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The number 5^867 is between 2^2013 and 2^2014. How many pairs of integ

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The number 5^867 is between 2^2013 and 2^2014. How many pairs of integ  [#permalink]

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New post 18 Mar 2019, 22:13
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The number 5^867 is between 2^2013 and 2^2014. How many pairs of integers (m,n) are there such that\(1 ≤ m ≤ 2012\) and \(5^n < 2^m\) < 2^ m+2 < 5^n+1?

(A) 278
(B) 279
(C) 280
(D) 281
(E) 282

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The number 5^867 is between 2^2013 and 2^2014. How many pairs of integ  [#permalink]

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New post 26 Mar 2019, 10:03
1
Archit3110 wrote:
Noshad wrote:
The number 5^867 is between 2^2013 and 2^2014. How many pairs of integers (m,n) are there such that\(1 ≤ m ≤ 2012\) and \(5^n < 2^m\) < 2^ m+2 < 5^n+1?

(A) 278
(B) 279
(C) 280
(D) 281
(E) 282


chetan2u ; please advise on solution to this question.. :(



Hi Archit,

Let us first understand the question..
\(5^n < 2^m< 2^ {m+2} < 5^{n+1}\) .. So we are looking for consecutive power of 5( n to n+1) that contains 3 consecutive powers of 2 ( m to m+2)
If you want to do it, it can have two solutions..
(I) Calculation intensive....
Try to get a pattern ..
\(5^0,2^1,2^2,5^1,2^3,2^4,5^2,2^5,2^6,5^3,2^7,2^8,2^9,5^3,2^{10},2^{11}, 5^4\)
I have not gone beyond this, but you likely to see a pattern after a certain time

(II) A more elegant way
The series \(5^0,2^1,2^2,5^1,2^3,2^4,5^2,2^5,2^6,5^3,2^7,2^8,2^9,5^3,2^{10},2^{11}, 5^4\) shows that
we can have 2 powers of 2( as between \(5^1\) and \(5^2\), we have \(2^3, 2^4\)) or
3 powers of 2 between consecutive powers of 5( as between \(5^3\) and \(5^4\), we have \(2^7,2^8,2^9\)).
Let there be x gaps that has 2 powers of 2 and y gaps that have 3 powers of 2.
SO when we add these GAPS, they should be equal to the total power of 5, which is 867.=>\(x+y=867\)
But in x gaps there are two powers of 2, that is 2x powers of 2, and y gaps have three power of 2, that is 3y, so when we add them we should get 2013 => \(2x+3y=2013\)
Multiply \(x+y=867\)by 2 and subtract from \(2x+3y=2013..=> 2x+3y-2(x+y)=2013-2*867....y=2013-1734=279\)

B
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Re: The number 5^867 is between 2^2013 and 2^2014. How many pairs of integ  [#permalink]

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New post 26 Mar 2019, 09:14
Noshad wrote:
The number 5^867 is between 2^2013 and 2^2014. How many pairs of integers (m,n) are there such that\(1 ≤ m ≤ 2012\) and \(5^n < 2^m\) < 2^ m+2 < 5^n+1?

(A) 278
(B) 279
(C) 280
(D) 281
(E) 282


chetan2u ; please advise on solution to this question.. :(
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Re: The number 5^867 is between 2^2013 and 2^2014. How many pairs of integ  [#permalink]

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New post 26 Mar 2019, 10:12
chetan2u wrote:
Archit3110 wrote:
Noshad wrote:
The number 5^867 is between 2^2013 and 2^2014. How many pairs of integers (m,n) are there such that\(1 ≤ m ≤ 2012\) and \(5^n < 2^m\) < 2^ m+2 < 5^n+1?

(A) 278
(B) 279
(C) 280
(D) 281
(E) 282


chetan2u ; please advise on solution to this question.. :(



Hi Archit,

Let us first understand the question..
\(5^n < 2^m< 2^ {m+2} < 5^{n+1}\) .. So we are looking for consecutive power of 5( n to n+1) that contains 3 consecutive powers of 2 ( m to m+2)
If you want to do it, it can have two solutions..
(I) Calculation intensive....
Try to get a pattern ..
\(5^0,2^1,2^2,5^1,2^3,2^4,5^2,2^5,2^6,5^3,2^7,2^8,2^9,5^3,2^{10},2^{11}, 5^4\)
I have not gone beyond this, but you likely to see a pattern after a certain time

(II) A more elegant way
The series \(5^0,2^1,2^2,5^1,2^3,2^4,5^2,2^5,2^6,5^3,2^7,2^8,2^9,5^3,2^{10},2^{11}, 5^4\) shows that
we can have 2 powers of 2( as between \(5^1\) and \(5^2\), we have \(2^3, 2^4\)) or
3 powers of 2 between consecutive powers of 5( as between \(5^3\) and \(5^4\), we have \(2^7,2^8,2^9\)).
Let there be x gaps that has 2 powers of 2 and y gaps that have 3 powers of 2.
SO when we add these GAPS, they should be equal to the total power of 5, which is 867.=>\(x+y=867\)
But in x gaps there are two powers of 2, that is 2x powers of 2, and y gaps have three power of 2, that is 3y, so when we add them we should get 2013 => \(2x+3y=2013\)
Multiply \(x+y=867\)by 2 and subtract from \(2x+3y=2013..=> 2x+3y-2(x+y)=2013-2*867....y=2013-1734=279\)

B


chetan2u ; appreciate and thanks for the solution..
honestly its a question which has really gone beyond my scope of understanding of gmat quant course... :( ... do such questions really come in actual exam :dazed is there really a boundary or limit to which gmat quant questions can be restricted to :|
also how did you arrive on the highlighted part .. gaps would be 867?
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Re: The number 5^867 is between 2^2013 and 2^2014. How many pairs of integ  [#permalink]

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New post 26 Mar 2019, 10:20
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[quote="Archit3110"]

You are NOT likely to see such questions in GMAT..
now we are looking at \(5^0...5^1...5^2...5^3........5^{866}...5^{867}\), so 867 gaps
Now \((5^0-2^1,2^2-5^1)...(5^1-2^3,2^4-5^2)...(5^2-2^5,2^6-5^3)....\) are the x gaps that have 2 power of 2
\(5^3-2^7,2^8,2^9-5^4.......\) are the y gaps that have 3 power of 2
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Re: The number 5^867 is between 2^2013 and 2^2014. How many pairs of integ  [#permalink]

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New post 26 Mar 2019, 10:23
chetan2u wrote:
Archit3110 wrote:

You are NOT likely to see such questions in GMAT..
now we are looking at \(5^0...5^1...5^2...5^3........5^{866}...5^{867}\), so 867 gaps
Now \((5^0-2^1,2^2-5^1)...(5^1-2^3,2^4-5^2)...(5^2-2^5,2^6-5^3)....\) are the x gaps that have 2 power of 2
\(5^3-2^7,2^8,2^9-5^4.......\) are the y gaps that have 3 power of 2


really glad to see the bold part :)
ok understood ; thanks a a lot ..
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Re: The number 5^867 is between 2^2013 and 2^2014. How many pairs of integ   [#permalink] 26 Mar 2019, 10:23
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