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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
+549
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$$x^2-ax+2016=0$$

(it was kind of weird to read)

Jun 13, 2020
#2
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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