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Posts: 2506
Given Kudos: 1051
GPA: 3.39
If 2^{20} = 2^{15}x+y, where x and y are non-negative integ
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28 Oct 2020, 01:20
28 Oct 2020, 01:20
Expert Reply
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Question Stats:
42% (00:45) correct
57% (01:09) wrong based on 21 sessions
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If \(2^{20} = 2^{15}x+y,\) where x and y are non-negative integers, what is the minimum possible value of \(|x − y|\)?
(A) 0
(B) 1
(C) \(2^5\)
(D) \(2^{10}\)
(E) \(2^{15}\)
(A) 0
(B) 1
(C) \(2^5\)
(D) \(2^{10}\)
(E) \(2^{15}\)
Re: If 2^{20} = 2^{15}x+y, where x and y are non-negative integ
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21 Nov 2020, 01:20
21 Nov 2020, 01:20
GeminiHeat wrote:
If \(2^{20} = 2^{15}x+y,\) where x and y are non-negative integers, what is the minimum possible value of \(|x − y|\)?
(A) 0
(B) 1
(C) \(2^5\)
(D) \(2^{10}\)
(E) \(2^{15}\)
(A) 0
(B) 1
(C) \(2^5\)
(D) \(2^{10}\)
(E) \(2^{15}\)
I am getting option C = 2^5 when |x+(y/2^15)|
I got that by dividing both the sides of stem equation by 2^15.
2^20/2^15 = x2^15/2^15 + y/2^15.
2^5 = x + (y/2^15).
Will be happy to know how you arrived at minimum possible value of |x−y| as 2^5.
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Joined: 10 Apr 2015
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Re: If 2^{20} = 2^{15}x+y, where x and y are non-negative integ
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22 Nov 2020, 06:27
22 Nov 2020, 06:27
2
GeminiHeat wrote:
If \(2^{20} = 2^{15}x+y,\) where x and y are non-negative integers, what is the minimum possible value of \(|x − y|\)?
(A) 0
(B) 1
(C) \(2^5\)
(D) \(2^{10}\)
(E) \(2^{15}\)
(A) 0
(B) 1
(C) \(2^5\)
(D) \(2^{10}\)
(E) \(2^{15}\)
Take: \(2^{20} = 2^{15}x+y\)
Subtract \(2^{15}x\) from both sides to get: \(2^{20} - 2^{15}x=y\)
Factor: \(2^{15}(2^{5}- x)=y\)
Rewrite as: \(y=2^{15}(32- x)\)
Since x and y are non-negative integers, we can conclude that \(0<x≤32\) (i.e, if x > 32, y is negative)
Notice that, if \(x=1\), \(y=2^{15}(32- 1)=2^{15}(31)\), which is a little bit less than \(2^{20}\)
In other words, the maximum value of y is a little bit less than \(2^{20}\)
Also, if \(x=32\), \(y=2^{15}(32-32)=2^{15}(0)=0\)
So, the minimum value of y is \(0\) AND \(x=32\) is the maximum value of x
As such, the minimum value \(|x − y|\) is \(|32 − 0|=32=2^5\)
Answer: C
Cheers,
Brent
