Let n be a positive integer not divisible by 6.

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

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

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.

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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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

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.

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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