Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Thanks Bunuel - but in the solution that's there in the link which is presented by Atish, I am struggling to understand why he did

Now if we add the two numbers above we end up double counting the factors of 5^2*7^2 = (2+1)*(2+1) = 9

Can you please explain?

This part is from another approach (direct counting), which is in my solution there too. Maybe it will answer your question:

How many numbers that are not divisible by 6 divide evenly into 264,600?

\(264,600=2^3*3^3*5^2*7^2\)

We should find the factor which contain no 2 and 3 together, so not to be divisible by 6.

Clearly, the factors which contain only 2, 5, 7 and 3, 5, 7 won't be divisible by 6. So how many such factors are there? \(2^3*5^2*7^2\) --> \((3+1)*(2+1)*(2+1)=36\);

\(3^3*5^2*7^2\) --> \((3+1)*(2+1)*(2+1)=36\);

36+36=72.

Here comes the part you have a problem with. This number (72) contains duplicates, (some factors which are not divisible by 6 are counted twice): both 36'es count the factors which have ONLY 5's and/or 7's. (5*7=35, 5*7^2=245, 5^2*7=175, 5*7^0=5, 5^0*7=7....), so basically factors of 5^2*7^2 are counted twice. How, many such factors does \(5^2*7^2\) have? (2+1)*(2+1)=9.

So we should subtract this 9 duplicated factors from 72 --> 72-9=63.

How many numbers that are not divisible by 6 divide evenly into 264,600? (A) 9 (B) 36 (C) 51 (D) 63 (E) 72

Any idea how to solve this please?

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: How many numbers that are not divisible by 6 divide evenly into 264,600? (A) 9 (B) 36 (C) 51 (D) 63 (E) 72

\(264,600=2^3*3^3*5^2*7^2\), thus it has total of \((3+1)(3+1)(2+1)(2+1)=144\) differernt positive factors, including 1 and the number itself.

# of factors that ARE divisible by 6 will be \(3*3*(2+1)(2+1)=81\): we are not adding 1 to the powers of 2 and 3, this time, to exclude all the cases with 2^0*... and 3^0*... (thus to exclude all the factors which are not divisible by 2 or 3), hence ensuring that at least one 2 and at least one 3 are present to get at least one 6 in the factors we are counting.

So, # of factors that ARE NOT divisible by 6 is 144-81=63.

How many numbers that are not divisible by 6 divide evenly into 264,600? (A) 9 (B) 36 (C) 51 (D) 63 (E) 72

hi,

please explain the above question. I am unable to understand the question

" 6 divide evenly into 264,600" wat does it mean.

"divide evenly into" means "is a factor of". Divides evenly means leaves no remainder.

Let's first find out the number of factors of 264,600.

\(264,600 = 2646 * 10 * 10 = 2^3*3^3*5^2*7^2\)

Total number of factors are (3+1)*(3+1)*(2+1)*(2+1) = 144

Now let's find the number of these 144 factors which are divisible by 6. No of factors which are divisible by 6 - To make a 6, you need a 2 and a 3. So keep a 2 and a 3 aside and find the factors you can make with the rest of the primes.

No of factors of 2^2*3^2*5^2*7^2 = (2+1)*(2+1)*(2+1)*(2+1) = 81. You can make 81 factors such that they will have a 6 in them i.e. will be divisible by 6.

So 81 of the 144 factors are divisible by 6. So the other 144 - 81 = 63 factors are not.

Re: How many numbers that are not divisible by 6 divide evenly [#permalink]

Show Tags

09 May 2013, 10:31

4

This post received KUDOS

2

This post was BOOKMARKED

How many numbers that are not divisible by 6 divide evenly into 264,600? (A) 9 (B) 36 (C) 51 (D) 63 (E) 72

264600 = (2^3) * (3^3) * (5^2) * (7^2)

Numbers not divisible by 6 ----->(factors that are not multiples of 3) + (factors that are not multiples of 2) - (factors that are multiples of neither 2 nor 3) ( number of factors of 2^3 * 5^2 *7^2 ) + ( number of factors of 3^3 * 5^2 *7^2 ) - ( number of factors of 5^2 *7^2 ) = [(3+1)(2+1)(2+1)] + [(3+1)(2+1)(2+1)] - (2+1)(2+1)] =36 + 36 -9 = 63 Note : (Subtract number of factors of 5^2 *7^2 , because you have counted them twice .)

I am really sorry and I will understand if you don't reply.

Okay so I got that 264600 has 144 factors.

Just for my understanding lets say we have to find out how many numbers out of 144 are divisible by 6. After this step I am unable to understand why are we finding factors of 144?

