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by gmatmachoman » Mon Mar 08, 2010 6:06 am
Is it E
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by bpgen » Mon Mar 08, 2010 6:24 am
Hello gmatmachoman, could you please explain in any algebraic way..or just to try putting values and check..?
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by gmatmachoman » Mon Mar 08, 2010 6:35 am
Bro, try to expand 10 in terms of (2 *5) with its powers and try to get all the multiples in terms of 2 raised to some power.

Now try to match the LHS&RHS

at n=90 we will equal the value on both sides.
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by ramyaravindran » Mon Mar 08, 2010 6:40 am
So you are saying 2^90 > 2^15 * 5 ^15. How did you deduce that 2^75 does not satisfy the above inequality..How did you evaluate 2^75 < 2^15*5^15...Can someone provide a better explanation of how to compare numbers like these.
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by bpgen » Mon Mar 08, 2010 6:41 am
Sorry, I'm stuck at 2^(n-15)>5^15...then?
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by rohan_vus » Mon Mar 08, 2010 6:53 am
2^10>1000 ,as 2^10 = 1024

so (2^10)^5 > (1000)^5
=> 2^50 > (10^3)^5
=> 2^50 > 10^15

n >=50 satisfies the inequality for sure

So IMO C
Last edited by rohan_vus on Mon Mar 08, 2010 7:04 am, edited 1 time in total.
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by firdaus117 » Mon Mar 08, 2010 6:57 am
We need the minimum value of n at which the above equality holds.
We can solve it using options.
OptionA n=30=2*15
2^n=2^(2*15)=4^15 < 10^15 Rejected
Option B n=45=3*15
2^n=2^(3*15)=8^15 < 10^15 Rejected
Option C n=60=4*15
2^n=2^(4*15)=16^15 > 10^15 Accepted
[spoiler]Hence,option C
Note that we are to choose the minimum n among the given options and not the minimum real value at which the inequality holds true.The situation would have changed if "none of the above" would have featured among the options.[/spoiler]
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by kstv » Mon Mar 08, 2010 7:10 am
If 2^n > 10^15 then what is the minimum value of n at which the equality holds.

2^n > 10^15 = 2^15*5^15
Express 5 as mulitple of 2 = 2³/x where x is 8/5
10^15 = 2^15*(2³/x)^15 = 2^(15+45) * (1/x) ^15
2^n > 2^60*(5/8)^15
the expression (5/8)^15 is a very small value
so when n = 60 , 2^n will be = 2^60
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by harshavardhanc » Mon Mar 08, 2010 7:30 am
firdaus117 wrote:We need the minimum value of n at which the above equality holds.
We can solve it using options.
OptionA n=30=2*15
2^n=2^(2*15)=4^15 < 10^15 Rejected
Option B n=45=3*15
2^n=2^(3*15)=8^15 < 10^15 Rejected
Option C n=60=4*15
2^n=2^(4*15)=16^15 > 10^15 Accepted
[spoiler]Hence,option C
Note that we are to choose the minimum n among the given options and not the minimum real value at which the inequality holds true.The situation would have changed if "none of the above" would have featured among the options.[/spoiler]
I too got C.

My approach.

10^15 is 1 followed by 15 Zeros, i.e 16 digits in all.

So, basically you are looking for a power of two which is a 16 digit #.

Now, we all know that 2^10 is 1024.

If you multiply any # with 1024, you get a # which has 3 digits more than the original one ( as you get 3 zeros added to a number when you multiply a number with 1000).


so, 2^10 * 1024 = 2 ^ 20 will have 4 + 3 = 7 digits

similarly, 2^30 will have 7+3 = 10 digits

2^40 will have 10 + 3 = 13 digits

and 2^50 will have 16 digits.

now, we don't have 50 as an answer choice. So, taking the next higher : 60.
Last edited by harshavardhanc on Mon Mar 08, 2010 7:33 am, edited 1 time in total.
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by gmatmachoman » Mon Mar 08, 2010 7:32 am
hey ramya..

sorry for that blunder i made..

10 ^ 15=(2*5)^15

=(2 ^15) * (4+1)^15

==(2 ^15) *{(2^2) ^15[1+0.25]^15}----( plz use a pencil & paper if u r confused here with my brackets i used, I am basically trying to reduce in terms of 2 or say multimles of 2)

=(2 ^15) *(2 ^30){(1.25)^15}

= =(2 ^45) *{(2 ^15)(0.625) ^15} -----------------------------( 15 th power of 0.625 will be less than 1....)

== 2^ 60

So the minimum value of n= 60

C is the right answer...

Earlier in my previous post i said 90..sorry guys..that was a big blunder while do addintion!!
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by Fiver » Mon Mar 08, 2010 8:36 am
rohan_vus wrote:2^10>1000 ,as 2^10 = 1024

so (2^10)^5 > (1000)^5
=> 2^50 > (10^3)^5
=> 2^50 > 10^15

n >=50 satisfies the inequality for sure

So IMO C
I like this approach, but i'd like to take it forward.
With the above apprach we arrive at the conclusion:
2^50 > 10^15.
This means n>=50
But the question is what is the minimum value for n,
so just to ensure we do not fall into a trap, it is good to rule out the possibility of n=45

Hence i suggest that we compliment Rohan's method with the following:
2^n > 10^15
2^n > 2^15 * 5^15
and 4^15 < 5^15
hence 2^30 < 5^15
hence n> 15+30
n>45.
Now we know that of the given options 60 undoubtedly qualifies as the minimum value of n.
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