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Exponent

by Aman verma » Sat Feb 20, 2010 2:09 am
Q : If 2 ^n ( 2 raised to the power n) can exactly divide z ! such that the quotient is an odd positive integer , then the value of n which is NOT possible is :

I. 43 II. 44 III. 45

a) I only

b) III only

c)Both I and II

d)Both II and III

e)I, II and III
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by Aman verma » Sat Feb 20, 2010 3:14 am
Ans. [spoiler]e) I,II and III. [/spoiler]
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by harsh.champ » Sat Feb 20, 2010 3:33 am
Aman verma wrote:Q : If 2 ^n ( 2 raised to the power n) can exactly divide z ! such that the quotient is an odd positive integer , then the value of n which is NOT possible is :

I. 43 II. 44 III. 45

a) I only

b) III only

c)Both I and II

d)Both II and III

e)I, II and III
z! = z(z-1)(z-2)................(3)(2)(1)

If z is odd,

Z! = z(z-2)(z-4)..................(3)(1) x [2 x (1 x 2 x 3 x 4 x5 x..............x (z-1)/2 ) ]

Now,carefully notice the bold-faced no.s.
Each of them generate sum exponent of 2.
The above transformation of z! can be carried out further and further and in each such transformation a factor of 2 will come out.

Now, my doubt is since we don't know whether z>n or n<z how can we determine how many exponents of 2 can be formed??
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by shashank.ism » Sat Feb 20, 2010 8:38 am
harsh.champ wrote:
Aman verma wrote:Q : If 2 ^n ( 2 raised to the power n) can exactly divide z ! such that the quotient is an odd positive integer , then the value of n which is NOT possible is :

I. 43 II. 44 III. 45

a) I only

b) III only

c)Both I and II

d)Both II and III

e)I, II and III
z! = z(z-1)(z-2)................(3)(2)(1)

If z is odd,

Z! = z(z-2)(z-4)..................(3)(1) x [2 x (1 x 2 x 3 x 4 x5 x..............x (z-1)/2 ) ]

Now,carefully notice the bold-faced no.s.
Each of them generate sum exponent of 2.
The above transformation of z! can be carried out further and further and in each such transformation a factor of 2 will come out.

Now, my doubt is since we don't know whether z>n or n<z how can we determine how many exponents of 2 can be formed??
you can form it till you get (z-1)/2 as a n integer...as soon as u get it as fraction...there is ur ans... though ur process of solving the question is not feasible according to me..
Aman could u please give Original solution here/.....thanks
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by Ian Stewart » Sat Feb 20, 2010 11:10 am
Aman verma wrote:Ans. [spoiler]e) I,II and III. [/spoiler]
In this question, we want to find the highest power of 2 which will divide z!. That is, we want to count how many 2's there are 'inside' z!. I've explained in more detail in earlier posts how this can be done, but hopefully the brief explanation below will make sense. If we take, for example, 48!, and write out the product 1*2*3*...*46*47*48, we will find there are:

24 multiples of 2
12 multiples of 4, each of which gives us one additional 2
6 multiples of 8, each of which gives us one additional 2
3 multiples of 16, each of which gives us one additional 2
1 multiple of 32, which gives us one additional 2

Thus 48! is divisible by 2^46.

If we did the same for 46!, we would find that it was only divisible by 2^42 (since we are leaving out 48 from the product, and 48 is divisible by 2^4). Thus, the highest power of 2 which divides 46! is 2^42, and the highest power of 2 which divides 48! is 2^46; it is never the case that 2^43, 2^44 or 2^45 is the highest power of 2 dividing z!.

This question seems a bit too involved for the GMAT, though there are certainly simpler questions which test a similar concept. Discovering that 46! and 48! are the factorials to investigate might require some trial and error, which makes this question too time consuming for the test. The wording of the question is also bad, since it makes it sounds as though there is only one impossible value for n; there are infinitely many impossible values of n, and the question should ask 'which of the following could not be equal to n?'
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by Aman verma » Sat Feb 20, 2010 11:38 am
Unfortunately, I didn't had the solution ,only had the net answer which is E.

But fortunately, Ian has provided the solution and the underlying logic.

Thanks to all the people for posting their views. I hope you all will continue to provide support to my queries.

If I have solutions to the questions I will definitely post them.
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