No. We are not finding the factors of 144.

264600 has 144 factors. They are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 15, 18, ... , 264600

Now some of them are divisible by 6 and some are not.

Divisible by 6: 6, 12, 18 ... Not divisible by 6: 1, 2, 3, 4, 5, ...

Now how do you split them - how many are multiples of 6 and how many are not.

The common thing about the multiple of 6 is that they have 6 in them i.e. they have a 2 and a 3.

\(264600 = 2^3 * 3^3 * 5^2 * 7^2\)

If you have understood the method of calculating 144, you should easily be able to understand the way we calculate multiples of 6. To get a multiple of 6, we need at least one 2 and at least one 3. So you can select 2 in 3 ways (either one 2, two 2s or three 2s. You cannot have zero 2s since you need to make a 6) You can select 3 in 3 ways (either one 3, two 3s or three 3s. You cannot have zero 3s since you need to make a 6) You can select 5 in 3 ways (either zero 5, one 5 or two 5s) You can select 7 in 3 ways (either zero 7, one 7 or two 7s)

Total number of multiples of 6 = 3*3*3*3 = 81

Total number of factors which are not multiples of 6 = 144 - 81 = 63
_________________

You say keep 2 and 3 aside. I dont get it what do you mean exactly?

You need to find the number of factors that are divisible by 6. So you certainly need 6. You can do it in two ways. You pick a 2 and 3 and then choose what and whether you want to pick other factors too.

Or we can say that given 2^3*3^3*5^2*7^2, you can select a 2 in only 3 ways because you cannot have zero 2s. You must have one 2 or two 2s or three 3s. Similarly, you can select a 3 in only 3 ways (one 3 or two 3s or three 3s) and you can select 5 and 7 in 3 ways each (zero 5, one 5, two 5s) etc. So number of factors that must have 6 are 3*3*3*3.

Again, the link given above discusses this along with another detailed discussion in the comments below the post.
_________________

How many numbers that are not divisible by 6 divide evenly into 264,600? (A) 9 (B) 36 (C) 51 (D) 63 (E) 72

Any idea how to solve this please?

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: How many numbers that are not divisible by 6 divide evenly into 264,600? (A) 9 (B) 36 (C) 51 (D) 63 (E) 72

\(264,600=2^3*3^3*5^2*7^2\), thus it has total of \((3+1)(3+1)(2+1)(2+1)=144\) differernt positive factors, including 1 and the number itself.

# of factors that ARE divisible by 6 will be \(3*3*(2+1)(2+1)=81\): we are not adding 1 to the powers of 2 and 3, this time, to exclude all the cases with 2^0*... and 3^0*... (thus to exclude all the factors which are not divisible by 2 or 3), hence ensuring that at least one 2 and at least one 3 are present to get at least one 6 in the factors we are counting.

So, # of factors that ARE NOT divisible by 6 is 144-81=63.

So if I was interested in knowing the number of factors \(264,600=2^3*3^3*5^2*7^2\) that are divisible by 35=5*7

Answer: \((3+1)(3+1)(2)(2)=64\) Your saying I shouldn't add 1 to powers the primes 5 and 7?

How about how many factors of \(264,600=2^3*3^3*5^2*7^2\) are divisible by 3675=3*(5^2)*(7^2)?

Would it be \((3+1)(3)(2)(2)=48?\)

NO. It should be \((3+1)(3)(1)(1)=12\). 2 can be at any power between 0 and 3; 3 can be at any power between 1 and 3 - we need at least a factor of 3; 5 and 7 are already at power 2 in 3675, so just one choice for each.

You have to ensure that you have each prime factor of 3675 at least at the power it shows in the factorization of 3675, but not greater than the power of that factor in the decomposition of 264,600.
_________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

NO. It should be \((3+1)(3)(1)(1)=12\). 2 can be at any power between 0 and 3; 3 can be at any power between 1 and 3 - we need at least a factor of 3; 5 and 7 are already at power 2 in 3675, so just one choice for each.

You have to ensure that you have each prime factor of 3675 at least at the power it shows in the factorization of 3675, but not greater than the power of that factor in the decomposition of 264,600.[/quote]

EvaJager,

Correct me if I am wrong but is this the method your using: How many factors of \(264,600=2^3*3^3*5^2*7^2\) are divisible by \(3675=3*(5^2)*(7^2)\)?

You are essentially: \(264,600/3675=(2^3*3^3*5^2*7^2)/(3*5^2*7^2)=2^3*3^2*5^0*7^0\). Now we find the number of factors of \(2^3*3^2*5^0*7^0\) which is (3+1)(2+1)(0+1)(0+1). Do I have your method down correctly? Is this what your doing in your head??

