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

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CEO
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x is the number of distinct positive divisors of X/ What is [#permalink] New post 03 Nov 2007, 14:11
@x is the number of distinct positive divisors of X/ What is the value of @@90?

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SVP
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 [#permalink] New post 03 Nov 2007, 14:26
(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
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 [#permalink] New post 03 Nov 2007, 14:47
What is the "@" symbol referring to?
VP
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Re: Challenge - function [#permalink] New post 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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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
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 [#permalink] New post 03 Nov 2007, 14:48
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
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 [#permalink] New post 03 Nov 2007, 15:07
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
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 [#permalink] New post 03 Nov 2007, 15:08
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
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 [#permalink] New post 03 Nov 2007, 17:49
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
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 [#permalink] New post 03 Nov 2007, 17:58
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
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 [#permalink] New post 03 Nov 2007, 18:03
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
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Re: Challenge - function [#permalink] New post 03 Nov 2007, 19:06
bmwhype2 wrote:
@x is the number of distinct positive divisors of X/ What is the value of @@90?

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5
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7


Answer is D.

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

then @12= 1,2,3,4,6,12 ---> 6
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 [#permalink] New post 03 Nov 2007, 23:51
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? ;) :)....
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Re: Challenge - function [#permalink] New post 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


Your approach is excellent.
CEO
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 [#permalink] New post 18 Nov 2007, 21:21
Excellent answers. OA is D 6.




For those "not in the know" about distinct divisors...
http://www.gmatclub.com/forum/t55240
  [#permalink] 18 Nov 2007, 21:21
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