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 Skip odd numbers

\(12: 2*2*3\)

\(10: 2*5\)

\(8: 2*2*2\)

\(6: 2*3\)

\(4: 2*2\)

\(2: 2*1\)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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