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18 Mar 2019, 22:13
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
33% (02:11) correct
67%
(02:28)
wrong
based on 27
sessions
History
Date
Time
Result
Not Attempted Yet
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
(A) 278
(B) 279
(C) 280
(D) 281
(E) 282
Archit3110
26 Mar 2019, 09:14
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26 Mar 2019, 10:03
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Archit3110
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
is there really a boundary or limit to which gmat quant questions can be restricted to 
