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Math Revolution and GMAT Club Contest! If a and b are positive integer

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Re: Math Revolution and GMAT Club Contest! If a and b are positive integer  [#permalink]

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New post 19 Dec 2015, 02:39
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.
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Re: Math Revolution and GMAT Club Contest! If a and b are positive integer  [#permalink]

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New post 19 Dec 2015, 04:44
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
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Re: Math Revolution and GMAT Club Contest! If a and b are positive integer  [#permalink]

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New post 20 Dec 2015, 09:45
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Bunuel wrote:

Math Revolution and GMAT Club Contest Starts!



QUESTION #10:

If a and b are positive integers, let \(n = a^3*b^4\), how many different factors n has?

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


Check conditions below:



Math Revolution and GMAT Club Contest

The Contest Starts November 28th in Quant Forum


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MATH REVOLUTION OFFICIAL SOLUTION:

Since we have 3 variables (n, a, b) and 1 equation (\(n=a^3b^4\)) in the original condition, we need 2 equations to match the number of variables and the number of equations. Since we need both 1) and 2), the correct answer is likely C.

Using con 1) & 2), we get \(n=5^37^4\) or \(n=7^35^4\). The number of different factors is (3+1)(4+1)=20. This is unique and sufficient. Therefore, the correct answer is C.

However, since this is an “integer” question, which is one of the key questions, we should apply Common Mistake Type 4(A).

In case of con 1), if a=b, \(n=a^7\) → (7+1)=8. However, if a≠b, \(n=a^3b^4\) → (3+1)(4+1)=20. So this is not unique and sufficient.

In case of con 2), \(n=5^37^4\) → (3+1)(4+1)=20. However, \(n=1^335^4=5^47^4\) → (4+1)(4+1)=25. This is not unique and sufficient. Therefore the correct answer is C.

Note : For cases where we need 2 more equations, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
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Re: Math Revolution and GMAT Club Contest! If a and b are positive integer  [#permalink]

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New post 20 Dec 2015, 09:59
Ans is indeed C.
thanks for the great explanation Bunuel. :)
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Re: Math Revolution and GMAT Club Contest! If a and b are positive integer  [#permalink]

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New post 20 Dec 2015, 10:04
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?
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Re: Math Revolution and GMAT Club Contest! If a and b are positive integer  [#permalink]

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New post 03 Oct 2018, 06:39
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.

Correct Answer : A

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 &nbs [#permalink] 03 Oct 2018, 06:39

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