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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...

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?

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....

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

Expert's post

1

This post was BOOKMARKED

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: \((2+1)(1+1)=6\).

Re: If @x is the number of distinct positive divisors of x, what is the va [#permalink]

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12 Apr 2016, 03:51

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