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Re: How many of the factors of 72 are divisible by 2? [#permalink]

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19 Nov 2012, 23:01

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kapsycumm wrote:

How many of the factors of 72 are divisible by 2? A. 4 B. 5 C. 6 D. 8 E. 9

I got it right, but I would like to know if my method is efficient.

72 = 2*2*2*3*3 therefore, different numbers that can be found from the above = 5!/3!*2! = 10 Out of these 10, only one number (3*3) is odd....hence, the answer is 10-1 = 9.

you were lucky there I'm afraid the method is incorrect.

Sol:

The number of factors of 72 will be 12 and not 10. The best way to find out is if X= a^b * c^d then number of factors are (b+1) * (d+1)

here 72= 2^3*3^2 so number of factors will be (3+1) * (2+1) = 12 --> this includes 1 and the number itself.

so for finding factors not divisible by 2, remove all the 2s from the prime factorization. You will be left with 3^2.

so factors will be 3 and 3^2(=9). Also, we need to include the number 1 to this list, as it is odd and not divisible by 2.

so 12-3=9

the flaw in your approach: 1.) 5!/2!*3! will give you the number of ways you can arrange (permute) 22233. essentially it gives you the following list: 22233,22323,33222,32322 etc. As you can see this is a mere representation of how you can write three 2s and two 3s. This does NOT give you the list of factors of 72. 2.) not only 3*3 is odd but 3 and 3*3 both are odd factors. Include 1 to this list and you get the 3 odd factors.

Re: m12 #19 How many of the factors of 72 [#permalink]

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08 Sep 2010, 23:09

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gtr022001 wrote:

How many of the factors of 72 are divisible by 2? a. 4 b. 5 c. 6 d. 8 e. 9

What is the quickest and fastest way to find all factors of 72? I drew a prime factor tree but missed some factors in the process.

FIRST:

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 as \(72=2^3*3^2\), then # of factors of 72 is \((3+1)(2+1)=12\). Out of which only 3 are odd 1, 3, and 9, so rest or 12-3=9 are even.

OR: as \(72=2^3*3^2\) then even factors MUST have 2 either in power of 1, 2, or 3 so 3 options and 3 either in power 0, 1, or 2 again 3 options --> \(3*3=9\).

Re: How many of the factors of 72 are divisible by 2? [#permalink]

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19 Nov 2012, 23:19

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kapsycumm wrote:

How many of the factors of 72 are divisible by 2? A. 4 B. 5 C. 6 D. 8 E. 9

I got it right, but I would like to know if my method is efficient.

72 = 2*2*2*3*3 therefore, different numbers that can be found from the above = 5!/3!*2! = 10 Out of these 10, only one number (3*3) is odd....hence, the answer is 10-1 = 9.

I'm with the poster above - think you were lucky on this one.

A quick check (good thing listing factors of 72 doesn't take long): Factors - (12) 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72 Factors NOT divisible by 2 - (3) 1, 3, 9

Re: How many of the factors of 72 are divisible by 2? [#permalink]

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25 Mar 2014, 09:03

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ANOTHER METHOD Just count the no of different factors 72 has

1x72 2x36 3x24 4x18 6x12 9x8 Now of of these - 2,4,6,72,36,24,18,12 and 8 are factors divisible by 2 i.e a total of 9 factors. Did not take more than a minute.

How many of the factors of 72 are divisible by 2? [#permalink]

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19 Nov 2012, 22:21

How many of the factors of 72 are divisible by 2? A. 4 B. 5 C. 6 D. 8 E. 9

I got it right, but I would like to know if my method is efficient.

72 = 2*2*2*3*3 therefore, different numbers that can be found from the above = 5!/3!*2! = 10 Out of these 10, only one number (3*3) is odd....hence, the answer is 10-1 = 9.

Re: How many of the factors of 72 are divisible by 2? [#permalink]

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19 Nov 2012, 23:27

jumsumtak wrote:

kapsycumm wrote:

How many of the factors of 72 are divisible by 2? A. 4 B. 5 C. 6 D. 8 E. 9

I got it right, but I would like to know if my method is efficient.

