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If @x is the number of distinct positive divisors of x, what is the va [#permalink]
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Updated on: 27 Sep 2014, 05:56
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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 M0135
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Originally posted by suntaurian on 28 Feb 2008, 09:23.
Last edited by Bunuel on 27 Sep 2014, 05:56, edited 1 time in total.
Renamed the topic, edited the question and added the OA.



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Re: If @x is the number of distinct positive divisors of x, what is the va [#permalink]
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28 Feb 2008, 09:30
suntaurian wrote: \(@x\) is the number of distinct positive divisors of \(x\) . What is the value of \(@@90\) ?
* 3 * 4 * 5 * 6 * 7 D. Factors of 90: 1,2,3,5,6,9,10,15,18,30,45,90 thus @90 = 12. @12= 6



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Re: If @x is the number of distinct positive divisors of x, what is the va [#permalink]
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28 Feb 2008, 13:38
is there a quick way to find all the factors of 90, or do you just have to go through all the numbers ?



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Re: If @x is the number of distinct positive divisors of x, what is the va [#permalink]
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28 Feb 2008, 14:46
pmenon wrote: is there a quick way to find all the factors of 90, or do you just have to go through all the numbers ? You can find out all the prime factors. But you probably wont save too much time here. might be a better approach on bigger numbers though. 2,3,3,5 dont forget 1 at the end. multiply all the possible combinations by. Ex/ 2*3, 3*3, etc... I cannot come up with an actual combinatorics solution yet, but im done w/ my work for the day and got bout 20min til i go home so il try and see if i can think of something...



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Re: If @x is the number of distinct positive divisors of x, what is the va [#permalink]
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28 Feb 2008, 16:48
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The quick way to get # of factors is to find prime factor the number:
90 = 3^2 * 5^1 * 2^1
Then you add the number 1 to exponent and multiply exponents together.
so (2+1) (1+1) (1+1) = 3*2*2 = 12
12 is number of factors of 90
do the same to get number of factors of 12
12 = 2^2 * 3^1
(2+1) * (1+1) = 6



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Re: If @x is the number of distinct positive divisors of x, what is the va [#permalink]
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28 Feb 2008, 18:37
terp26 wrote: The quick way to get # of factors is to find prime factor the number:
90 = 3^2 * 5^1 * 2^1
Then you add the number 1 to exponent and multiply exponents together.
so (2+1) (1+1) (1+1) = 3*2*2 = 12
12 is number of factors of 90
do the same to get number of factors of 12
12 = 2^2 * 3^1
(2+1) * (1+1) = 6 right, thats the trick i remember reading somewhere on this forum !! thanks !!



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Re: If @x is the number of distinct positive divisors of x, what is the va [#permalink]
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29 Feb 2008, 09:13
I got D a different way. I found the primes of 90, distinct, which came out to 2,3,5. Since there are two OO, i multiplyed 3 * 2 = 6. Did I do it correctly?



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Re: If @x is the number of distinct positive divisors of x, what is the va [#permalink]
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27 Sep 2014, 05:52
terp26 wrote: The quick way to get # of factors is to find prime factor the number:
90 = 3^2 * 5^1 * 2^1
Then you add the number 1 to exponent and multiply exponents together.
so (2+1) (1+1) (1+1) = 3*2*2 = 12
12 is number of factors of 90
do the same to get number of factors of 12
12 = 2^2 * 3^1
(2+1) * (1+1) = 6 Is this how to calculate the number of distinct factors? In the number 12, we have 2,3,4,6... but 6 is made up from 2 and 3, so does it count in the distinct? what about 216? it's 6*6*6... so does 6*6 count as distinct, or do we just count 6? I got confused here....



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Re: If @x is the number of distinct positive divisors of x, what is the va [#permalink]
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27 Sep 2014, 06:00
ronr34 wrote: terp26 wrote: The quick way to get # of factors is to find prime factor the number:
90 = 3^2 * 5^1 * 2^1
Then you add the number 1 to exponent and multiply exponents together.
so (2+1) (1+1) (1+1) = 3*2*2 = 12
12 is number of factors of 90
do the same to get number of factors of 12
12 = 2^2 * 3^1
(2+1) * (1+1) = 6 Is this how to calculate the number of distinct factors? In the number 12, we have 2,3,4,6... but 6 is made up from 2 and 3, so does it count in the distinct? what about 216? it's 6*6*6... so does 6*6 count as distinct, or do we just count 6? I got confused here.... First of all, the factors of 12 are 1, 2, 3, 4, 6, and 12. Next, how is 6 and 6 different from one another? You should count it once. 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 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: 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 \(@90=12\). Next, \(@(@90)=@12\). Now, since \(12=2^2*3\), then the number of factors of 12 is: \((2+1)(1+1)=6\). Answer: D
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Re: If @x is the number of distinct positive divisors of x, what is the va [#permalink]
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29 Sep 2014, 01:08
suntaurian wrote: 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
M0135 \(90 = 2^1 * 3^2 * 5^1\) \(@90 = (1+1) (2+1) (1+1) = 3*4 = 12\) \(@90 = 2^2 * 3^1\) \(@(@90) = (2+1) (1+1) = 6\) Answer = D
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If @x is the number of distinct positive divisors of x, what is the va [#permalink]
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12 May 2017, 03:00
pmenon wrote: is there a quick way to find all the factors of 90, or do you just have to go through all the numbers ? The quickest mathematical way to solve this problem is 90 = 3^2 x 2 x 5 So the no. of factors of 90 = (2+1)(1+1)(1+1) .. The values 2,1,1 is power of the prime factors. so @90 = 12 Now 12 = 2^2 x 3 So \(@12\) = (2+1)(1+1) = 6.
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