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A number N^2 has 35 factors. How many factors can N have?

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A number N^2 has 35 factors. How many factors can N have? [#permalink]

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A number \(N^2\) has 35 factors. How many factors can \(N\) have?

A. 6 or 10 factors
B. 8 or 14 factors
C. 10 or 16 factors
D. 12 or 18 factors
E. 14 or 20 factors
[Reveal] Spoiler: OA

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A number N^2 has 35 factors. How many factors can N have? [#permalink]

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Lets take an example 12. Factors of 12 are 2- 2time and 3- 1 time. Total no of factor - (2+1) * (1+1) =6.
For 12*12 - it has 2 - 4times and 3 - 2 timed. Total no of factor - 5*3=15

Case 1 :-

N=x.y
For N*N
Trying reverse approach. 35 = 5*7.
Factors are x - 4times and y- 6times.
FOR N - x -2 times and y- 3 times.
Total no of factor is (2+1)*(3+1)=12

Case 2:-

If N^2 is raised to the power of 34.
N^2 will have 34 factors.
And hence for N we have 17 factors.
Hence total no of factor as 17+1 = 18

Ans is D

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A number N^2 has 35 factors. How many factors can N have? [#permalink]

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Mahmud6 wrote:
A number \(N^2\) has 35 factors. How many factors can \(N\) have?

A. 6 or 10 factors
B. 8 or 14 factors
C. 10 or 16 factors
D. 12 or 18 factors
E. 14 or 20 factors



Hi...

Factors of 35 are 1,5,7,35 .. 35 = 1*35= 5*7..

So we can look for two cases..

1) where N or \(N^2\) contains ONLY one type of prime factor, a, that is \(N=a^x\)
Number of factors =(power of factor +1)=35.......
In this case power of the prime factor will be TWO times that of N as
\(N^2=a^2x....... So 2x+1=35....x=17..
N=a^17\)
So factor of N =17+1=18

2) where N^2 contains TWO prime factors...a and b.
So \(N=a^x*b^y..........N^2=a^2x*b^2y..\)
Number of factors=\((2x+1)(2y+1)=35=5*7\).....
Only possibility is when 2x+1=5....x=2
And \(2y+1=7...y=3\)
So \(N=a^2*b^3\)..
Number of factors=(2+1)(3+1)=3*4=12

Ans 12 and 18
D
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Re: A number N^2 has 35 factors. How many factors can N have? [#permalink]

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New post 20 Oct 2017, 08:28
Case 1: N^2 is composed by two different primes
A number with 35 factors = a^6*b^4 (to count the number of factors we must add one to the power and multiply them, for instance, (6+1)*(4+1) = 7*5 = 35 factors.

So N = a^3*b^2. The number of factors of N can be obtained in the same way that we obtained the 35 factors of N^2.
Number of factors of N =(3+1)*(2+1) = 12 factors

Case 2: N^2 is composed by one prime
Using the same rule to count factors mentioned above, N^2 = a^34
so N = a^17 then N has 18 factors

Answer D

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A number N^2 has 35 factors. How many factors can N have? [#permalink]

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New post 21 Oct 2017, 03:59
Mahmud6 wrote:
A number \(N^2\) has 35 factors. How many factors can \(N\) have?

A. 6 or 10 factors
B. 8 or 14 factors
C. 10 or 16 factors
D. 12 or 18 factors
E. 14 or 20 factors


D is the answer as follows.

Sometime in GMAT where we have time crunch we have to consider specific case based on the answer options provided.

Here, particularly in this question just by considering a single prime factor will be sufficient to answer the question.

Lets assume \(a^p = N^2\) => since it has 35 factors including N^2, hence p = 34
For N the number of factors will be 17 (\(\frac{34}{2}\)). Considering N as one of the factor the number of factors will become 18.

Only D satisfies the condition, hence no need to consider the case of multiple prime factors.

Hope, I am clear.
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A number N^2 has 35 factors. How many factors can N have?   [#permalink] 21 Oct 2017, 03:59
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A number N^2 has 35 factors. How many factors can N have?

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