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a, b and c are three distinct integers, greater than 1, such that ....

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a, b and c are three distinct integers, greater than 1, such that ....  [#permalink]

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New post 19 Dec 2018, 10:19
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Question Stats:

78% (01:00) correct 22% (02:12) wrong based on 50 sessions

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a, b and c are three distinct integers, greater than 1, such that the product of these integers is 150. If the greatest common divisor of any two numbers, among the three integers, is 1, then what is the sum of all the three integers?

    A. 18
    B. 22
    C. 30
    D. 32
    E. 54

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a, b and c are three distinct integers, greater than 1, such that ....  [#permalink]

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New post 19 Dec 2018, 11:07
EgmatQuantExpert wrote:
a, b and c are three distinct integers, greater than 1, such that the product of these integers is 150. If the greatest common divisor of any two numbers, among the three integers, is 1, then what is the sum of all the three integers?

A. 18
B. 22
C. 30
D. 32
E. 54


if only one of the integers can be a multiple of 5,
the other two integers can be 2 and 3
2*3*5*5=150
2+3+25=30
C
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Re: a, b and c are three distinct integers, greater than 1, such that ....  [#permalink]

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New post 20 Dec 2018, 00:28
150 = 2*3*5^2

so two no. whose GCD is 1 means they are prime

so sum 2+3+25 = 30 IMO C



EgmatQuantExpert wrote:
a, b and c are three distinct integers, greater than 1, such that the product of these integers is 150. If the greatest common divisor of any two numbers, among the three integers, is 1, then what is the sum of all the three integers?

    A. 18
    B. 22
    C. 30
    D. 32
    E. 54

To read all our articles:Must read articles to reach Q51

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a, b and c are three distinct integers, greater than 1, such that ....  [#permalink]

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New post 01 Jan 2019, 21:43

Solution


Given:
We are given that,
    • a, b and c are three distinct integers, greater than 1
    • abc = 150
    • GCD(a, b) = GCD(b, c) = GCD(c, a) = 1

To find:
    • We are asked to find the value of a + b + c

Approach and Working:
    • We have, abc = 150
      o And, 150 can be prime factorized as \(2 * 3 * 5^2\)
      o Thus, abc = \(2 * 3 * 5^2\)

    • And we are given that GCD(a, b) = GCD(b, c) = GCD(c, a) = 1
      o This is possible only when a, b and c do not have any common prime factor

    • And if we observe 150 has three distinct prime factors, {2, 3, 5}
      o So, a, b and c must have exactly one of these three prime factors in them

Therefore, a + b + c = \(2 + 3 + 5^2 = 30\)

Hence the correct answer is Option C.

Answer: C

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Articles and Question to reach Q51 | Question of the week

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Number Properties – Even Odd | LCM GCD | Statistics-1 | Statistics-2 | Remainders-1 | Remainders-2
Word Problems – Percentage 1 | Percentage 2 | Time and Work 1 | Time and Work 2 | Time, Speed and Distance 1 | Time, Speed and Distance 2
Advanced Topics- Permutation and Combination 1 | Permutation and Combination 2 | Permutation and Combination 3 | Probability
Geometry- Triangles 1 | Triangles 2 | Triangles 3 | Common Mistakes in Geometry
Algebra- Wavy line | Inequalities

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