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If p is completely divided by the number 17, and [m]p = x^2 y[/m],
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02 Sep 2022, 08:13
02 Sep 2022, 08:13
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39% (02:04) correct
60% (01:40) wrong based on 28 sessions
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If “p” is completely divided by the number 17, and \(p = x^2 ∗ y\), where x and y are distinct prime numbers, which of these numbers must be divisible by 289?
A. \(x^2\)
B. \(y^2\)
C. \(xy\)
D. \(x^2y^2\)
E. \( x^3y\)
A. \(x^2\)
B. \(y^2\)
C. \(xy\)
D. \(x^2y^2\)
E. \( x^3y\)
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Re: If p is completely divided by the number 17, and [m]p = x^2 y[/m],
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03 Sep 2022, 01:36
03 Sep 2022, 01:36
3
Given that “p” is completely divided by the number 17, and \(p = x^2 ∗ y\), where x and y are distinct prime numbers and we need to find which of these numbers must be divisible by 289
p is divisible by 17 => p is a multiple of 17 => p can be written as 17*k = 17k (where k is an integer)
Now, \(p = x^2 ∗ y\) = 17k
=> \(x^2 ∗ y\) = 17k
Now, 17 is a prime number => either x is 17 or y is 17 (as both are distinct prime numbers)
In, any case x*y will be a multiple of 17 as either x or y is equal to 17
=> xy will be a multiple of 17 => xy = 17t (where t is an integer)
Squaring both sides we get
\((xy)^2\) = \((17t)^2\)
=> \(x^2 * y^2\) = 289 * \(t^2\)
=> \(x^2 * y^2\) will be divisible by 289
So, Answer will be D
Hope it helps!
p is divisible by 17 => p is a multiple of 17 => p can be written as 17*k = 17k (where k is an integer)
Now, \(p = x^2 ∗ y\) = 17k
=> \(x^2 ∗ y\) = 17k
Now, 17 is a prime number => either x is 17 or y is 17 (as both are distinct prime numbers)
In, any case x*y will be a multiple of 17 as either x or y is equal to 17
=> xy will be a multiple of 17 => xy = 17t (where t is an integer)
Squaring both sides we get
\((xy)^2\) = \((17t)^2\)
=> \(x^2 * y^2\) = 289 * \(t^2\)
=> \(x^2 * y^2\) will be divisible by 289
So, Answer will be D
Hope it helps!
Re: If p is completely divided by the number 17, and [m]p = x^2 y[/m],
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10 May 2024, 04:42
10 May 2024, 04:42
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