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# x is the number of distinct positive divisors of x . What is

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Senior Manager
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x is the number of distinct positive divisors of x . What is [#permalink]  28 Feb 2008, 08:23
@x is the number of distinct positive divisors of x . What is the value of @@90 ?

* 3
* 4
* 5
* 6
* 7
CEO
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Re: Positive Divisors of X [#permalink]  28 Feb 2008, 08:30
suntaurian wrote:
@x is the number of distinct positive divisors of x . What is the value of @@90 ?

* 3
* 4
* 5
* 6
* 7

D.

Factors of 90: 1,2,3,5,6,9,10,15,18,30,45,90

thus @90 = 12. @12= 6
SVP
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Re: Positive Divisors of X [#permalink]  28 Feb 2008, 12:38
is there a quick way to find all the factors of 90, or do you just have to go through all the numbers ?
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Re: Positive Divisors of X [#permalink]  28 Feb 2008, 13:46
pmenon wrote:
is there a quick way to find all the factors of 90, or do you just have to go through all the numbers ?

You can find out all the prime factors. But you probably wont save too much time here. might be a better approach on bigger numbers though.

2,3,3,5 dont forget 1 at the end. multiply all the possible combinations by.

Ex/ 2*3, 3*3, etc...

I cannot come up with an actual combinatorics solution yet, but im done w/ my work for the day and got bout 20min til i go home so il try and see if i can think of something...
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Re: Positive Divisors of X [#permalink]  28 Feb 2008, 15:48
The quick way to get # of factors is to find prime factor the number:

90 = 3^2 * 5^1 * 2^1

Then you add the number 1 to exponent and multiply exponents together.

so (2+1) (1+1) (1+1) = 3*2*2 = 12

12 is number of factors of 90

do the same to get number of factors of 12

12 = 2^2 * 3^1

(2+1) * (1+1) = 6
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Re: Positive Divisors of X [#permalink]  28 Feb 2008, 17:37
terp26 wrote:
The quick way to get # of factors is to find prime factor the number:

90 = 3^2 * 5^1 * 2^1

Then you add the number 1 to exponent and multiply exponents together.

so (2+1) (1+1) (1+1) = 3*2*2 = 12

12 is number of factors of 90

do the same to get number of factors of 12

12 = 2^2 * 3^1

(2+1) * (1+1) = 6

right, thats the trick i remember reading somewhere on this forum !! thanks !!
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Re: Positive Divisors of X [#permalink]  29 Feb 2008, 08:13
I got D a different way. I found the primes of 90, distinct, which came out to 2,3,5. Since there are two OO, i multiplyed 3 * 2 = 6.
Did I do it correctly?
Re: Positive Divisors of X   [#permalink] 29 Feb 2008, 08:13
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