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.
Answer choice (a).
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!