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

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BTGModeratorVI wrote:
Fri May 29, 2020 6:38 am
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

Answer: D
Source: Veritas Prep
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

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We know that 2 is a factor of 4 (2^2 = 4)

So we can use the equation \(\frac{n}{2^1}+\frac{n}{2^2}+\frac{n}{2^3}...\frac{n}{2^x}\ where\ x\ is\ \inf\)

\(\frac{12}{2}+\frac{12}{2^2}+\frac{12}{2^3}\ =\ 6\ +\ 3\ +\ 1\ =\ 10\)

So we know that there are ten 2s in 12!, meaning that 2^10 is factor of 12! However, we are asked for 4 which is equal to 2^2. 2^10 = (2^2)^5 = 4^5

Answer choice D.

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BTGModeratorVI wrote:
Fri May 29, 2020 6:38 am
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

Answer: D
Solution:

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

To determine the number of 2s in 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 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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