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Bunuel
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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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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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Bunuel
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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Bunuel
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

12! = (12)(11)(10)(9)(8)(7)(6)(5)(4)(3)(2)(1)
= (2)(2)(3)(11)(2)(5)(9)(2)(2)(2)(7)(2)(3)(5)(2)(2)(3)(2)(1)
= (2¹⁰)(3)(11)(5)(9)(7)(3)(5)(3)(1)
= (2²)⁵(3)(11)(5)(9)(7)(3)(5)(3)(1)
= (4)⁵(3)(11)(5)(9)(7)(3)(5)(3)(1)

So, 4⁵ is a factor of 12!, which means the correct answer is D

Cheers,
Brent
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