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Re: M01Q35 Please help with this question. thanks [#permalink]
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04 Apr 2009, 06:39
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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: M01Q35 Please help with this question. thanks [#permalink]
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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: M01Q35 Please help with this question. thanks [#permalink]
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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: M01Q35 Please help with this question. thanks [#permalink]
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22 Sep 2010, 05:34
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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: M01Q35 Please help with this question. thanks [#permalink]
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22 Sep 2010, 06:00
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Work from the inside out. Work @90, and get 12. then @12.



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Re: M01Q35 Please help with this question. thanks [#permalink]
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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: M01Q35 Please help with this question. thanks [#permalink]
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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:( http://gmatclub.com/forum/mathcombinatorics87345.html#p785179Read the example at the end, it derives this result and explains where it comes from
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Re: M01Q35 Please help with this question. thanks [#permalink]
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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: M01Q35 Please help with this question. thanks [#permalink]
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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:( http://gmatclub.com/forum/mathcombinatorics87345.html#p785179Read 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: M01Q35 Please help with this question. thanks [#permalink]
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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: M01Q35 Please help with this question. thanks [#permalink]
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22 Sep 2010, 08:57
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: \(450=2^1*3^2*5^2\) 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: mathnumbertheory88376.htmlHope it helps.
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Re: M01Q35 Please help with this question. thanks [#permalink]
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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: M01Q35 Please help with this question. thanks [#permalink]
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Re: M01Q35 Please help with this question. thanks [#permalink]
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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: M01Q35 Please help with this question. thanks [#permalink]
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14 Oct 2010, 09:11
What level question do you think this is?



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Re: M01Q35 Please help with this question. thanks [#permalink]
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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: M01Q35 Please help with this question. thanks [#permalink]
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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: M01Q35 Please help with this question. thanks [#permalink]
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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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