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Re: M01-Q35 Please help with this question. thanks

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Re: M01-Q35 Please help with this question. thanks [#permalink] New post 04 Apr 2009, 05:39
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Difficulty:

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Question Stats:

55% (01:55) correct 45% (01:09) wrong based on 195 sessions
If @x is the number of distinct positive divisors of x , what is the value of @@90 ?

(A) 3
(B) 4
(C) 5
(D) 6
(E) 7

[Reveal] Spoiler: OA
D

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mxb908 wrote:
Not sure how the solution works itself out. Could someone please explain it. thanks.


1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45 and 90 divide 90.
So @90 = 12.

@@90 = @12 = 6
[Reveal] Spoiler: OA

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Re: M01-Q35 Please help with this question. thanks [#permalink] New post 04 Apr 2009, 20:09
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Prime factorize 90
90= 2 * 3^2*5

Add 1 to each of the distinct prime factors exponents and multiply since we just need to find the number of distinct factors and not the actual factors.

(1+1) * (2+1)*(1+1)

12
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Re: M01-Q35 Please help with this question. thanks [#permalink] New post 30 Jan 2010, 12:43
cramya wrote:
Prime factorize 90
90= 2 * 3^2*5

Add 1 to each of the distinct prime factors exponents and multiply since we just need to find the number of distinct factors and not the actual factors.

(1+1) * (2+1)*(1+1)

12



I don't have 12 in my answer choices and i answered incorrectly :( Although i understood the question perfectly

If @x is the number of distinct positive divisors of x, what is the value of @@90 ?



* 3
* 4
* 5
* 6
* 7

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Re: M01-Q35 Please help with this question. thanks [#permalink] New post 22 Sep 2010, 04:34
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still need to @12, prime factors = 2^2 x 3

(2+1) x (1+1 ) = 6
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Re: M01-Q35 Please help with this question. thanks [#permalink] New post 22 Sep 2010, 05:00
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Work from the inside out. Work @90, and get 12. then @12.
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Re: M01-Q35 Please help with this question. thanks [#permalink] New post 22 Sep 2010, 05:13
cramya wrote:
Prime factorize 90
90= 2 * 3^2*5

Add 1 to each of the distinct prime factors exponents and multiply since we just need to find the number of distinct factors and not the actual factors.

(1+1) * (2+1)*(1+1)

12

thanks:)

but could you please explain why addition of each of the exponents can produce the result? I cannot understand this:(
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Re: M01-Q35 Please help with this question. thanks [#permalink] New post 22 Sep 2010, 05:18
cnrnld wrote:
cramya wrote:
Prime factorize 90
90= 2 * 3^2*5

Add 1 to each of the distinct prime factors exponents and multiply since we just need to find the number of distinct factors and not the actual factors.

(1+1) * (2+1)*(1+1)

12

thanks:)

but could you please explain why addition of each of the exponents can produce the result? I cannot understand this:(


http://gmatclub.com/forum/math-combinatorics-87345.html#p785179

Read the example at the end, it derives this result and explains where it comes from

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Re: M01-Q35 Please help with this question. thanks [#permalink] New post 22 Sep 2010, 07:05
Divisors of 90 are: 1 90, 2 45, 3 30, 5 18, 6 15, 9 10. Add em up and you get 12. Divisors of 12 are 1 12, 3 4, 2 6... Answer is D...
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Re: M01-Q35 Please help with this question. thanks [#permalink] New post 22 Sep 2010, 07:43
shrouded1 wrote:
cnrnld wrote:
cramya wrote:
Prime factorize 90
90= 2 * 3^2*5

Add 1 to each of the distinct prime factors exponents and multiply since we just need to find the number of distinct factors and not the actual factors.

(1+1) * (2+1)*(1+1)

12

thanks:)

but could you please explain why addition of each of the exponents can produce the result? I cannot understand this:(


http://gmatclub.com/forum/math-combinatorics-87345.html#p785179

Read the example at the end, it derives this result and explains where it comes from

thank you a lot for referring me to the post and the example. it's still complicated, and i need to read it and the theory more deeply.
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Re: M01-Q35 Please help with this question. thanks [#permalink] New post 22 Sep 2010, 07:46
cnrnld wrote:
thank you a lot for referring me to the post and the example. it's still complicated, and i need to read it and the theory more deeply.


