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CEO
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x is the number of distinct positive divisors of X/ What is [#permalink]
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03 Nov 2007, 15:11
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@x is the number of distinct positive divisors of X/ What is the value of @@90?
3
4
5
6
7



SVP
Joined: 01 May 2006
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(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
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What is the "@" symbol referring to?



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Re: Challenge  function [#permalink]
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03 Nov 2007, 15:48
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



Director
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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!



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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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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?



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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 ?



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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.



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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!



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Re: Challenge  function [#permalink]
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03 Nov 2007, 20:06
bmwhype2 wrote: @x is the number of distinct positive divisors of X/ What is the value of @@90?
3 4 5 6 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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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]
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04 Nov 2007, 00:56
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
Joined: 21 Jan 2007
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Excellent answers. OA is D 6.
For those "not in the know" about distinct divisors...
http://www.gmatclub.com/forum/t55240










