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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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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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Posts: 59075
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
15
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).

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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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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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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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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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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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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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01 Oct 2012, 06:13
4
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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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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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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15 Oct 2012, 08:24
mindmind wrote:
The tricky thing is limiting factor : 3...
Bunuel you made it look simple..
But is it really sub 600?

It's ~700 level question. Tag changed.
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Re: If 6^y is a factor of (10!)^2, What is the greatest possible  [#permalink]

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10 Jul 2014, 01:06
1
1
$$(10!)^2 = 10^2 * 9^2 * 8^2 * 7^2 * 6^2 * 5^2 * 4^2 * 3^2 * 2^2$$

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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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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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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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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15 Apr 2015, 07:04
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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Re: If 6^y is a factor of (10!)^2, What is the greatest possible  [#permalink]

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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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17 Jan 2016, 05:48
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 $$\frac{100}{6}= 6$$?
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Re: If 6^y is a factor of (10!)^2, What is the greatest possible  [#permalink]

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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 $$15^4$$ ?
Thanks
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Re: If 6^y is a factor of (10!)^2, What is the greatest possible  [#permalink]

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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
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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