Re: How many numbers that are not divisible by 6 divide evenly [#permalink]

Show Tags

27 Sep 2013, 13:51

1

This post received KUDOS

I did it a slightly different way that I think is slightly more brute force. First I found the prime factorization: 2^3 * 3^3 * 5^2 * 7^2.

Since any multiple of 6 must contain at least one 2 and one 3, I turned it into a combinatorics problem:

The factors will all be of the form: 7^x * 5^y * (3 or 2)^z

3 possibilities for 7: 7^2, 7^1, 7^0 3 possibilities for 5: 5^2, 5^1, 5^0 7 possibilities for the number that can be 2 OR 3 and not both: 2^3, 2^2, 2^1, 3^3, 3^2, 3^1 and 2^0 or 3^0 (don't count twice because they both equal 1)

How many numbers that are not divisible by 6 divide evenly into 264,600? (A) 9 (B) 36 (C) 51 (D) 63 (E) 72

Any idea how to solve this please?

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: How many numbers that are not divisible by 6 divide evenly into 264,600? (A) 9 (B) 36 (C) 51 (D) 63 (E) 72

\(264,600=2^3*3^3*5^2*7^2\), thus it has total of \((3+1)(3+1)(2+1)(2+1)=144\) differernt positive factors, including 1 and the number itself.

# of factors that ARE divisible by 6 will be \(3*3*(2+1)(2+1)=81\): we are not adding 1 to the powers of 2 and 3, this time, to exclude all the cases with 2^0*... and 3^0*... (thus to exclude all the factors which are not divisible by 2 or 3), hence ensuring that at least one 2 and at least one 3 are present to get at least one 6 in the factors we are counting.

So, # of factors that ARE NOT divisible by 6 is 144-81=63.

Right, that was silly mistake... Now I have another question. I tried to just take the factors and calculate how many numbers do not include both 2 and 3....

3*3*4 + 3*3*4 (the powers of: 7,5,3 + 7,5,2) but I do not get the right result. What is wrong with this logic?

You are double counting here. 3*3*4 + 3*3*4 = 72 Actual answer is 63. Here is the difference:

\(264600 = 2^3 * 3^3 * 5^2 * 7^2\)

3*3*4 (powers of 7, 5 and 3) include cases where power of 3 could be 0 e.g. \(3^0 * 5^1 * 7^1 = 35, 3^0 * 5^2 * 7^0 = 25\) etc

3*3*4 (powers of 7, 5 and 2) include cases where power of 2 could be 0 e.g. \(2^0 * 5^1 * 7^1 = 35, 2^0 * 5^2 * 7^0 = 25\) etc

So when you add the two cases, you double count the scenario where power of 2 or 3 is 0. You need to subtract those once. 3*3 (powers of 7 and 5) = 9

When you subtract this 9 from your 72, you get 63, the actual answer.
_________________

# of factors that ARE divisible by 6 will be \(3*3*(2+1)(2+1)=81\): we are not adding 1 to the powers of 2 and 3, this time, to exclude all the cases with 2^0*... and 3^0*... (thus to exclude all the factors which are not divisible by 2 or 3), hence ensuring that at least one 2 and at least one 3 are present to get at least one 6 in the factors we are counting.

Hi, when you are not adding 1 to the power of 2 and 3 what does it mean? Does it mean that we are excluding 2^0=1 and 3^0=1 because they can't increase the number of factors by multiplying by 1 or that we are including one power to include that 2^(a power) and 3^(a power) to add the possibility that in the final output multiples of 2 and 3 exists? Hope, I have expressed my doubts clearly.

First, understand - why do we add 1 to find the number of factors. We say that given N = 2^p * 3^q * 5^r ... Total number of factors of N = (p + 1) * (q + 1) * (r + 1) * ...

Why the +1s? Because say p is 4, then you can have 2^1 / 2^2 / 2^3 / 2^4 in the factor but what about 2^0 (i.e. no 2 in the factor)? So in all you have 5 ways to take 2s - you can take no 2, one 2, two 2s, three 2s and all four 2s.

Now, if you want all factors which must have a 2, the first case (no 2s) is not acceptable to you. So you don't add 1 to p.

So in the generic case, If you want factors that are divisible by 6 (i.e. they have both a 2 and 3), you will not add 1s to their powers. You will say that number of factors = p * q * (r + 1)...
_________________

How many numbers that are not divisible by 6 divide evenly into 264,600? (A) 9 (B) 36 (C) 51 (D) 63 (E) 72

Any idea how to solve this please?

