How many numbers that are not divisible by 6 divide evenly : Problem Solving (PS)

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Bunuel

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

28 Jan 2012, 16:22

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enigma123 wrote:

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.

Hope it's clear.

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

28 Jan 2012, 15:59

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?

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Re: Numbers divisible by 6 [#permalink]

15 Sep 2012, 14:36

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Bunuel wrote:

enigma123 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

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.

Answer: D.

Another solution here: new-set-of-good-ps-85440.html#p642384

Hope now it's clear.

Hey Bunuel,

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 \((3+1)(3)(2)(2)=48?\)

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Re: Numbers divisible by 6 [#permalink]

15 Sep 2012, 16:23

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alphabeta1234 wrote:

Bunuel wrote:

enigma123 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

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.

Answer: D.

Another solution here: new-set-of-good-ps-85440.html#p642384

Hope now it's clear.

Hey Bunuel,

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.

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Re: Numbers divisible by 6 [#permalink]

15 Sep 2012, 17:19

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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??

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Re: Numbers divisible by 6 [#permalink]

16 Sep 2012, 00:24

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alphabeta1234 wrote:

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!

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

21 Aug 2013, 22:08

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Responding to a pm:

Quote:

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.

For more on this, check: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2010/12 ... ly-number/
Check out the discussion below the post as well.

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

21 Aug 2013, 22:58

Edit:

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

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

21 Aug 2013, 23:17

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theGame001 wrote:

How are you calculating total number of factors? Unable to understand, please help

Total number of factors of \(2^a*3^b*5^c...\) (prime factorization) is given by (a+1)(b+1)(c+1)...

Check out the link given above. It has a detailed discussion on this concept.

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

21 Aug 2013, 23:21

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theGame001 wrote:

Edit:

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.

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

22 Aug 2013, 21:27

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theGame001 wrote:

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

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Re: Numbers divisible by 6 [#permalink]

18 Oct 2013, 02:45

Bunuel wrote:

enigma123 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

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.

Answer: D.

Another solution here: new-set-of-good-ps-85440.html#p642384

Hope now it's clear.

Hi,
Any pointers about the correct way of factorization of a number?
I tried and got: 2^6*3*5^3*11

How can this be true? isn't prime factorization unique for every number?

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Re: Numbers divisible by 6 [#permalink]

18 Oct 2013, 02:53

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ronr34 wrote:

Bunuel wrote:

enigma123 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

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.

Answer: D.

Another solution here: new-set-of-good-ps-85440.html#p642384

Hope now it's clear.

Hi,
Any pointers about the correct way of factorization of a number?
I tried and got: 2^6*3*5^3*11

How can this be true? isn't prime factorization unique for every number?

Yes, prime factorization is unique: 2^6*3*5^3*11=264,000 not 264,600.

264,600 = 2646*100
2646 is divisible by 9: \(2646*100 = 294*9*100 = 98*3*9*100 = 49*2*3*9*100 = 7^2*2*3^3*2^2*5^2 = 2^3*3^3*5^2*7^2\).

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Re: Numbers divisible by 6 [#permalink]

02 Nov 2013, 01:29

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?

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Re: Numbers divisible by 6 [#permalink]

06 Nov 2013, 21:32

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ronr34 wrote:

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.

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

19 Nov 2013, 07:20

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I found this explanation easier to understand.

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

28 Aug 2016, 08:36

Bunuel wrote:

enigma123 wrote:

# 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.

gmatclubot

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

28 Aug 2016, 08:36

GMAT Problem Solving (PS) Questions

How many numbers that are not divisible by 6 divide evenly