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What is the greatest value y for which 4^y is a factor of 12!?

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What is the greatest value y for which 4^y is a factor of 12!?  [#permalink]

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New post 30 Jan 2018, 00:11
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Question Stats:

57% (00:49) correct 43% (00:43) wrong based on 143 sessions

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Re: What is the greatest value y for which 4^y is a factor of 12!?  [#permalink]

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New post 30 Jan 2018, 00:17
D

2^(2*y) is factor of 12!

There are 10 2s in 12! Hence five 4s.


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Re: What is the greatest value y for which 4^y is a factor of 12!?  [#permalink]

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New post 30 Jan 2018, 00:57
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D

4^5 is the biggest factor in terms of 4^y of 12!. Therefore, y's maximum value can be 5.
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Re: What is the greatest value y for which 4^y is a factor of 12!?  [#permalink]

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New post 31 Jan 2018, 17:09
1
Bunuel wrote:
What is the greatest value y for which 4^y is a factor of 12!?

A. 2
B. 3
C. 4
D. 5
E. 6


Let’s first determine the number of 2’s in 12!.

To determine the number of 2s within 12!, we can use the following shortcut in which we divide 12 by 2, then divide the quotient of 12/2 by 2 and continue this process until we no longer get a nonzero quotient.

12/2 = 6

6/2 = 3

3/2 = 1 (we can ignore the remainder)

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

The next step is to add up our quotients; that sum represents the number of factors of 2 within 12!.

6 + 3 + 1 = 10

We see that there are ten 2’s in 12!; however, since 4 = 2^2, and since 10/2 = 5, we see that there are five 4’s in 12!.

Answer: D
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Re: What is the greatest value y for which 4^y is a factor of 12!?  [#permalink]

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New post 08 Jul 2018, 05:16
Expand 12! and find how many 4s are there in the expansion. Make sure to split even numbers into 2*(half of that number). Because two of those numbers combine to form a new 4 which is not readily visible in the expansion of the number 12!
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What is the greatest value y for which 4^y is a factor of 12!?  [#permalink]

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New post 10 Jul 2018, 09:09
1
Bunuel wrote:
What is the greatest value y for which 4^y is a factor of 12!?

A. 2
B. 3
C. 4
D. 5
E. 6

I. Another shortcut.*
The question basically asks how many 4s are in (12*11*10*...). Shortcut steps:

(1) Prime factorize \(4 = 2^2\)

(2) Divide \(n=12\) (without !) by increasing powers of \(2\) until you cannot get an integer as a quotient. Ignore remainders.
\(\frac{12}{2^1}=6\)
\(\frac{12}{2^2}=3\)
\(\frac{12}{2^3}=1\) (ignore remainder)
\(\frac{12}{2^4}\) => will not work. 12/16 is not an integer

(3) Add the quotient results from above: \((6+3+1) = 10\)
There are 10 factors of 2 in 12! (There are ten 2s.)

(4) How many 4s (= \(y\))?
\(2^{10}=(2^2)^5=4^5\)
Or, \(2*10=4*5\)
\((4*4*4*4*4)=4^5=4^{y}\)
\(y=5\)

The greatest value of \(y\) is 5

Answer D

II. List factors of 2 in the factorial
If we need to find 4s, then we need to find 2s. \(4=2*2\)

If a number is not prime, list its prime factors and search for those.
If we check for 4s only, we will miss the 2 in 10, for example.

Skip odd numbers (we know that odd numbers have no factors of 2)
\(12: 2*2*3\)
\(10: 2*5\)
\(8: 2*2*2\)
\(6: 2*3\)
\(4: 2*2\)
\(2: 2*1\)


Count the 2s. 10 total.
10 factors of 2 = 5 factors of 4
\((2*10)= 20 = (4*5)\)

Answer D

*Shortcut - For the theory and an example, See Bunuel , Number Theory, Factorials, here
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