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Re: If 6^y is a factor of (10!)^2, What is the greatest possible [#permalink]

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19 Mar 2012, 23:47

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essarr wrote:

If 6^y is a factor of (10!)^2, What is the greatest possible value of y ?

A. 2 B. 4 C. 6 D. 8 E. 10

... I was hoping to get a better explanation, as I'm still confused about the explanation provided. Thanks!

6=2*3. Now, there will be obviously less 3's than 2's in (10!)^2, so maximum power of 3 will be limiting factor for maximum power of 6, which is y.

Finding the maximum powers of a prime number 3, in 10!: \(\frac{10}{3}+\frac{10}{3^2}=3+1=4\) (take only quotients into account). So, we have that the maximum power of 3 in 10! is 4, thus maximum power of 3 in (10!)^2 will be 8: \((3^4)^2=3^8\). As discussed 8 is the maximum power of 6 as well.

Re: If 6^y is a factor of (10!)^2, What is the greatest possible [#permalink]

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23 Mar 2012, 02:49

Expert's post

pappueshwar wrote:

hi bunuel,

i did nt understand how did u get 3's more than 2's in 10!.

i am of the view that 10! = 10*9*8*7*6*5*4*3*2*1

if we expand this we get = 2*5*3*3*2*2*2 *7 *3*2* 5* 2*2* 3* 2 *1

so there are 8 2's and 4 3's in the expansion above.

so how to interpret this and proceed

It's: "there will be obviously LESS 3's than 2's in (10!)^2, so maximum power of 3 will be limiting factor for maximum power of 6, which is y." _________________

Re: If 6^y is a factor of (10!)^2, What is the greatest possible [#permalink]

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01 Oct 2012, 06:13

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Factorial of 10 can be written as 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1

or 2 x 5 x 3 x 3 x 2 x 2 x 2 x 7 x 2 x 3 x 5 x 2 x 2 x 3 x 2 x 1

So in 10 factorial we have 08 2's and 4 three's ... Square of 10 factorial will give us 16 2's and 8 three's

to get six we know that we would have to take one 2 and one three ..the maximum number of three's we can take is 8 therefore 8 different 6's can be formed therefore max possible value of y can be 6 ..

2 x 3 2 x 3 2 x 3 2 x 3 2 x 3 2 x 3 2 x 3 2 x 3

D.. _________________

"When you want to succeed as bad as you want to breathe, then you’ll be successful.” - Eric Thomas

Re: If 6^y is a factor of (10!)^2, What is the greatest possible [#permalink]

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01 Oct 2012, 06:13

Factorial of 10 can be written as 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1

or 2 x 5 x 3 x 3 x 2 x 2 x 2 x 7 x 2 x 3 x 5 x 2 x 2 x 3 x 2 x 1

So in 10 factorial we have 08 2's and 4 three's ... Square of 10 factorial will give us 16 2's and 8 three's

to get six we know that we would have to take one 2 and one three ..the maximum number of three's we can take is 8 therefore 8 different 6's can be formed therefore max possible value of y can be 8 ..

2 x 3 2 x 3 2 x 3 2 x 3 2 x 3 2 x 3 2 x 3 2 x 3

D.. _________________

"When you want to succeed as bad as you want to breathe, then you’ll be successful.” - Eric Thomas

Re: If 6^y is a factor of (10!)^2, What is the greatest possible [#permalink]

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09 Jul 2014, 06:52

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Re: If 6^y is a factor of (10!)^2, What is the greatest possible [#permalink]

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15 Apr 2015, 06:33

H:

I was doing trailing zero questions and I got the concept but I have some difficulty in understanding a concept in which sometimes we factorize the denominator but sometimes we don't factorize. For example

If 6^y is a factor of (10!)^2, What is the greatest possible value of y ?. We are not factorizing 6 as 2 and 3 because higher power of 3 is sufficient but in this question... If N is the product of all positive integers less than 31, than what is the greatest integer k for which N/18^k is an integer? we are factoring 18 as 2^2 and 3.

Could you please help me to understand when we have to factorize and when we do not.

Re: If 6^y is a factor of (10!)^2, What is the greatest possible [#permalink]

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15 Apr 2015, 06:41

Expert's post

harrisadiq wrote:

H:

I was doing trailing zero questions and I got the concept but I have some difficulty in understanding a concept in which sometimes we factorize the denominator but sometimes we don't factorize. For example

If 6^y is a factor of (10!)^2, What is the greatest possible value of y ?. We are not factorizing 6 as 2 and 3 because higher power of 3 is sufficient but in this question... If N is the product of all positive integers less than 31, than what is the greatest integer k for which N/18^k is an integer? we are factoring 18 as 2^2 and 3.

Could you please help me to understand when we have to factorize and when we do not.

I shall be looking for your reply.

Aren't we make prime factorization in both cases? 6=2*3 and 18=2*3^2. Sorry, but your question is not very clear... _________________

Re: If 6^y is a factor of (10!)^2, What is the greatest possible [#permalink]

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15 Apr 2015, 06:45

in question number 1, we are not considering the power of 2's but in q2 we are considering the power of 2 and 3 both. I am unable to understand when we consider all the bases as we did in q2 but not in q1

Re: If 6^y is a factor of (10!)^2, What is the greatest possible [#permalink]

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15 Apr 2015, 07:04

Expert's post

harrisadiq wrote:

in question number 1, we are not considering the power of 2's but in q2 we are considering the power of 2 and 3 both. I am unable to understand when we consider all the bases as we did in q2 but not in q1

Yes, we could count only 3's in the second question too and then divide that by 2 (because of 3^2) to get the power of 18. _________________

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