Don't worry too much about the details, because it's not tested on the GMAT :) Make yourself a note to look it up one day after you finish studying!
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Re: M01-Q35 Please help with this question. thanks [#permalink] New post 22 Sep 2010, 07:57
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cnrnld wrote:
thanks:)

but could you please explain why addition of each of the exponents can produce the result? I cannot understand this:(


MUST KNOW FOR GMAT:

Finding the Number of Factors of an Integer

First make prime factorization of an integer n=a^p*b^q*c^r, where a, b, and c are prime factors of n and p, q, and r are their powers.

The number of factors of n will be expressed by the formula (p+1)(q+1)(r+1). NOTE: this will include 1 and n itself.

Example: Finding the number of all factors of 450: 450=2^1*3^2*5^2

Total number of factors of 450 including 1 and 450 itself is (1+1)*(2+1)*(2+1)=2*3*3=18 factors.


Back to the original question:

If @x is the number of distinct positive divisors of x , what is the value of @@90?
(A) 3
(B) 4
(C) 5
(D) 6
(E) 7

The question defines @x as the number of distinct positive divisors of x. Say @6=4, as 6 have 4 distinct positive divisors: 1, 2, 3, 6.

Question: @@90=?

90=2*3^2*5, which means that the number of factors of 90 is: (1+1)(2+1)(1+1)=12. So @90=12 --> @@90=@12 --> 12=2^2*3, so the number of factors of 12 is: (2+1)(1+1)=6.

Answer: D (6).

For more on these issues check Number Theory chapter of Math Book: math-number-theory-88376.html

Hope it helps.

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Re: M01-Q35 Please help with this question. thanks [#permalink] New post 22 Sep 2010, 08:08
suppose n=a^p*b^q, i guess +1 means picking 0 out of p or q, right?

thank you, Bunuel, for your detailed instruction, and jpr200012 for your encouragement:)
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Re: M01-Q35 Please help with this question. thanks [#permalink] New post 22 Sep 2010, 08:09
Yep, +1 refers to the case where you pick the "0 powe" or p^0 = 1

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Re: M01-Q35 Please help with this question. thanks [#permalink] New post 22 Sep 2010, 23:39
clearly for n=(a^p)*(b^q)*(c^r)....prime factorized
and p, q, and r are exponents

no of factors = (p+1)*(q+1)*(r+1)
90 = (2^1)*(3^2)*(5^1)...
no of factors = (1+1)(2+1)*(1+1) = 2*3*2 = 12

12 = (2^2)*(3^1)
no of factors = (2+1)*(1+1) = 3*2 = 6

OA = 6 (D)

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Re: M01-Q35 Please help with this question. thanks [#permalink] New post 14 Oct 2010, 08:11
What level question do you think this is?
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Re: M01-Q35 Please help with this question. thanks [#permalink] New post 26 Sep 2011, 08:00
I picked D as well.

@90 = (1x90,2x45,3x30,5x18,6x15,10x9) = 12 divisors
@12 = (1x12,2x6,3x4) = 6 divisors Correct answer!!!
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Re: M01-Q35 Please help with this question. thanks [#permalink] New post 26 Sep 2011, 08:45
To get number of positive factors of any number follow below 4 steps:

1. The prime factorization raised to the power: 90= (3^2)(2^1)(5^1)(1^0)
2. The powers: 2, 1, 1, 0
3. Add one to the powers: 3,2,2,1
4. Multiply the results: (3)(2)(2)(1)=12

and repeat the steps for 12. thats it you are done!

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Re: M01-Q35 Please help with this question. thanks [#permalink] New post 26 Sep 2012, 05:57
Use the Fundamental Counting Theory...

Prime factorization of 90= 2*3^2*5
Add 1 to each exponent and multiply= (2)(3)(2)=12 distinct positive divisors of 90
Prime factorizatino of 12=3*2^2
(2)(3)=6 distinct factors of 12

Answer is D
Re: M01-Q35 Please help with this question. thanks   [#permalink] 26 Sep 2012, 05:57
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