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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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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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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]
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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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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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]
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.
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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]
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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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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ScottTargetTestPrep wrote:
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.

"We see that there are 3 options for k and 5 options for j" - Why are we adding 1 to the power of 2 but not to the power of 5 ?
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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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rishabhshreyas14 wrote:
ScottTargetTestPrep wrote:
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.

"We see that there are 3 options for k and 5 options for j" - Why are we adding 1 to the power of 2 but not to the power of 5 ?

Since the factor we are looking for must be divisible by 5, there has to be at least one factor of 5; i.e. the exponent of 5 cannot be 0. For 2, on the other hand, the exponent can be 0; that simply means the number contains no factor of 2 (which just means that the number is not divisible by 2). That’s the reason we add 1 to the exponent of 2 but not to the exponent of 5; 0 is included in the former but not the latter.
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Re: If N = 2^23^35^5,how many factors of N are divisible by 5 but not di [#permalink]
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Re: If N = 2^23^35^5,how many factors of N are divisible by 5 but not di [#permalink]
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