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

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

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New post 19 Mar 2012, 23:15
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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
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Re: If 6^y is a factor of (10!)^2, What is the greatest possible  [#permalink]

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New post 19 Mar 2012, 23:47
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24
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.

Answer: D.

For more on this subject check: everything-about-factorials-on-the-gmat-85592.html (explanation of this concept in details).

Similar questions to practice:
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how-many-zeros-does-100-end-with-100599.html
if-n-is-the-product-of-integers-from-1-to-20-inclusive-106289.html
what-is-the-greatest-value-of-m-such-that-4-m-is-a-factor-of-105746.html
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if-10-2-5-2-is-divisible-by-10-n-what-is-the-greatest-106060.html

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

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New post 23 Mar 2012, 02:45
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
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Re: If 6^y is a factor of (10!)^2, What is the greatest possible  [#permalink]

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New post 23 Mar 2012, 02:49
1
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."
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Re: If 6^y is a factor of (10!)^2, What is the greatest possible  [#permalink]

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New post 23 Mar 2012, 22:44
wow, that link really helped thanks soooo much; it's so much simpler now that I understand the concept
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Re: If 6^y is a factor of (10!)^2, What is the greatest possible  [#permalink]

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New post 11 Apr 2012, 02:19
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!


10!= 10*9*8*7*6*5*4*3*2*1 = 2*5*3*3*2*2*2*7*2*3*5*2*2*3*2 = 2^8*3^4*5^2*7
6= 2*3

Therefore only the exponents of 2 and 3 are relevant, 2^8 or 3^4 -> higher number counts = 8 -> Answer Choice D
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Re: If 6^y is a factor of (10!)^2, What is the greatest possible  [#permalink]

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New post 11 Apr 2012, 02:22
Thank Bunuel for very clear and concise answer.
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Re: If 6^y is a factor of (10!)^2, What is the greatest possible  [#permalink]

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

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

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New post 15 Oct 2012, 08:21
The tricky thing is limiting factor : 3...
Bunuel you made it look simple..
But is it really sub 600?
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Re: If 6^y is a factor of (10!)^2, What is the greatest possible  [#permalink]

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

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New post 10 Jul 2014, 01:06
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\((10!)^2 = 10^2 * 9^2 * 8^2 * 7^2 * 6^2 * 5^2 * 4^2 * 3^2 * 2^2\)

Just concentrate on the power of 3 (Power of 2's would be more as compared to 3; so it can be ignored)

\(9^2\) = 3^4

\(6^2\) = 3^2 * 2^2

\(3^2\) = 3^2

Total powers of 3 = 8

Answer = 8
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Re: If 6^y is a factor of (10!)^2, What is the greatest possible  [#permalink]

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New post 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.

I shall be looking for your reply.
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Re: If 6^y is a factor of (10!)^2, What is the greatest possible  [#permalink]

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New post 15 Apr 2015, 06:41
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...
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Re: If 6^y is a factor of (10!)^2, What is the greatest possible  [#permalink]

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

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

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New post 17 Jan 2016, 05:42
Hi Bunuel,

Is it ok to square the 10 before computing or did I just get lucky?
\(\frac{100}{6}+ \frac{100}{36}= 6 + 2\)
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Re: If 6^y is a factor of (10!)^2, What is the greatest possible  [#permalink]

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

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New post 17 Jan 2016, 06:22
Bunuel wrote:
Icecream87 wrote:
Hi Bunuel,

Is it ok to square the 10 before computing or did I just get lucky?
\(\frac{100}{6}+ \frac{100}{36}= 6 + 2\)


Why do you want to square?
What do you square?
How is 10/3+10/3^2 squared equal to \(\frac{100}{6}+ \frac{100}{36}\)?
How do you get \(\frac{100}{6}= 6\)?


Good question. I somehow managed to omit the 1 (from 16) on 100/6 to only maintain 6 to my liking. And I also just squared the 10 initially not the whole fraction and I dind't get why 3 was used instead of 6. Anyway, I am having trouble understanding all the powers.

Does this mean that if we had 15 instead of 6 we would have taken the biggest of the primes of 15 to find the powers: \(\frac{10!^2}{15^x}\)then we would use \(\frac{10}{5}= 2*2\) so as to get \(15^4\) ?
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Re: If 6^y is a factor of (10!)^2, What is the greatest possible  [#permalink]

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New post 24 Jan 2017, 20:18
1) The problem is asking us to find the number of 2*3 prime factors in the sequence of consecutive integers
2) 10/3=3; 10/9=1; 3+1=4
3) Since the sequence is squared 4*2=8

The greatest possible value of y is 8
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Re: If 6^y is a factor of (10!)^2, What is the greatest possible   [#permalink] 24 Jan 2017, 20:18

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