Kiana has two older twin brothers. The product of their three ages is 128. What is the sum of their three ages?

(A) 10
(B) 12
(C) 16
(D) 18
(E) 24

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Joined: 06 Dec 2017
Posts: 14
Re: Kiana has two older twin brothers. The product of their three ages is  [#permalink]

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28 Mar 2019, 22:43
1
The product of ages = 128 = $$2^7 ;i.e. 2^3 * 2^3 * 2$$ ;

8+8+2 = 18
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Joined: 31 Oct 2013
Posts: 1491
Concentration: Accounting, Finance
GPA: 3.68
WE: Analyst (Accounting)
Re: Kiana has two older twin brothers. The product of their three ages is  [#permalink]

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28 Mar 2019, 23:27
Bunuel wrote:
Kiana has two older twin brothers. The product of their three ages is 128. What is the sum of their three ages?

(A) 10
(B) 12
(C) 16
(D) 18
(E) 24

prime factorization will help to determine the age of the twins.

128 = 64*2 = 32 * 2 * 2 = 4 * 8 * 2 * 2 = $$2^2 *2^3 *2 *2$$ = $$2^3 2^3 * 2$$

age of the twins = 2^3.

8 + 8 + 2 = 18.

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Re: Kiana has two older twin brothers. The product of their three ages is  [#permalink]

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29 Mar 2019, 01:48
Bunuel wrote:
Kiana has two older twin brothers. The product of their three ages is 128. What is the sum of their three ages?

(A) 10
(B) 12
(C) 16
(D) 18
(E) 24

a*b^2= 128
128 = 2*8^2
so sum = 18
IMO D
Re: Kiana has two older twin brothers. The product of their three ages is   [#permalink] 29 Mar 2019, 01:48