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Let n be a positive integer not divisible by 6.

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Let n be a positive integer not divisible by 6. [#permalink] Let n be a positive integer not divisible by 6.   11 Apr 2020, 19:31
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GMATBusters’ Quant Quiz Question -8


Let n be a positive integer not divisible by 6. Suppose n has 6 positive divisors. The number of positive divisors of 9n could be
A) 54
B) 36
C) 18
D) 12
E) More than one of the above
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E
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Raxit85
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Re: Let n be a positive integer not divisible by 6. [#permalink] Let n be a positive integer not divisible by 6.   11 Apr 2020, 19:45
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Tricky one, still i'll try.
Here, n has 6 positive divisors and n is not divisible by 6.
So, it implies that n is not divisible by 2 and 3
Therefore, 9n is not divisibe by 2 and 3 but is divisible by 6 positive divisors which divide n.
It also has 6 positive divisors at least.

As per options, 54 and 36 both have more than 6 divisors.
So. Ans. E
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Re: Let n be a positive integer not divisible by 6. [#permalink] Let n be a positive integer not divisible by 6.   11 Apr 2020, 20:24
Given:
1. Let n be a positive integer not divisible by 6.
2. Suppose n has 6 positive divisors.

Asked: The number of positive divisors of 9n could be
A) 54
B) 36
C) 18
D) 12
E) More than one of the above

6 = 3*2 = 1*6
n can be of the form = p1^2*p2; where p1 & p2 are prime numbers
p1 & p2 can not be 2 & multiple of 3
or n can be of the form = p^5

Case 1: n is of the form = p1^2*p2
9n = 9*p1^2*p2 = 3^2*p1^2*p2
Either p1 or p2 can be 3.

If p1=3
9n = 3^4*p2
Number of positive divisors of 9n = 5*2 = 10

If p2 = 3
9n = 3^3*p1^2
Number of positive divisors of 9n = 4*3 = 12

If p1 & p2 are different than 3
9n = 9*p1^2*p2 = 3^2*p1^2*p2
Number of positive divisors of 9n = 3*3*2 = 18

Maximum number of divisors of 9n = 18

Case 1: n is of the form = p^5
9n = 3^2*p^5; where p is different than 3
Number of positive divisors = 3*6 = 18

But if p=3
9n = 3^7
Number of positive divisors = 8

Since 12 & 18 both are possible number of positive divisors

IMO E
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Re: Let n be a positive integer not divisible by 6. [#permalink] Let n be a positive integer not divisible by 6.   11 Apr 2020, 21:04
given,n>0,n not equal to 3x2xk
Also given,n has 6 positive divisor means 6 factors,then n could be in the form: 5^1X7^2 OR 3^1X5^2 OR 5^5 OR 7^5
Then 9n=3^2 X n=
Let check for 5^1X7^2X3^2, total number of positive divisor=18
Now check for n=3^1X5^2 then 9n=3^2X3^1X5^2X=3^3X5^2=total number of divisor=12
Hence Answer E
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Re: Let n be a positive integer not divisible by 6. [#permalink] Let n be a positive integer not divisible by 6.   11 Apr 2020, 21:21
Answer : E

Per prompt, n is not divisible by 6
and the number of divisors of n is 6.

Meaning 9n = 3^2 × n --- (1)

Total number of positive divisors = (2 + 1) * 6 = 18

If n=147=7^2 ×3^1
Substituting in (1)

9n=7^2 ×3^3
No. of positive divisors = d(pn)=(2+1) * (3+1)=12

If n = 243 = 3^5
Substituting in (1)
9n = 3^7
Number of positive divisors = 7+1=8

We have several other possibilities, hence E.
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Re: Let n be a positive integer not divisible by 6. [#permalink] Let n be a positive integer not divisible by 6.   11 Apr 2020, 21:57
Let N =7^5
So factors of 9n = 3^2×7^5
=3×6
=18
Let N = 5×7^2
So factors of 9N = 3^2×5×7^2
=3×2×3
=18
So answer is option C

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Re: Let n be a positive integer not divisible by 6. [#permalink] Let n be a positive integer not divisible by 6.   12 Apr 2020, 00:35
n not divisible by 6
and n has 6 + ve divisors
value of n has to multiple square of an integer and another integer except 6 ;
eg 7^2*8^1 ; total factors ; 3*2 ; 6
now 9*n factors ; 3^2 ; 3
and factors of n= 6
9n factors ; 6*3 ; 18
IMO C

Let n be a positive integer not divisible by 6. Suppose n has 6 positive divisors. The number of positive divisors of 9n could be
A) 54
B) 36
C) 18
D) 12
E) More than one of the above
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Re: Let n be a positive integer not divisible by 6. [#permalink] Let n be a positive integer not divisible by 6.   12 Apr 2020, 06:23
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Let n be a positive integer not divisible by 6. Suppose n has 6 positive divisors. The number of positive divisors of 9n could be
A) 54
B) 36
C) 18
D) 12
E) More than one of the above

Since n has 6 positive factors not divisible by 6 it can be in the form of \(3^5, 3^2*5^1, 2^2*5^1\)

2^2*5^1 => Any 2 co primes would do other than 3.

9n = 3^2 * n

Case i) 3^5 * 3^2 = 3^7 => 8 factors

Case ii) 3^2 * 3^2 * 5^1 = 3^4 * 5^1 => 10 factors

Case iii) 3^2 * 2^2 * 5^1 => 3*3*2 = 18 factors (These are the maximum number of factors 9n can have)

Hence Answer - C
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Re: Let n be a positive integer not divisible by 6. [#permalink] Let n be a positive integer not divisible by 6.   12 Apr 2020, 09:37
Solution, not sure if its correct.

Suppose n has 6 positive divisors - as given in the question.
Lets take for example: 245 - the factors include = 1,5,7,35,49,245 - 6 divisors and also 245 is not divisible by 6. I got this number through a little trial and error.
Now we need to find the number of divisors of 9n. = 9*245 = 2205.

It can be found by the normal factors method or you can just multiply, each of the divisors with 3 and 9 separately, so earlier we had 6 divisors,
multiplying all 6 by 3 gives us 12 and by 9 gives us a total of 18 divisors.

So my answer is 18. C.

It could be more than that, if we take a number greater than 245 and multiples of prime.
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Re: Let n be a positive integer not divisible by 6. [#permalink] Let n be a positive integer not divisible by 6.   12 Apr 2020, 10:56
32 has factors 1,2,4,8,16,32.
6 positive divisors. 32 is not divisible by 6.
So let n=32, then 9n=\(9.2^5 = 2^5.3^2\)
Therefore, number of positive factors = (5+1).(2+1)=18
Answer: C
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Re: Let n be a positive integer not divisible by 6. [#permalink] Let n be a positive integer not divisible by 6.   12 Apr 2020, 19:16
AnirudhaS wrote:
32 has factors 1,2,4,8,16,32.
6 positive divisors. 32 is not divisible by 6.
So let n=32, then 9n=\(9.2^5 = 2^5.3^2\)
Therefore, number of positive factors = (5+1).(2+1)=18
Answer: C

Let n =3×5^2
So 9n =3^3×5^2
=4×3=12 factors

n still is not divisible by 6

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Re: Let n be a positive integer not divisible by 6. [#permalink]
12 Apr 2020, 19:16
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