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In the equation n^2 – kn + 16 = 0, n is an integer.

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In the equation n^2 – kn + 16 = 0, n is an integer. [#permalink] New post 14 Mar 2019, 10:50
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In the equation \(n^2 – kn + 16 = 0\), \(n\) is an integer. Which of the following could be the value of \(k\)?

Indicate all such values

A. 8

B. 15

C. –17
[Reveal] Spoiler: OA

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Re: In the equation n^2 – kn + 16 = 0, n is an integer. [#permalink] New post 20 Mar 2019, 16:19
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This is a great one for picking numbers.

With just a bit of fiddling, it's not too difficult to make 8 work. When n = 4, then the equation works.

Now here's the thing with the bigger numbers. When k and n are equal, then the left is equal to 16: too large. If n is larger than k, then the left will also be too large. But if n is too much less than k, then kn will become more than 16 larger than n². So we really only have to check one or two values.

For example, when n = 14 and k = 15, then we get: 14² - (14*15) + 16 = 2, not quite.

But when we go down further, to n = 13 and k = 15, we get: 13² - (13*15) + 16 = -10.

And then as n decreases, the answer will just get lower and lower. So the equation never works.

We'll follow the same process with -17. If n = -16, then: (-16)² - (-16*-17) + 16 = 0. Boom.
Re: In the equation n^2 – kn + 16 = 0, n is an integer.   [#permalink] 20 Mar 2019, 16:19
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In the equation n^2 – kn + 16 = 0, n is an integer.

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