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

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by Jay@ManhattanReview » Mon Mar 13, 2017 11:52 pm
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
Hi ziyuenlau,

We have 200! divisible by 33^n. We see that 33^n = (3*11)^n = 3^n*11^n, where 3 and 11 are prime factors of 33.

So, 200! is divisible by 3^n*11^n.

We want the greatest possible value of integer n. We see that 3 << 11, so there would be far more 3s than 11s, so the greatest possible value of integer n would be pegged by 11 than by 3.

Let's find out how many 11s are there in 200! = 1*2*3*....... 198*199*200.

The first integer divisible by 11 is 11, and the last one would be the quotient of 200/11 = 18th.

We see that a square of 11 = 121 is also there in 200!, so there are in fact two 11s in 121. Thus, total number of 11s = 18 + 1 = 19.

The greatest possible value of integer n = 19.

The correct answer: D

Hope this helps!

Relevant book: Manhattan Review GMAT Number Properties Guide

-Jay
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by [email protected] » Tue Mar 14, 2017 10:24 am
Hi ziyuenlau,

This question can be solved with PRIME FACTORIZATION.

200! is a really big number: (200)(199)(198).....(2)(1). The GMAT would NEVER expect you to calculate that, but you should be able to break it down into relevant prime factors.

33 can be prime factored into (3)(11). There are LOTS of 3s in 200! (3, 6, 9, 12, etc.) so we just have to find all of the 11s in 200! to answer this question. While you could list them all out (re: 11, 22, 33, etc.), that's not really necessary as long as you don't miss the fact that 121 = (11)(11). The largest multiple of 11 in this calculation is 198 = (11)(18). So there are eighteen 11s + one extra 11 (in 121) - for a total of nineteen 11s.

Final Answer: D

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by DavidG@VeritasPrep » Tue Mar 14, 2017 10:35 am
For practice with a similar question, see here: https://www.beatthegmat.com/if-x-is-the- ... 70022.html
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by Jeff@TargetTestPrep » Thu Mar 16, 2017 12:13 pm
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

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by Matt@VeritasPrep » Thu Mar 16, 2017 8:04 pm
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.
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