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# Find the largest power of 12 that can divide 200! a. 97 b.

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Senior Manager
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Find the largest power of 12 that can divide 200! a. 97 b. [#permalink]

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07 Jan 2004, 13:34
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Find the largest power of 12 that can divide 200!

a. 97
b. 98
c. 17
e. None of these

***I got answer choice (e), but official answer disagrees. Just making sure here. The answer should be 88. RIght?
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07 Jan 2004, 13:54
I can't think of an easy way to do this question.

In one through 200

100 numbers are multiples of two
50 numbers are multiples of four
25 numbers are multiples of eight
12 numbers are multiples of 16
6 numbers are multiples of 32
3 numbers are multiples of 64
1 number is a multiple of 128

2^197

66 numbers are multiples of three
22 numbers are multiples of nine
7 numbers are multiples of 27
3 numbers are multiples of 51
1 number is a multiple of 153

3^98

since you need two twos and a three to make 12, and there are more than twice as many twos as threes, the threes are the limiting parameter.

98?

I could be terribly wrong...
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07 Jan 2004, 14:09
An simpler method of attacking this question:

Since 12 is an even #, take the prime factorization: 2^2 *3
Now, 200/2 leaves 'Q'uotient 100, 100/2 leaves Q 50, 50/2 leaves Q 25, 25/2 leaves Q 12, 12/2 leaves Q 6, 6/2 leaves Q 3, 3/2 leaves Q 1
Now, add up all the quotients: 100+50+25+12+6+3+1 = 197
2^2 is two twoos, so divide 197 by 2 = Qoutient 98

Now do the same with 200/3. Add up all quotients. If the total is larger than 98. 98 is the answer. If the total is smaller than 98, the smaller # is the answer. I realized my mistake. I made an error with computing the quotients above when I did it before. The official answer is correct. You will find that when you get the quotients from 200/3 the total will yield 97.
97 is smaller than 98, so it is the answer.
_________________

Pls include reasoning along with all answer posts.
****GMAT Loco****
Este examen me conduce jodiendo loco

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07 Jan 2004, 16:02
sunniboy007 wrote:
An simpler method of attacking this question:

Since 12 is an even #, take the prime factorization: 2^2 *3
Now, 200/2 leaves 'Q'uotient 100, 100/2 leaves Q 50, 50/2 leaves Q 25, 25/2 leaves Q 12, 12/2 leaves Q 6, 6/2 leaves Q 3, 3/2 leaves Q 1
Now, add up all the quotients: 100+50+25+12+6+3+1 = 197
2^2 is two twoos, so divide 197 by 2 = Qoutient 98

Now do the same with 200/3. Add up all quotients. If the total is larger than 98. 98 is the answer. If the total is smaller than 98, the smaller # is the answer. I realized my mistake. I made an error with computing the quotients above when I did it before. The official answer is correct. You will find that when you get the quotients from 200/3 the total will yield 97.
97 is smaller than 98, so it is the answer.

Is that really a simple approach? I'm not trying to be sarcastic but I still don't understand your approach.

This is how i view the problem:
200! assume - it's a big number!! obviously - you need something close to a super computer to calculate this number.
but we do know that it's even, it's greater than 12^98 (the largest answer choice) and it ends in lots of zeros (thus, it's even!).

so any number that ends with 0,2,5,4 and 8 will divide evenly into this number.

just like powers of 2 we can do the same for powers of 12

2^1=2 2^5=32 12^1=12
2^2=4 2^6=64 12^2=144
2^3=8 2^7=128 etc.
2^4=16 2^8=256

Thus use the powers of 2 as a proxy for the powers of 12 since the pattern repeats itself after the 4th power

thus the 97th power ends in 8 - thus 12^97 can divide evenly into 200! the 98th power ends in 6 so that would end in .6666667

Thus 12^97 is your answer - one of those problems that you can solve under 1.5 mins. if you see the patterns!
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07 Jan 2004, 16:12
Nice work, Titleist.
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07 Jan 2004, 16:22
stoolfi wrote:
Nice work, Titleist.

Thanks Stoolfi. You haven't changed your avatar yet!
07 Jan 2004, 16:22
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