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M12-19

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New post 16 Sep 2014, 00:47
1
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A
B
C
D
E

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60% (00:44) correct 40% (00:35) wrong based on 118 sessions

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Re M12-19  [#permalink]

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New post 16 Sep 2014, 00:47
1
2
Official Solution:

How many of the factors of 72 are divisible by 2?

A. 4
B. 5
C. 6
D. 8
E. 9


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:

According to the above, since \(72=2^3*3^2\), the # of factors of 72 is \((3+1)(2+1)=12\). Out of which only 3 are odd (1, 3, and 9), so the remaining \(12-3=9\) factors must be even.


Answer: E
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Re: M12-19  [#permalink]

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New post 23 Aug 2016, 06:33
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Bunuel

Here's how I did it. I'm not sure if this can be generalized, but I think it can be:

If we want all factors of 72 that are divisible by 2, then the factor itself could be thought of as 2 * x = factor of 72. Therefore, x must be a factor of 36. If you wanted a list of factors, all you'd have to do is find the factors of 36 and multiply by 2

with that I get 2^2 * 3^2 = (2 + 1)(2 + 1) = 9.

while it was easy to find factors that are odd, what if the question were what are all factors of 72 that are divisible by 3, or what are all factors of 252 that are divisible by 7?

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Re: M12-19  [#permalink]

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New post 11 Sep 2016, 04:26
72=2^3 * 3^2 this gives a tota of 12 factor.

here we have 2^0 =1

all the multiple of 2^0 will be odd.

there will be three multiple of 2^0 with the other prime number in this question which are:
3^0, 3^1, 3^2

Hence 3 numbers will have odd factor.

hence 12-3 = 9 even factor.
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New post 19 Sep 2016, 06:24
Bunuel wrote:
Official Solution:Out of which only 3 are odd (1, 3, and 9)

Answer: E


Can someone let me know me how we find out so ?

It sounds like it's supposed to be obvious but I really don't see how

Thank you very much
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New post 19 Sep 2016, 06:26
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octaviuscaius wrote:
Bunuel wrote:
Official Solution:Out of which only 3 are odd (1, 3, and 9)

Answer: E


Can someone let me know me how we find out so ?

It sounds like it's supposed to be obvious but I really don't see how

Thank you very much


72 = 2^3*3^2. Only 3^2 can give odd factors. The number of factors of 3^2 is (2 + 1) = 3, namely - 1, 3, and 9.
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Re: M12-19  [#permalink]

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New post 20 Sep 2016, 07:02
octaviuscaius wrote:
Bunuel wrote:
Official Solution:Out of which only 3 are odd (1, 3, and 9)

Answer: E


Can someone let me know me how we find out so ?

It sounds like it's supposed to be obvious but I really don't see how

Thank you very much



Hi,

When you are asked about factors divides into 2, it implied you are asked about even factors. There is not ready formula to find. So you you can find total factors and then find odd factors which could be obtained by choosing other all prime factors other than 2. Factors other than 2 are 3 as shown by Bunuel.

Total factors = 12
Even factors =12-3 =9
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New post 20 Sep 2016, 07:34
bsattin1 wrote:
Bunuel

Here's how I did it. I'm not sure if this can be generalized, but I think it can be:

If we want all factors of 72 that are divisible by 2, then the factor itself could be thought of as 2 * x = factor of 72. Therefore, x must be a factor of 36. If you wanted a list of factors, all you'd have to do is find the factors of 36 and multiply by 2

with that I get 2^2 * 3^2 = (2 + 1)(2 + 1) = 9.

while it was easy to find factors that are odd, what if the question were what are all factors of 72 that are divisible by 3, or what are all factors of 252 that are divisible by 7?

Cheers,
Ben



Hi Ben,

You can do the same concept every time. Let's take your example what are all factors of 252 that are divisible by 7?

Divide 252 by 7 = 36 = 2^2 * 3^2

Total factors in 36 are (2+1) (2+1)= 9

252 = 2^2 * 3^2 * 7. Total factors in 252 are 18.

1 , 2, 3, 4, 6, 7, 9, 12, 14, 18, 21, 28, 36, 42, 63, 84, 126, 252

We 9 factors that is divisible by 7: 7, 14, 21, 28, 42, 63, 84, 126, 252

Hope it helps
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Re: M12-19  [#permalink]

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New post 09 Nov 2016, 00:23
Now can some one help me find the no. of even factors of 504.

My process:
504=2^3*3^2*7
Total no. of factors = 4*3*2=24
Odd factors= (2+1)*(1+1)=6

Even factors=24-6= 18.

Is this process correct?
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New post 09 Nov 2016, 02:57
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New post 05 Feb 2018, 04:46
Bunuel wrote:
Sumanth8492 wrote:
Now can some one help me find the no. of even factors of 504.

My process:
504=2^3*3^2*7
Total no. of factors = 4*3*2=24
Odd factors= (2+1)*(1+1)=6

Even factors=24-6= 18.

Is this process correct?

________________
Yes, that's correct.


But Does this methodology apply to calculating number of even factors ... like number of even factors = (4+1) ?? Not right .
So the algorithm should be to calculate total number of factors and then arrive at number of odd factors ( obtained by odd prime factors , there is no even prime other than 2) and then subtract the two to obtain number of even factors.
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New post 09 Oct 2018, 13:59
This is how I did it.
For a moment, we consider that 72=(2^x)*(3^y). which x and y are 3 and 2 respectively.
In order to find all even factors, we can use different amounts of x and y to make different factors of 72. It's obvious that even factors have at least one 2 in them.

x could be one of these numbers: 1,2,3
x could be one of these numbers: 0,1,2
you see that there are 3*3=9 different values for x and y.
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Re: M12-19 &nbs [#permalink] 09 Oct 2018, 13:59
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