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Re: M01-Q35 Please help with this question. thanks [#permalink]
 04 Apr 2009, 06:39
Question Stats:
62% (01:40) correct
37% (01:10) wrong based on 1 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 Source: GMAT Club Tests - hardest GMAT questions 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
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Re: M01-Q35 Please help with this question. thanks [#permalink]
04 Apr 2009, 21:09
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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]
30 Jan 2010, 13:43
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]
22 Sep 2010, 05:34
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]
22 Sep 2010, 06:00
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]
22 Sep 2010, 06:13
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]
22 Sep 2010, 06:18
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:( 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]
22 Sep 2010, 08:05
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]
22 Sep 2010, 08:43
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:( 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]
22 Sep 2010, 08:46
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]
22 Sep 2010, 08:57
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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 IntegerFirst 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*5^2Total 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.htmlHope it helps.
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Re: M01-Q35 Please help with this question. thanks [#permalink]
22 Sep 2010, 09:08
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]
22 Sep 2010, 09:09
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Re: M01-Q35 Please help with this question. thanks [#permalink]
23 Sep 2010, 00:39
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]
14 Oct 2010, 09:11
What level question do you think this is?
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Re: M01-Q35 Please help with this question. thanks [#permalink]
26 Sep 2011, 09:00
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]
26 Sep 2011, 09:45
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]
26 Sep 2012, 06:57
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
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Re: M01-Q35 Please help with this question. thanks
[#permalink]
26 Sep 2012, 06:57
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