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The number of even factors of 21600 is

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Intern
Joined: 18 Nov 2013
Posts: 28

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Location: India
WE: Engineering (Transportation)
The number of even factors of 21600 is [#permalink]

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29 Dec 2013, 06:20
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Difficulty:

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Question Stats:

64% (01:51) correct 36% (01:46) wrong based on 194 sessions

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The number of even factors of 21600 is

A. 32
B. 42
C. 60
D. 25
E. 52
[Reveal] Spoiler: OA

Last edited by Bunuel on 29 Dec 2013, 06:43, edited 1 time in total.
Renamed the topic and edited the question.

Kudos [?]: 19 [2], given: 11

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Re: The number of even factors of 21600 is [#permalink]

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29 Dec 2013, 06:55
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Klenex wrote:
The number of even factors of 21600 is

A. 32
B. 42
C. 60
D. 25
E. 52

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.
For more on number properties check: math-number-theory-88376.html

Back to the original question:

Make a prime factorization of the number: $$21,600 =2^5*3^3*5^2$$.

According to the above the number of factors is $$(5+1)(3+1)(2+1)=72$$.

Now, get rid of powers of 2 as they give even factors --> you'll have $$3^3*5^2$$ which has (3+1)(2+1)=12 factors. All the remaining factors will be odd, therefore 21,600 has 72-12=60 even factors.

Similar questions to practice:
how-many-odd-positive-divisors-does-540-have-106082.html
how-many-of-the-factors-of-72-are-divisible-by-100708.html
how-many-even-different-factors-does-the-integer-p-have-132835.html

P.S. Please do not reword or shorten questions when posting. Thank you.
_________________

Kudos [?]: 139210 [4], given: 12779

Intern
Joined: 18 Nov 2013
Posts: 28

Kudos [?]: 19 [0], given: 11

Location: India
WE: Engineering (Transportation)
Re: The number of even factors of 21600 is [#permalink]

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29 Dec 2013, 07:08
Bunuel wrote:
Klenex wrote:
The number of even factors of 21600 is

A. 32
B. 42
C. 60
D. 25
E. 52

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:

Make a prime factorization of the number: $$21,600 =2^5*3^3*5^2$$.

According to the above the number of factors is $$(5+1)(3+1)(2+1)=72$$.

Now, get rid of powers of 2 as they give even factors --> you'll have $$3^3*5^2$$ which has (3+1)(2+1)=12 factors. All the remaining factors will be odd, therefore 21,600 has 72-12=60 even factors.

P.S. Please do not reword or shorten questions when posting. Thank you.

Thank you for the answer. But I am not able to figure out the 60 even factors

I have approached like this

As Even*Even=Even & Even*Odd=Even , I considered 2^5*3^3 which has (5+1)(3+1)=24 factors and 2^5*5^2 which has (5+1)(2+1)=18 factors.

So, total 24+18=42 factors. What am I missing here?

Kudos [?]: 19 [0], given: 11

Math Expert
Joined: 02 Sep 2009
Posts: 43296

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Re: The number of even factors of 21600 is [#permalink]

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29 Dec 2013, 10:24
Klenex wrote:
Bunuel wrote:
Klenex wrote:
The number of even factors of 21600 is

A. 32
B. 42
C. 60
D. 25
E. 52

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:

Make a prime factorization of the number: $$21,600 =2^5*3^3*5^2$$.

According to the above the number of factors is $$(5+1)(3+1)(2+1)=72$$.

Now, get rid of powers of 2 as they give even factors --> you'll have $$3^3*5^2$$ which has (3+1)(2+1)=12 factors. All the remaining factors will be odd, therefore 21,600 has 72-12=60 even factors.

P.S. Please do not reword or shorten questions when posting. Thank you.

Thank you for the answer. But I am not able to figure out the 60 even factors

I have approached like this

As Even*Even=Even & Even*Odd=Even , I considered 2^5*3^3 which has (5+1)(3+1)=24 factors and 2^5*5^2 which has (5+1)(2+1)=18 factors.

So, total 24+18=42 factors. What am I missing here?

This approach is not correct. Out of 24 factors of 2^5*3^3 not all are even, for example, 3 is a factor of it and is odd, The same for 2^5*5^2. Correct approach is given in my post above.
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Re: The number of even factors of 21600 is [#permalink]

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03 Sep 2017, 02:58
Total number of factors- Total number of odd factors= Total number of even factors

72-12= 60
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Re: The number of even factors of 21600 is [#permalink]

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02 Jan 2018, 23:51
Is it possible to do the question this way :

1. 21,600 =2^5*3^3*5^2
2. Take one 2 out. so the powers left are 2^{4}*3^{3}*5{2}
3. Number of factors now are : (4+1)*(3+1)*(2+1)
4. 5*4*3 = 60

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Math Expert
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Re: The number of even factors of 21600 is [#permalink]

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03 Jan 2018, 00:05
arhitsharma wrote:
Is it possible to do the question this way :

1. 21,600 =2^5*3^3*5^2
2. Take one 2 out. so the powers left are 2^{4}*3^{3}*5{2}
3. Number of factors now are : (4+1)*(3+1)*(2+1)
4. 5*4*3 = 60

No. You got correct answer just by luck. The proper way is given here: https://gmatclub.com/forum/the-number-o ... l#p1311013
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Re: The number of even factors of 21600 is   [#permalink] 03 Jan 2018, 00:05
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