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: How many numbers that are not divisible by 6 divide evenly into 264,600? (A) 9 (B) 36 (C) 51 (D) 63 (E) 72

\(264,600=2^3*3^3*5^2*7^2\), thus it has total of \((3+1)(3+1)(2+1)(2+1)=144\) differernt positive factors, including 1 and the number itself.

# of factors that ARE divisible by 6 will be \(3*3*(2+1)(2+1)=81\): we are not adding 1 to the powers of 2 and 3, this time, to exclude all the cases with 2^0*... and 3^0*... (thus to exclude all the factors which are not divisible by 2 or 3), hence ensuring that at least one 2 and at least one 3 are present to get at least one 6 in the factors we are counting.

So, # of factors that ARE NOT divisible by 6 is 144-81=63.

NO. It should be \((3+1)(3)(1)(1)=12\). 2 can be at any power between 0 and 3; 3 can be at any power between 1 and 3 - we need at least a factor of 3; 5 and 7 are already at power 2 in 3675, so just one choice for each.

You have to ensure that you have each prime factor of 3675 at least at the power it shows in the factorization of 3675, but not greater than the power of that factor in the decomposition of 264,600.

EvaJager,

Correct me if I am wrong but is this the method your using: How many factors of \(264,600=2^3*3^3*5^2*7^2\) are divisible by \(3675=3*(5^2)*(7^2)\)?

You are essentially: \(264,600/3675=(2^3*3^3*5^2*7^2)/(3*5^2*7^2)=2^3*3^2*5^0*7^0\). Now we find the number of factors of \(2^3*3^2*5^0*7^0\) which is (3+1)(2+1)(0+1)(0+1). Do I have your method down correctly? Is this what your doing in your head??

Yes, exactly!
_________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: How many numbers that are not divisible by 6 divide evenly [#permalink]

Show Tags

21 Jul 2013, 14:57

gmacforjyoab wrote:

How many numbers that are not divisible by 6 divide evenly into 264,600? (A) 9 (B) 36 (C) 51 (D) 63 (E) 72

264600 = (2^3) * (3^3) * (5^2) * (7^2)

Numbers not divisible by 6 ----->(factors that are not multiples of 3) + (factors that are not multiples of 2) - (factors that are multiples of neither 2 nor 3) ( number of factors of 2^3 * 5^2 *7^2 ) + ( number of factors of 3^3 * 5^2 *7^2 ) - ( number of factors of 5^2 *7^2 ) = [(3+1)(2+1)(2+1)] + [(3+1)(2+1)(2+1)] - (2+1)(2+1)] =36 + 36 -9 = 63 Note : (Subtract number of factors of 5^2 *7^2 , because you have counted them twice .)

HTH -Jyothi

Not divisible by 6 is to say not divisible by (2n3 = 2 and 3)

Why does ~(2n3)=~(2)+~(3)-~(2u3). Where does this expression come from?

Can I say ~(AnB)=(~A)+(~B)-~(AuB).

Because I have always dealt with not statments "~" such that ~(AnB)=Total-(AnB)

I know (AuB)=(A)+(B)-(AnB) which when rearranged can give us (AnB)=(A)+(B)-(AuB). But what happens when you have a not "~". Does the same formula hold true simply with the not's "~" distributed to each component?

Re: How many numbers that are not divisible by 6 divide evenly [#permalink]

Show Tags

22 Aug 2013, 11:20

Bunuel since past 2 hours I have probably visited 5 sites to find a solution to this problem.

I will try to put it in the best way.

1) Numbers not divisible by 6 ----->(factors that are not multiples of 3) + (factors that are not multiples of 2) - (factors that are multiples of neither 2 nor 3)

Where is the duplication occurring?

2) # of factors that ARE divisible by 6 will be : 3*3* (2+1)*(2+1) =81 we are not adding 1 to the powers of 2 and 3, this time, to exclude all the cases with 2^0*... and 3^0*

We’ve given one of our favorite features a boost! You can now manage your profile photo, or avatar , right on WordPress.com. This avatar, powered by a service...

Sometimes it’s the extra touches that make all the difference; on your website, that’s the photos and video that give your content life. You asked for streamlined access...

A lot has been written recently about the big five technology giants (Microsoft, Google, Amazon, Apple, and Facebook) that dominate the technology sector. There are fears about the...

Post today is short and sweet for my MBA batchmates! We survived Foundations term, and tomorrow's the start of our Term 1! I'm sharing my pre-MBA notes...