The value of 12! is closest...

[BTGmoderatorLU]

The value of 12! is closest to:

A. (10^6)
B. 3(10^7)
C. 5(10^8)
D. 7(10^9)
E. 9(10^11)

The OA is C.

How can I determine this? Can eny expert help me with this PS question please? Thanks.

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Hi LUANDATO,

The prompt does NOT ask us for an EXACT calculation, so we can use a bit of estimation to determine which of the answers is relatively close to 12!. It's also worth noting that these answers are NOT 'close together' - each answer is at least 10 times GREATER than the answer that precedes it, so we have to be careful about noting all of the 'powers of 10' that we're dealing with.

To start, 12! = (12)(11)(10)(9)(8)(7)(6)(5)(4)(3)(2)(1)

Let's starting 'grouping' together numbers that form a product that is 'close' to a multiple of 10:

(12)(8) = 96 .... about 100 = 10^2
(11)(9) = 99 .... about 100 = 10^2
10 = 10 .... = 10^1
(7)(5)(3) = 105 .... about 100 = 10^2
(6)(4)(2)(1) = 48 ... about 50 = 5(10)

Thus, the product is approximately: (10^2)(10^2)(10^1)(10^2)(5)(10) = 5(10^8)

Final Answer: C

GMAT assassins aren't born, they're made,
Rich

[EconomistGMATTutor]

The value of 12! is closest to:

A. (10^6)
B. 3(10^7)
C. 5(10^8)
D. 7(10^9)
E. 9(10^11)

The OA is C.

How can I determine this? Can eny expert help me with this PS question please? Thanks.

Hi LUANDATO,
Let's take a look at your question.

We are asked to estimate the value of 12!.

$$12!\ =\ 12\times11\times10\times9\times8\times7\times6\times5\times4\times3\times2\times1$$
Let's now make groups of products that are aproximately equal to 10 or 100,
$$=\ \left(12\times8\right)\times\left(11\times9\right)\times\left(10\times1\right)\times\left(7\times6\times2\right)\times\left(4\times3\right)\times5$$
$$=\ \left(96\right)\times\left(99\right)\times\left(10\right)\times\left(84\right)\times\left(12\right)\times5$$
$$\approx\left(100\right)\times\left(100\right)\times\left(10\right)\times\left(100\right)\times\left(10\right)\times5$$
$$\approx\left(10^2\right)\times\left(10^2\right)\times\left(10^1\right)\times\left(10^2\right)\times\left(10^1\right)\times5$$
$$\approx\left(10^{2+2+1+2+1}\right)\times5$$
$$\approx\left(10^8\right)\times5$$
$$\approx5\left(10^8\right)$$

Therefore, Option D is correct.

Hope this helps.
I am available if you'd like any follow up.