How many numbers that are not divisible by 6 divide evenly

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

Bunuel · Accepted Answer

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.

gmacforjyoab · Answer

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 ) 
 =   +   - (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

Bunuel · Answer

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.

enigma123 · Answer

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?

alphabeta1234 · Answer

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?

EvaJager · Answer

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.

alphabeta1234 · Answer

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??

EvaJager · Answer

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!

KarishmaB · Answer

Responding to a pm: "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.

theGame001 · Answer

Edit:

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

KarishmaB · Answer

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.

KarishmaB · Answer

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.

KarishmaB · Answer

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

ronr34 · Answer

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?

Bunuel · Answer

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 .

ronr34 · Answer

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?

KarishmaB · Answer

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.

mohnish104 · Answer

I found this explanation easier to understand. https://gmatclub.com/forum/download/file.php?id=40736 btg-1.jpg

NaeemHasan · Answer

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.

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

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