72 = 2*2*2*3*3 therefore, different numbers that can be found from the above = 5!/3!*2! = 10 Out of these 10, only one number (3*3) is odd....hence, the answer is 10-1 = 9.

you were lucky there I'm afraid the method is incorrect.

Sol:

The number of factors of 72 will be 12 and not 10. The best way to find out is if X= a^b * c^d then number of factors are (b+1) * (d+1)

here 72= 2^3*3^2 so number of factors will be (3+1) * (2+1) = 12 --> this includes 1 and the number itself.

so for finding factors not divisible by 2, remove all the 2s from the prime factorization. You will be left with 3^2.

so factors will be 3 and 3^2(=9). Also, we need to include the number 1 to this list, as it is odd and not divisible by 2.

so 12-3=9

the flaw in your approach: 1.) 5!/2!*3! will give you the number of ways you can arrange (permute) 22233. essentially it gives you the following list: 22233,22323,33222,32322 etc. As you can see this is a mere representation of how you can write three 2s and two 3s. This does NOT give you the list of factors of 72. 2.) not only 3*3 is odd but 3 and 3*3 both are odd factors. Include 1 to this list and you get the 3 odd factors.

hope this helps.

BTW - jumsumtak actually shows you how to find the number of factors for a number, and will be a time saver in the exam. Note the example here works with only two prime factors. For another example with three prime factors: "How many factors in 360?", 360 = 5 x 8 x 9 --> (1+1) x (3+1) x (2+1) = 24.

Re: How many of the factors of 72 are divisible by 2? [#permalink]

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19 Nov 2012, 23:59

jcaine wrote:

BTW - jumsumtak actually shows you how to find the number of factors for a number, and will be a time saver in the exam. Note the example here works with only two prime factors. For another example with three prime factors: "How many factors in 360?", 360 = 5 x 8 x 9 --> (1+1) x (3+1) x (2+1) = 24.

That is correct. This works with every number not with just 2 prime factors. you can have 'n' PRIME factors and it will still hold true.

360 = 5 x 8 x 9 = 5 x 2^3 x 3^2. So the factors will be (1+1) x (3+1) x ( 2+1) = 2 x 4 x 3= 24

Re: How many of the factors of 72 are divisible by 2? [#permalink]

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20 Nov 2012, 03:30

Expert's post

kapsycumm wrote:

How many of the factors of 72 are divisible by 2? A. 4 B. 5 C. 6 D. 8 E. 9

I got it right, but I would like to know if my method is efficient.

72 = 2*2*2*3*3 therefore, different numbers that can be found from the above = 5!/3!*2! = 10 Out of these 10, only one number (3*3) is odd....hence, the answer is 10-1 = 9.

Merging similar topics. Please refer to the solutions above. _________________

Re: m12 #19 How many of the factors of 72 [#permalink]

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20 Nov 2012, 08:03

Bunuel wrote:

gtr022001 wrote:

How many of the factors of 72 are divisible by 2? a. 4 b. 5 c. 6 d. 8 e. 9

What is the quickest and fastest way to find all factors of 72? I drew a prime factor tree but missed some factors in the process.

FIRST:

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 as \(72=2^3*3^2\), then # of factors of 72 is \((3+1)(2+1)=12\). Out of which only 3 are odd 1, 3, and 9, so rest or 12-3=9 are even.

OR: as \(72=2^3*3^2\) then even factors MUST have 2 either in power of 1, 2, or 3 so 3 options and 3 either in power 0, 1, or 2 again 3 options --> \(3*3=9\).

Answer: E.

Hope it helps.

number of factors and prime factors is fine.... but out of those number of factors... how did you conclude on as only 3 being odd?

Re: m12 #19 How many of the factors of 72 [#permalink]

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24 Nov 2012, 21:20

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Amateur wrote:

number of factors and prime factors is fine.... but out of those number of factors... how did you conclude on as only 3 being odd?

(Will try to explain this using an easier but slightly slower approach since many have difficulties grasping perms & combs)

Using the number 72 from the original question; [1] Find number of factors: 1. 72 = 2^3 x 3^2; 2. therefore number of factors = (3+1) x (2+1) = 12

[2] Find number of ODD factors: *Number property, N1: We know that all primes, except 2, are odd. *Number property, N2: We know that ODD x ODD = ODD. *Number property, N3: Multiplying any number by 2 (an Even Number) will yield an EVEN number.

