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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
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!