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If @x is the number of distinct positive divisors of x, what is the va

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

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28 Feb 2008, 09:23
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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

M01-35
[Reveal] Spoiler: OA

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

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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

M01-35

$$90 = 2^1 * 3^2 * 5^1$$

$$@90 = (1+1) (2+1) (1+1) = 3*4 = 12$$

$$@90 = 2^2 * 3^1$$

$$@(@90) = (2+1) (1+1) = 6$$

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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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Re: If @x is the number of distinct positive divisors of x, what is the va   [#permalink] 12 Apr 2016, 03:51
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