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

Question

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

rohit8865 · Answer

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

rakaisraka · Answer

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?

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Abhishek009 · Answer

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

rohit8865 · Answer

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

JeffTargetTestPrep · Answer

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

Paras96 · Answer

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

4^y = (2^2)^y = 2^2y

Successively dividing 9 by 2

9/2= 4 /2= 2 /2= 1

Power of 2 in 9! = 4+2+1=7

But we need 2y and 7 is odd, therefore 2y=6=> y = 3

Hence C

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! ?

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