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

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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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12 Feb 2017, 04:25
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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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12 Feb 2017, 04:33
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,
Saquib
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Re: If 6^y is a factor of (10!)^2, What is the greatest possible [#permalink]

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12 Feb 2018, 16:08
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Re: If 6^y is a factor of (10!)^2, What is the greatest possible   [#permalink] 12 Feb 2018, 16:08

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