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Retired Moderator V
Joined: 22 Jun 2014
Posts: 1087
Location: India
Concentration: General Management, Technology
GMAT 1: 540 Q45 V20
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WE: Information Technology (Computer Software)
Re: Math Revolution and GMAT Club Contest! If a and b are positive integer  [#permalink]

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n = a^3 * b^4

if a number is written in the form of the powers of its prime factors then total factors for that number are calculated by increasing powers of each prime number by 1 and multiplying all of them.

(1) a and b are prime numbers - SUFFICIENT
n will have (3+1)*(4+1) = 4*5 = 20 factors.

(2) n has only prime factors 5 and 7 - SUFFICIENT
here it wont matter if a = 5 or 7 or b = 5 or 7. we only need powers of prime numbers. hence n will have 20 factors.

Option D is the correct answer.
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Intern  B
Joined: 05 Jun 2013
Posts: 35
Re: Math Revolution and GMAT Club Contest! If a and b are positive integer  [#permalink]

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n=a^3∗b^4, To find the number of factors of n.

Lets take choice 1.
(1) a and b are prime numbers

Since both a and b are prime numbers, then number of factors of n will be (3+1)(4+1)=4*5=20 factors.
This choice is sufficient.

(2) n has only prime factors 5 and 7.
Again, since both a and b are prime numbers, then number of factors of n will be (3+1)(4+1)=4*5=20 factors.
This choice is sufficient.

Each statement Alone is sufficient to answer the question and therefore choice D.
Intern  Joined: 10 Sep 2015
Posts: 31
Re: Math Revolution and GMAT Club Contest! If a and b are positive integer  [#permalink]

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N=a^3*b^4

1) Sufficient, (3+1)*(4+1) = 20 factors

2) n has only 5, 7 as prime factors.. again we have 20 factors.. so D ans
Current Student B
Joined: 20 Jan 2015
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Re: Math Revolution and GMAT Club Contest! If a and b are positive integer  [#permalink]

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Ans is indeed C.
thanks for the great explanation Bunuel. Intern  B
Joined: 21 Jan 2013
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Schools: LBS
GPA: 3.82
WE: Engineering (Computer Software)
Re: Math Revolution and GMAT Club Contest! If a and b are positive integer  [#permalink]

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Bunuel, I think the explanation is clear on why C is correct answer, however curious on where I can get to see these error types 4(A) and 4(B) spoken about in the solution?
Intern  B
Joined: 11 Apr 2018
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Re: Math Revolution and GMAT Club Contest! If a and b are positive integer  [#permalink]

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rockthegmat007 wrote:
If a and b are positive integers, let n=a3∗b4, how many different factors n has?

(1) a and b are prime numbers
(2) n has only prime factors 5 and 7

the number of different factors n will have depends on the prime factorization of a and b ; and the resultant powers of all the prime factors of a and b together. it follows that we need to know the powers of all different prime factors to calculate the total number of factors . we can then apply the formula of (x+1)(y+1)(z+1).. so on where x , y and z are the powers of distinct prime factors that constitute n.

1) says that a and b are prime factors themselves - therefore subsequent factorization is not possible for both a and b. so the number of total factors for n = (3+1)(4+1) - sufficient.
2) says n has only prime factors 5 and 7 . it could very well be the case that a is a multiple of 5 and 7 (each raised to any power) and so is the case with b. it could also be the case that a is 5 raised to power of any positive integer and b is 7 raised to thepower of any positive integer. (or vice versa)
the fact that the product of a^3 and b^7 could be ANY power of 5 or 7 raised subsequently by power of 3 and 7 generates multiple possible powers to both 5 and 7. so a unique set of powers for prime numbers 5 and 7 is not possible

e.g.,
one possibility : a = 5 ; b = 7 : n = 5^3 * 7^4 and the # of factors = (3+1)*(4+1)
another possibility : a = 125 ; b = 49 ; n = 5^9 * 7^8 and the # of factors = (9+1)*(8+1)

both the values are clearly different without needing further calculations - B is clearly insufficient.

what if we have m=n den??
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Re: Math Revolution and GMAT Club Contest! If a and b are positive integer  [#permalink]

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_________________ Re: Math Revolution and GMAT Club Contest! If a and b are positive integer   [#permalink] 09 Dec 2019, 18:49

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