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# x is the number of distinct positive divisors of X/ What is

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CEO
Joined: 21 Jan 2007
Posts: 2760
Location: New York City
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x is the number of distinct positive divisors of X/ What is [#permalink]  03 Nov 2007, 14:11
@x is the number of distinct positive divisors of X/ What is the value of @@90?

3
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5
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7
SVP
Joined: 01 May 2006
Posts: 1798
Followers: 8

Kudos [?]: 102 [0], given: 0

(C) for me

@@90

90 = 9*10 = 3*3*5*2 = 15*6 = 18*5 = 30*3 = 45*2

We have thus : @90 = 12

Then

12 = 3*4 = 3*2*2 = 6*2

@@90 = @12 = 5
Current Student
Joined: 03 Oct 2006
Posts: 85
Followers: 1

Kudos [?]: 0 [0], given: 0

What is the "@" symbol referring to?
VP
Joined: 10 Jun 2007
Posts: 1463
Followers: 6

Kudos [?]: 143 [0], given: 0

Re: Challenge - function [#permalink]  03 Nov 2007, 14:48
bmwhype2 wrote:
@x is the number of distinct positive divisors of X/ What is the value of @@90?

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4
5
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7

D for me

90 = 9*5*2 = 2^1 * 3^2 * 5^1
@90 = Total number of factors of 90 = (1+1)*(2+1)*(1+1) = 2*3*2 = 12
@@90 = @12 = 2^2 * 3^1 = (2+1)*(1+1) = 3*2 = 6
Director
Joined: 11 Jun 2007
Posts: 932
Followers: 1

Kudos [?]: 80 [0], given: 0

Fig wrote:
(C) for me

@@90

90 = 9*10 = 3*3*5*2 = 15*6 = 18*5 = 30*3 = 45*2

We have thus : @90 = 12

Then

12 = 3*4 = 3*2*2 = 6*2

@@90 = @12 = 5

Fig, this is exactly the way I solved it but @12 should be 6 and not 5!
SVP
Joined: 01 May 2006
Posts: 1798
Followers: 8

Kudos [?]: 102 [0], given: 0

beckee529 wrote:
Fig wrote:
(C) for me

@@90

90 = 9*10 = 3*3*5*2 = 15*6 = 18*5 = 30*3 = 45*2

We have thus : @90 = 12

Then

12 = 3*4 = 3*2*2 = 6*2

@@90 = @12 = 5

Fig, this is exactly the way I solved it but @12 should be 6 and not 5!

Complete me here ... I'm only counting 5 ... 2,3,4,6 and 12
Director
Joined: 11 Jun 2007
Posts: 932
Followers: 1

Kudos [?]: 80 [0], given: 0

Fig wrote:
beckee529 wrote:
Fig wrote:
(C) for me

@@90

90 = 9*10 = 3*3*5*2 = 15*6 = 18*5 = 30*3 = 45*2

We have thus : @90 = 12

Then

12 = 3*4 = 3*2*2 = 6*2

@@90 = @12 = 5

Fig, this is exactly the way I solved it but @12 should be 6 and not 5!

Complete me here ... I'm only counting 5 ... 2,3,4,6 and 12

doesn't 1 count as a factor/divisor as well?
SVP
Joined: 28 Dec 2005
Posts: 1579
Followers: 2

Kudos [?]: 91 [0], given: 2

So i have a silly question. Once we break 90 down into its prime factors of 2,3,3,5, how do we know how many divisors of 90 this gives us ?
VP
Joined: 10 Jun 2007
Posts: 1463
Followers: 6

Kudos [?]: 143 [0], given: 0

pmenon wrote:
So i have a silly question. Once we break 90 down into its prime factors of 2,3,3,5, how do we know how many divisors of 90 this gives us ?

Not silly at all. I don't know how many people know this, but remember this method and remember it well...

To find the number of divisors (or factors):
1) Prime factorize
2) Add 1 to the power of the factors
3) Multiply them together to get answer

For example, 90 = 2^1 * 3^2 * 5^1
So the powers are 1, 2, 1
Then add 1 to the powers and you get: 1+1=2, 2+1=3, 1+1=2
Then multiply them together: 2*3*2 = 12
So 12 is the the number of divisors (or factors) that 90 has

If you are in doubt, try it with any number...there is a theory behind this method, but I'm not going into that.
Director
Joined: 11 Jun 2007
Posts: 932
Followers: 1

Kudos [?]: 80 [0], given: 0

bkk145 wrote:
pmenon wrote:
So i have a silly question. Once we break 90 down into its prime factors of 2,3,3,5, how do we know how many divisors of 90 this gives us ?

Not silly at all. I don't know how many people know this, but remember this method and remember it well...

To find the number of divisors:
1) Prime factorize
2) Add 1 to the power of the factors
3) Multiply them together to get answer

For example, 90 = 2^1 * 3^2 * 5^1
So the powers are 1, 2, 1
Then add 1 to the powers and you get: 1+1=2, 2+1=3, 1+1=2
Then multiply then together: 2*3*2 = 12
So 12 is the the number of divisors (or factors) that 90 has

well explained bkk.. you had explained this in a much earlier post that i was able to pick up, so now the concept is nice and clear to me and i can easily apply this, thanks!
CEO
Joined: 29 Mar 2007
Posts: 2585
Followers: 17

Kudos [?]: 244 [0], given: 0

Re: Challenge - function [#permalink]  03 Nov 2007, 19:06
bmwhype2 wrote:
@x is the number of distinct positive divisors of X/ What is the value of @@90?

3
4
5
6
7

@90= 1,2,3,5,6,9,10,15,18,30,45,90

then @12= 1,2,3,4,6,12 ---> 6
SVP
Joined: 01 May 2006
Posts: 1798
Followers: 8

Kudos [?]: 102 [0], given: 0

beckee529 wrote:
Fig wrote:
beckee529 wrote:
Fig wrote:
(C) for me

@@90

90 = 9*10 = 3*3*5*2 = 15*6 = 18*5 = 30*3 = 45*2

We have thus : @90 = 12

Then

12 = 3*4 = 3*2*2 = 6*2

@@90 = @12 = 5

Fig, this is exactly the way I solved it but @12 should be 6 and not 5!

Complete me here ... I'm only counting 5 ... 2,3,4,6 and 12

doesn't 1 count as a factor/divisor as well?

Yes... Indeed, it is ... I see what I have forgotten now :D .... Silly mistake? ....
SVP
Joined: 29 Aug 2007
Posts: 2493
Followers: 59

Kudos [?]: 578 [0], given: 19

Re: Challenge - function [#permalink]  03 Nov 2007, 23:56
bkk145 wrote:
bmwhype2 wrote:
@x is the number of distinct positive divisors of X/ What is the value of @@90?

3
4
5
6
7

D for me

90 = 9*5*2 = 2^1 * 3^2 * 5^1
@90 = Total number of factors of 90 = (1+1)*(2+1)*(1+1) = 2*3*2 = 12
@@90 = @12 = 2^2 * 3^1 = (2+1)*(1+1) = 3*2 = 6

CEO
Joined: 21 Jan 2007
Posts: 2760
Location: New York City
Followers: 9

Kudos [?]: 390 [0], given: 4

Excellent answers. OA is D 6.

For those "not in the know" about distinct divisors...
http://www.gmatclub.com/forum/t55240
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