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06 Jan 2022, 06:00
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64% (02:16) correct
36%
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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) \(2^{15}\)
08 Jan 2022, 03:40
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\(2^{20} = 2^{15}x + y\)
\(2^5 = x + \frac{y}{2^{15}}\)
As \(y\) is non negative, \(y\) can be \(0\), then \(x = 2^5 = 32\)
Hence \(|x-y|\) minimum value can be \(|x-0| = 32\)
\(2^5 = x + \frac{y}{2^{15}}\)
As \(y\) is non negative, \(y\) can be \(0\), then \(x = 2^5 = 32\)
Hence \(|x-y|\) minimum value can be \(|x-0| = 32\)
06 Jan 2022, 11:15
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|x−y| represents the distance between points x and y on a number line.
Before we move onto the problem, lets take an example.
Let the distance between two points on a number line be 5.
Assume that one of the point is x and the other is y (also x > y). We know that the distance between these two points is 5. This statement can be represented mathematically as -
x - y = 5
Now if x is twice as far from y, we can denote the distance between them as 2x - y. And logically we know that the distance between them will now be 10.
In the form of an equation, this relationship is written as -
2x - y = 10
The distance between x and y is thus \(\frac{1}{2}\) * 10
Similarly if x is thrice as far from y, the distance is denoted by 3x - y
3x - y = 15
However as we have seen in the previous example, the distance between x and y can be denoted as \(\frac{1}{3}\) * 15
If we apply the same concept to the above problem we can now read the problem as
"The distance between \(2^{15}\) times x and y is \(2^{20}\)"
\(2^{20}\) = \(2^{15}\)x + y
Therefore the distance between x and y can denoted as -
|x - y| = \(\frac{1 }{ 2^{15} }* 2^{20}\)
|x - y| = \(2^5\)
IMO - C
Before we move onto the problem, lets take an example.
Let the distance between two points on a number line be 5.
Assume that one of the point is x and the other is y (also x > y). We know that the distance between these two points is 5. This statement can be represented mathematically as -
x - y = 5
Now if x is twice as far from y, we can denote the distance between them as 2x - y. And logically we know that the distance between them will now be 10.
In the form of an equation, this relationship is written as -
2x - y = 10
The distance between x and y is thus \(\frac{1}{2}\) * 10
Similarly if x is thrice as far from y, the distance is denoted by 3x - y
3x - y = 15
However as we have seen in the previous example, the distance between x and y can be denoted as \(\frac{1}{3}\) * 15
If we apply the same concept to the above problem we can now read the problem as
"The distance between \(2^{15}\) times x and y is \(2^{20}\)"
\(2^{20}\) = \(2^{15}\)x + y
Therefore the distance between x and y can denoted as -
|x - y| = \(\frac{1 }{ 2^{15} }* 2^{20}\)
|x - y| = \(2^5\)
IMO - C
