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Simple... I think

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Re: Simple... I think

by Stuart@KaplanGMAT » Thu Mar 20, 2008 12:46 pm
HarvardDreamin wrote:q3) What is the maximum integer value of M for which 12!/2^M is an integer?

A. 2 B. 4 C. 6 D. 10 E. 12
We need to figure out how many times "2" goes into the expression in the numerator.

12! = 12 * 11 * 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1

Let's start by ignoring the numbers that aren't multiples of 2. For this question, we only care about:

12 * 10 * 8 * 6 * 4 * 2

Next, let's factor into primes:

2*2*3 * 2*5 * 2*2*2 * 2*3 * 2*2 * 2

Count 'em up, we have 10 2s: choose (D).
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by HarvardDreamin » Fri Mar 21, 2008 8:51 am
Many Thanks
ON MY WAY TO HBS......
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by ritz » Fri Mar 21, 2008 6:25 pm
I dont know if you are still searching for the answer, but here is a much easier & faster way to answer these type of questions.
Q. q3) What is the maximum integer value of M for which 12!/2^M is an integer?
A. 2 B. 4 C. 6 D. 10 E. 12
Sol:-
find out how many times
Step 1:- 2 divides 12!
a) 6
Step 2:- 2 square divides 12!
b) 3
Step 3:- 2 cube divides 12!
c) 1
Step 4:- 2 to the power 4 divides 12!
d) 0
STOP
add
a b c d & that is your answer which is 10.
You can solve any such problems this way.
You can also solve that questions like how many zeroes are there at the end of 20!

regards
Ritz[/b]
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by mmukher » Sat Mar 22, 2008 1:43 pm
ritz wrote: You can also solve that questions like how many zeroes are there at the end of 20!

Ritz[/b]

Not sure I understand ritz, how would we find out how many zeroes at the end of 20! ?

Many thanks
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by ritz » Sat Mar 22, 2008 1:53 pm
20/5= 4
20/25= 0
As we will surely have more 2's than 5,(zeroes will be created by the multiplication of 5 & 2) so we can safely say that there will be 4 zeroes at the end of 20| (sorry, no factorial sign on my handheld)
Let me know if you have some doubt still...
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by mmukher » Sat Mar 22, 2008 2:02 pm
Nice!... thanks a lot pal.
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