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The quadratic equation x^2 - ax + 2016 = 0 has two postiive integer solutions.  Find the minimum value of a.

 Jun 13, 2020
 #1
avatar+549 
+1

\(x^2-ax+2016=0\)

fixed your LaTeX

(it was kind of weird to read)

 Jun 13, 2020
 #2
avatar+8342 
+2

We let the positive integer solutions to be \(\alpha\) and \(\beta\). Without loss of generality, assume \(\alpha < \beta\).

 

Then, 

\(x^2 - ax + 2016 = (x- \alpha)(x - \beta)\\ x^2 - ax + 2016 = x^2 - (\alpha + \beta)x + \alpha \beta\)

 

So, the question is actually asking for the minimum value of \(\alpha + \beta\) for \(\alpha \beta = 2016\). We can list the factors of 2016.

 

Factors of 2016:

\(\alpha\) 1 2 3 4 6 7 8 9 12 14 16 18 21 24 28 32 36 42
\(\beta\) 2016 1008 672 504 336 288 252 224 168 144 126 112 96 84 72 63 56 48

 

I have listed them in a unique way such that on each column, the numbers multiply to 2016.

 

We can see that a is the minimum sum of these pair of numbers, which is 42 + 48 = 90.

 Jun 13, 2020

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