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If N = 2^2∗3^3∗5^5,how many factors of N are divisible by 5 but not di

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If N = 2^2∗3^3∗5^5,how many factors of N are divisible by 5 but not di  [#permalink]

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New post 29 Jan 2017, 05:09
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If \(N = 2^{2}∗3^{3}∗5^{5}\) , how many factors of N are divisible by 5 but not divisible by 3?

A) 10
B) 15
C) 18
D) 24
E) 72

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Re: If N = 2^2∗3^3∗5^5,how many factors of N are divisible by 5 but not di  [#permalink]

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New post 29 Jan 2017, 05:25
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ziyuenlau wrote:
If \(N = 2^{2}∗3^{3}∗5^{5}\) , how many factors of N are divisible by 5 but not divisible by 3?

A) 10
B) 15
C) 18
D) 24
E) 72



The highest factor might be 2^2*5^5, which has (2+1)*(5+1) = 18 factors
According to the question the factors must have at least one 5
So highest factor without 5 and 3 will 2^2, which has (2+1)= 3 factors.
So, the answer is 18-3= 15
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Re: If N = 2^2∗3^3∗5^5,how many factors of N are divisible by 5 but not di  [#permalink]

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New post 29 Jan 2017, 05:25
ziyuenlau wrote:
If \(N = 2^{2}∗3^{3}∗5^{5}\) , how many factors of N are divisible by 5 but not divisible by 3?

A) 10
B) 15
C) 18
D) 24
E) 72



The highest factor might be 2^2*5^5, which has (2+1)*(5+1) = 18 factors
According to the question the factors must have at least one 5
So highest factor without 5 and 3 will 2^2, which has (2+1)= 3 factors.
So, the answer is 18-3= 15
_________________

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Re: If N = 2^2∗3^3∗5^5,how many factors of N are divisible by 5 but not di  [#permalink]

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New post 29 Jan 2017, 05:31
nahid78 wrote:
ziyuenlau wrote:
If \(N = 2^{2}∗3^{3}∗5^{5}\) , how many factors of N are divisible by 5 but not divisible by 3?

A) 10
B) 15
C) 18
D) 24
E) 72



The highest factor might be \(2^2*5^5\), which has (2+1)*(5+1) = 18 factors
According to the question the factors must have at least one 5
So highest factor without 5 and 3 will 2^2, which has (2+1)= 3 factors.
So, the answer is 18-3= 15


Dear nahid78, What does the sentence highlighted in red means? Could you briefly elaborate?
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“Strength doesn’t come from what you can do. It comes from overcoming the things you once thought you couldn’t.”

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Re: If N = 2^2∗3^3∗5^5,how many factors of N are divisible by 5 but not di  [#permalink]

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New post 29 Jan 2017, 05:50
I would like to, but my english isn't good enough to explain such terms. (my bad)
but i can recommend that you download the free math book from gmat club. I hope you'll understand it easily.
https://gmatclub.com/forum/gmat-math-book-87417.html

Happy learning :)
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Re: If N = 2^2∗3^3∗5^5,how many factors of N are divisible by 5 but not di  [#permalink]

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New post 29 Jan 2017, 05:53
nahid78 wrote:
ziyuenlau wrote:
If \(N = 2^{2}∗3^{3}∗5^{5}\) , how many factors of N are divisible by 5 but not divisible by 3?

A) 10
B) 15
C) 18
D) 24
E) 72



The highest factor might be 2^2*5^5, which has (2+1)*(5+1) = 18 factors
According to the question the factors must have at least one 5
So highest factor without 5 and 3 will 2^2, which has (2+1)= 3 factors.
So, the answer is 18-3= 15


Dear nahid78, Could you delete this post as it is duplicated. Thank you. :-D
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"Be challenged at EVERY MOMENT."

“Strength doesn’t come from what you can do. It comes from overcoming the things you once thought you couldn’t.”

"Each stage of the journey is crucial to attaining new heights of knowledge."

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Re: If N = 2^2∗3^3∗5^5,how many factors of N are divisible by 5 but not di  [#permalink]

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New post 29 Jan 2017, 09:13
ziyuenlau wrote:
If \(N = 2^{2}∗3^{3}∗5^{5}\) , how many factors of N are divisible by 5 but not divisible by 3?

A) 10
B) 15
C) 18
D) 24
E) 72


Similar questions:
https://gmatclub.com/forum/if-n-2-7-3-5 ... 52264.html
https://gmatclub.com/forum/how-many-int ... 09333.html
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Re: If N = 2^2∗3^3∗5^5,how many factors of N are divisible by 5 but not di  [#permalink]

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New post 30 Jan 2017, 07:35
3
ziyuenlau wrote:
If \(N = 2^{2}∗3^{3}∗5^{5}\) , how many factors of N are divisible by 5 but not divisible by 3?

A) 10
B) 15
C) 18
D) 24
E) 72


Quote:
factors of N are divisible by 5 but not divisible by 3

Here, we need at least one factor of 5 and no factors of 3..factors of 2 is not important.

So possible values for powers of 5= 1,2,3,4, and 5 (5 choices) = \(5^1,...,5^5\)
possible values for powers of 3 = 0 (1 choice) =\(3^0\)
possible values for powers of 2 = 0, 1, and 2 (3 choices) = \(2^0,2^1,2^2\)

Total choices = 1 * 5 * 3 = 15

Ans.: B
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Re: If N = 2^2∗3^3∗5^5,how many factors of N are divisible by 5 but not di  [#permalink]

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New post 23 Jul 2018, 17:24
1
1
hazelnut wrote:
If \(N = 2^{2}∗3^{3}∗5^{5}\) , how many factors of N are divisible by 5 but not divisible by 3?

A) 10
B) 15
C) 18
D) 24
E) 72


If a factor of N is of the form 2^k x 5^j where k is 0, 1 or 2 and j is 1, 2, 3, 4 or 5, then the factor is divisible by 5 but not by 3.

We see that there are 3 options for k and 5 options for j, therefore, there are a total of 3 x 5 = 15 factors of N that are divisible by 5 but not by 3.

Alternate Solution:

We can first eliminate 3^3 from the discussion, since any positive power of 3 will make N divisible by 3. This leaves us with 2^2 * 5^5. We know that these yield (2 + 1) * (5 + 1) = 18 total factors of N that are not divisible by 3.

We must further limit our answer to ensure that the factors of N that we are identifying must also be divisible by 5. We see that any factors involving only powers of 2 will be factors that are not divisible by 5. Therefore, we must eliminate the (2 + 1) = 3 factors that involve only powers of 2 so we must subtract them from the 18 that we previously found. Therefore, the total number of factors of N that are divisible by 5 but not by 3 will be 18 - 3 = 15.

Answer: B
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Re: If N = 2^2∗3^3∗5^5,how many factors of N are divisible by 5 but not di &nbs [#permalink] 23 Jul 2018, 17:24
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