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Re: If 6^y is a factor of (10!)^2, What is the greatest possible [#permalink]
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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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]
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]
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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\((10!)^2 = 10^2 * 9^2 * 8^2 * 7^2 * 6^2 * 5^2 * 4^2 * 3^2 * 2^2\)

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 6y is a factor of (10!)2 , what is the greatest possible value o [#permalink]
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gupta87 wrote:
If
6^y is a factor of (10!)^2, what is the greatest possible value of y?

Ans is 8......kindly explain.........i understood there are 8 3s but there are 14 2s


Hi,

Since you already know there are 8 3s and 14 2s, I'll start from thereon...
Each 6 is composed of 3 and 2...
So it requires equal number of 3 and 2..

But (14-8) that is 6 of 2s do not have a 3 to make a 6..
Therefore the number of 6s will depend on the integer (3or2) with lowest value thus answer is 8..
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If 6y is a factor of (10!)2 , what is the greatest possible value o [#permalink]
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gupta87 wrote:
If
6^y is a factor of (10!)^2, what is the greatest possible value of y?

Ans is 8......kindly explain.........i understood there are 8 3s but there are 14 2s



Hey,

PFB the solution.

    • \(6^y\) can be written as \(2^y * 3^y\)

    • To find the greatest possible value of y, we need to find out how many \(3\)'s are there in \((10!)^2\)

    • Now \(10! = 1 * 2 *3 * 4 * 5 * 6 * 7 * 8 * 9 * 10\)
    • Which can be written as -
      o \(10! = 2^8 * 3^4 * 5^2 * 7^1\)

    • Therefore \((10!)^2 = 2^{16} * 3^8 * 5^4 * 7^2\)
    • As we can see there are 16 2's but only 8 3's

    • But to make a \(6\) we need both one \(2\) and one \(3\).

    • Therefore, the maximum number of \(6\)'s that we can make is \(8\).

    Please note: that out of the 16 2's we can use only 8 of them and the rest 8 cannot be clubbed with any 3's, as there aren't any left.

    • Hence, the value of y = \(8\).


Thanks,
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Re: If 6^y is a factor of (10!)^2, What is the greatest possible [#permalink]
Bunuel wrote:
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: https://gmatclub.com/forum/everything-a ... 85592.html (explanation of this concept in details).

Similar questions to practice:
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Hope it helps.

­Hey what's a limiting factor and why are we considering only that number to find the 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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Akriti_Khetawat wrote:
Bunuel wrote:
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: https://gmatclub.com/forum/everything-a ... 85592.html (explanation of this concept in details).

Similar questions to practice:
https://gmatclub.com/forum/p-and-q-are- ... 09038.html
https://gmatclub.com/forum/question-abo ... 08086.html
https://gmatclub.com/forum/how-many-zer ... 00599.html
https://gmatclub.com/forum/if-n-is-the- ... 06289.html
https://gmatclub.com/forum/what-is-the- ... 05746.html
https://gmatclub.com/forum/find-the-num ... 08248.html
https://gmatclub.com/forum/find-the-num ... 08249.html
https://gmatclub.com/forum/if-d-is-a-po ... 26692.html
https://gmatclub.com/forum/if-m-is-the- ... 08971.html
https://gmatclub.com/forum/if-10-2-5-2- ... 06060.html

Hope it helps.

­Hey what's a limiting factor and why are we considering only that number to find the answer? 

­
There will be obviously less 3's than 2's in (10!)^2, so the power of 3 in (10!)^2 will determine the power of 6 in (10!)^2. Please follow the links given in that post for more.
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Re: If 6^y is a factor of (10!)^2, What is the greatest possible [#permalink]
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