1. Recalling from [1], we have identified 2 and 3 as the prime factors of 72. 2. We ignore the "2" remembering N3. 3. We can construct factors that consist of prime factor 3 only:3, 3 x 3 (since there are only two "3"s we stop here). 4. Let's not forget that "1" is also a non-even factor. 5. Total of ODD factors is 3.

[3] Find number of EVEN factors: Total of factors - total of odd factors = total of even factors = 12 - 3 = 9.

One could directly use combinations of 2, 2, 2, 3, 3 to list all EVEN factors but I've found it faster to find ODD factors first.

For example, in my explanation for counting factors for a number with three distinct primes: 1. 360 = 5 x 8 x 9 = 5^1 x 2^3 x 3^2 2. Number of primes = (1+1) x (3+1) x (2+1) = 24 3. Number of odd factors will be multiples of only up to 1 "5" and 2 "3"s. 4. List odd factors: 3 5 9 = 3 x 3 15 = 3 x 5 45 = 3 x 3 x 5 5. Do not forget that "1" is also a factor, therefore there are 6 ODD factors in 360. 6. Total of EVEN factors in 360 is 24 - 6 = 18.

*Quick Check with pairs indeed reveals 6 ODD factors: 1, 360 --> 1 is ODD 2, 180 3, 120 --> 3 is ODD 4, 90 5, 72 --> 5 is ODD 6, 60 8, 45 --> 45 is ODD 9, 40 --> 9 is ODD 10, 36 12, 30 15, 24 --> 15 is ODD 18, 20

Hope this clarifies things. * Do note that I would expect 750+ questions to involve combinatorics that involve the use of perms & combs to solve within the time limit.

Re: m12 #19 How many of the factors of 72 [#permalink]

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07 Mar 2014, 20:27

Bunuel wrote:

gtr022001 wrote:

How many of the factors of 72 are divisible by 2? a. 4 b. 5 c. 6 d. 8 e. 9

What is the quickest and fastest way to find all factors of 72? I drew a prime factor tree but missed some factors in the process.

FIRST:

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 as \(72=2^3*3^2\), then # of factors of 72 is \((3+1)(2+1)=12\). Out of which only 3 are odd 1, 3, and 9, so rest or 12-3=9 are even.

OR: as \(72=2^3*3^2\) then even factors MUST have 2 either in power of 1, 2, or 3 so 3 options and 3 either in power 0, 1, or 2 again 3 options --> \(3*3=9\).

Answer: E.

Hope it helps.

Bunuel,

how do we know the red part without writing all the factors of 72? _________________

"Where are my Kudos" ............ Good Question = kudos

Re: m12 #19 How many of the factors of 72 [#permalink]

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08 Mar 2014, 06:45

Expert's post

Mountain14 wrote:

Bunuel wrote:

gtr022001 wrote:

How many of the factors of 72 are divisible by 2? a. 4 b. 5 c. 6 d. 8 e. 9

What is the quickest and fastest way to find all factors of 72? I drew a prime factor tree but missed some factors in the process.

FIRST:

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 as \(72=2^3*3^2\), then # of factors of 72 is \((3+1)(2+1)=12\). Out of which only 3 are odd 1, 3, and 9, so rest or 12-3=9 are even.

OR: as \(72=2^3*3^2\) then even factors MUST have 2 either in power of 1, 2, or 3 so 3 options and 3 either in power 0, 1, or 2 again 3 options --> \(3*3=9\).

Answer: E.

Hope it helps.

Bunuel,

how do we know the red part without writing all the factors of 72?

It's not hard to find the number of odd factors of 72 manually but if you want more systematic approach, refer to the red part above or consider the following:

Get rid of all the 2’s which give even factors in 72, so divide 72 by 2^3=8: 72/2^3=9=3^2. Now, 9 will have all the odd factors of 72 and won’t have its even factors. The number of factors of 9 is (2+1)=3.

So, we know that 72 has total of 12 factors out of which 3 are odd. Therefore 72 has 12-3=9 even factors.

Re: How many of the factors of 72 are divisible by 2? [#permalink]

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