What is the greatest possible value of integer n if 200! is

[BTGmoderatorAT]

What is the greatest possible value of integer n if 200! is divisible by 33^n

A)6
B)12
C)18
D)19
E)66

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Hi ardz24,

We're told that 200! is divisible by 33^N. We're asked for the LARGEST possible value of N. This question is ultimately about 'prime factorization.'

33 = (3)(11), and 200! will have a LOT of 3s and a certain number of 11s within it, so we have to essentially 'count up' all of the 11s that exist within 200!

11 = (1)(11)
22 = (2)(11)
33 = (3)(11)
121 = (11)(11)
...
198 = (18)(11)

It's worth noting that 121 actually has TWO 11s in it, so the total is 18+1 = 19 and the largest possible value of N is 19.

Final Answer: D

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

[Scott@TargetTestPrep]

What is the greatest possible value of integer n if 200! is divisible by 33^n

A)6
B)12
C)18
D)19
E)66

We need to determine the maximum value of n such that 200!/(33^n) is an integer. We must remember that an integer is divisible by 33 if it's divisible by both 11 and 3. Thus, we must determine the number of factors of 11 and 3 in 200!. However, since we know there are fewer factors of 11 than factors of 3 in 200!, we can find the number of factors of 11 and thus be able to determine the maximum value of n.

To determine the number of factors of 11 in 200!, we can use the following shortcut in which we divide 200 by 11, and then divide the quotient of 200/11 by 11 and continue this process until we no longer get a nonzero quotient:

200/11 = 18 (we can ignore the remainder)

18/11 = 1 (we can ignore the remainder)

Since 1/11 does not produce a nonzero quotient, we can stop.

The final step is to add up our quotients; that sum represents the number of factors of 11 in 200!. Thus, there are 18 + 1 = 19 factors of 11 in 200!, and so the maximum value of n is 19.

Answer: D