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Math Expert V
Joined: 02 Sep 2009
Posts: 55228
M12-19  [#permalink]

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2 00:00

Difficulty:   35% (medium)

Question Stats: 61% (00:43) correct 39% (00:34) wrong based on 126 sessions

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

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

_________________
Math Expert V
Joined: 02 Sep 2009
Posts: 55228
Re M12-19  [#permalink]

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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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Intern  Joined: 22 Apr 2015
Posts: 11
GMAT 1: 760 Q50 V44 Re: M12-19  [#permalink]

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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?

Cheers,
Ben
Intern  B
Joined: 18 Jun 2015
Posts: 39
Re: M12-19  [#permalink]

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1
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.
Intern  Joined: 10 Aug 2016
Posts: 1
Re: M12-19  [#permalink]

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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
Math Expert V
Joined: 02 Sep 2009
Posts: 55228
Re: M12-19  [#permalink]

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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.
_________________
SVP  V
Joined: 26 Mar 2013
Posts: 2176
Re: M12-19  [#permalink]

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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

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
SVP  V
Joined: 26 Mar 2013
Posts: 2176
M12-19  [#permalink]

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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
Intern  B
Joined: 07 Feb 2015
Posts: 9
GMAT 1: 670 Q51 V28 GPA: 3.7
Re: M12-19  [#permalink]

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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?
Math Expert V
Joined: 02 Sep 2009
Posts: 55228
Re: M12-19  [#permalink]

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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.
_________________
Intern  B
Joined: 13 Sep 2016
Posts: 22
Location: India
GMAT 1: 640 Q48 V29 GPA: 3.5
M12-19  [#permalink]

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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.
Intern  B
Joined: 22 Jun 2014
Posts: 21
GMAT 1: 560 Q44 V24 GMAT 2: 660 Q49 V31 GPA: 2.5
Re: M12-19  [#permalink]

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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.
Intern  B
Joined: 16 Apr 2014
Posts: 16
Re: M12-19  [#permalink]

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Easy way,

1|72
2|36
3|24
4|18
6|12
8|9

Hence, factors of 72 divisible by 2:
2,4,6,8,12,18,24,36,72
Total= 9 Re: M12-19   [#permalink] 26 Nov 2018, 09:15
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