Orange08 wrote:

If 60! is written out as an integer, with how many consecutive 0’s will that integer end?

A. 6

B. 12

C. 14

D. 42

E. 56

60! = (1)(2)(3)(4)(5)(6) . . . (57)(58)(59)(60)

KEY CONCEPT: For every pair of one 2 and one 5, we get a product of 10, which accounts for one zero at the end of the integer. So, the question is "How many pairs of one 2 and one 5 are "hiding" in the product?"

Well, there is no shortage of 2's hiding in the product. In fact, there are FAR MORE 2's than 5's. So, all we need to do is determine how many

5's are hiding in the product.

60! = (1)(2)(3)(4)(5)(6) . . . (57)(58)(59)(60)

= (1)(2)(3)(4)(

5)(6)(7)(8)(9)(2)(

5)(11)...(3)(

5)...(4)(

5)...(

5)(

5)...(6)(

5)...(7)(

5)...(8)(

5)...(9)(

5)...(2)(

5)(

5)...(11)(

5)...(56)(57)(58)(59)(12)(

5)

In total, there are 14

5's hiding in the product.

And there are MORE THAN 14 2's hiding in the product.

So, there are

14 pairs of 2's and 5'2, which means the integer ends with 14 zeros

Answer:C

Cheers,

Brent

_________________

Brent Hanneson – GMATPrepNow.com

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