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What is the greatest value of y such that 4^y is a factor of 9! ?

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What is the greatest value of y such that 4^y is a factor of 9! ?  [#permalink]

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New post 08 Dec 2016, 11:57
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What is the greatest value of y such that 4^y is a factor of 9! ?  [#permalink]

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New post 08 Dec 2016, 19:02
1
Bunuel wrote:
What is the greatest value of y such that 4^y is a factor of 9! ?

A. 5
B. 4
C. 3
D. 1
E. 0



4^y=2^(2y)

no. of 2's in 9!
9/2=4
9/2^2=2
9/2^3=1
total= 4+2+1=7

so as 2y=7 we get y=3

Ans C
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Re: What is the greatest value of y such that 4^y is a factor of 9! ?  [#permalink]

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New post 09 Dec 2016, 02:25
1
The formula for such problems is like
9 /4= 2
9/4^2=0
Total = 2
However answer should be 3 if we actually count it. Where am I going wrong?

Posted from my mobile device
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Re: What is the greatest value of y such that 4^y is a factor of 9! ?  [#permalink]

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New post 09 Dec 2016, 10:18
2
Bunuel wrote:
What is the greatest value of y such that 4^y is a factor of 9! ?

A. 5
B. 4
C. 3
D. 1
E. 0


\(9! = 9*8*7*6*5*4*3*2*1\)

Or, \(9! = 3^2*2^3*7*2*3*5*2^2*3*2*1\)

Or, \(9! = 2^7*3^4*5*7\)

Now, \(2^7 = 4^3*2\)

Thus, we have the greatest value of y = 3 , hence answer will be (C)

rakaisraka hope its clear with you ...

Further I suggest you go through the concept once again to clear your doubts here math-number-theory-88376.html#p666609
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Re: What is the greatest value of y such that 4^y is a factor of 9! ?  [#permalink]

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New post 09 Dec 2016, 10:35
1
rakaisraka wrote:
The formula for such problems is like
9 /4= 2
9/4^2=0
Total = 2
However answer should be 3 if we actually count it. Where am I going wrong?

Posted from my mobile device


rakaisraka

when u r finding 4^y means u have to count every 2's ..
suppose if it was 10! then it must have 1*2*...*6...*10
then it has 6=2*3 && 10=2*5
where one no. 2 from 6 and one no. 2 from 10 also counted as a 4 in 10!

let me make more clear if u have to find 6^y in X!
as 6=2*3
then u have to count every 2 and every 3 in X!
and the minimum pair of 2&3 will make the answer

hope it is clear
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Re: What is the greatest value of y such that 4^y is a factor of 9! ?  [#permalink]

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New post 12 Dec 2016, 17:16
1
Bunuel wrote:
What is the greatest value of y such that 4^y is a factor of 9! ?

A. 5
B. 4
C. 3
D. 1
E. 0


Since 4 = 2^2, we are actually trying to determine the largest value y such that 2^(2y) is a factor of 9!.

Let’s first determine the number of factors of 2 within 9!. To do that, we can use the following shortcut in which we divide 9 by 2, and then divide the quotient of 9/2 by 2 and continue this process until we can no longer get a nonzero integer as the quotient.

9/2 = 4 (we can ignore the remainder)

4/2 = 2

2/2 = 1

Since 1/2 does not produce a nonzero quotient, we can stop.

The final step is to add up our quotients; that sum represents the number of factors of 2 within 9!.

Thus, there are 4 + 2 + 1 = 7 factors of 2 within 9!

However, we are not asked for the number of factors of 2; instead we are asked for the number of factors of 4. We see that 7 factors of 2 will produce 3 factors of 4.

Answer: C
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Re: What is the greatest value of y such that 4^y is a factor of 9! ?  [#permalink]

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Re: What is the greatest value of y such that 4^y is a factor of 9! ?   [#permalink] 17 Jan 2020, 02:54
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