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

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

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Updated on: 23 Jul 2013, 02:01
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How many numbers that are not divisible by 6 divide evenly into 264,600?

(A) 9
(B) 36
(C) 51
(D) 63
(E) 72

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E.

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Originally posted by enigma123 on 28 Jan 2012, 01:09.
Last edited by Bunuel on 23 Jul 2013, 02:01, edited 2 times in total.
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Re: Numbers divisible by 6  [#permalink]

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28 Jan 2012, 02:08
26
44
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.

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

Hope now it's clear.
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Re: How many numbers that are not divisible by 6 divide evenly  [#permalink]

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09 May 2013, 10:31
13
3
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
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Re: How many numbers that are not divisible by 6 divide evenly  [#permalink]

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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

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

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28 Jan 2012, 16:22
19
12
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

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: Numbers divisible by 6  [#permalink]

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15 Sep 2012, 14:36
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.

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]

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15 Sep 2012, 16:23
1
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.

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]

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15 Sep 2012, 17:19
1
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]

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16 Sep 2012, 00:24
1
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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Re: How many numbers that are not divisible by 6 divide evenly  [#permalink]

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

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21 Aug 2013, 22:08
9
1
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: http://www.veritasprep.com/blog/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]

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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]

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21 Aug 2013, 23:17
theGame001 wrote:

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]

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21 Aug 2013, 23:21
2
1
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]

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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*

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

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22 Aug 2013, 21:27
4
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: How many numbers that are not divisible by 6 divide evenly  [#permalink]

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22 Aug 2013, 22:06
Thank you for the reply Karishma, if you don't mind can we try with a smaller number? Say 18. Now these are the factors of 18

2 x 3^2 or

1 , 2 , 3 , 6 , 9 , 18

If I want to find out how many of them are divisible by 2, how do I do it using your method?

Note to self: If you dont understand after this then quit your job and stay home for GMAT.
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Re: How many numbers that are not divisible by 6 divide evenly  [#permalink]

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27 Sep 2013, 13:51
1
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)

Then just multiply: 3 * 3 * 7 = 63
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Re: Numbers divisible by 6  [#permalink]

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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.

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]

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18 Oct 2013, 02:53
1
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.

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] 18 Oct 2013, 02:53

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