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

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ziyuenlau wrote: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 the maximum value of n is 19.

Answer: D

An easier way, with proper superscripts to boot:

We want 200! / 33� to clear all the 33s in the denominator. With that in mind, we want to COUNT the 33s in the numerator.

There will always be more 3s than 11s, since we get a three in every third term in the factorial, but an 11 only every eleventh term.

With that in mind, we only need to count the 11s:

11, 22, 33, 44, 55, 66, 77, 88, 99, 110, 121, 132, 143, 154, 165, 176, 187, 198

We've got 11*1 to 11*18, and 11*11 counts TWICE, giving us 19 in all.