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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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18 Mar 2019, 22:13
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65% (hard)

Question Stats:

30% (01:52) correct 70% (02:12) wrong based on 20 sessions

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

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26 Mar 2019, 09:14
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

Math Expert
Joined: 02 Aug 2009
Posts: 8006
The number 5^867 is between 2^2013 and 2^2014. How many pairs of integ  [#permalink]

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26 Mar 2019, 10:03
1
Archit3110 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

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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26 Mar 2019, 10:12
chetan2u wrote:
Archit3110 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

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 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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Joined: 02 Aug 2009
Posts: 8006
Re: The number 5^867 is between 2^2013 and 2^2014. How many pairs of integ  [#permalink]

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26 Mar 2019, 10:20
1
[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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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 ..
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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