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Re: M01-Q35 Please help with this question. thanks [#permalink]
04 Apr 2009, 20:09
4
This post received KUDOS
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.
Re: M01-Q35 Please help with this question. thanks [#permalink]
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 ?
Re: M01-Q35 Please help with this question. thanks [#permalink]
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:(
Re: M01-Q35 Please help with this question. thanks [#permalink]
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:(
Re: M01-Q35 Please help with this question. thanks [#permalink]
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:(
Re: M01-Q35 Please help with this question. thanks [#permalink]
22 Sep 2010, 07:57
4
This post received KUDOS
Expert's post
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: \((2+1)(1+1)=6\).
Re: M01-Q35 Please help with this question. thanks [#permalink]
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! _________________
Re: M01-Q35 Please help with this question. thanks [#permalink]
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
gmatclubot
Re: M01-Q35 Please help with this question. thanks
[#permalink]
26 Sep 2012, 05:57