Waste less time on Facebook — follow Brilliant.
×

Pre-RMO 2014/6

What is the smallest possible natural number \(n\) for which the equation \(x^2 - nx + 2014 = 0\) has integer roots?


This note is part of the set Pre-RMO 2014

Note by Pranshu Gaba
2 years, 7 months ago

No vote yet
1 vote

Comments

Sort by:

Top Newest

Let the roots of the equation be \(\alpha\) and \(\beta\).

Then \(n = \alpha + \beta\) and \(2014 = \alpha \beta\).

The possible integral pairs of \((\alpha, \beta)\) are \((1, 2014), (2, 1007), (19, 106),\) and \((38, 53)\).

Therefore, the possible values of \(n\) are \(2015, 1009, 125\) and \(91\). The minimum value is \(91\), so \(n = \boxed{91}\) Pranshu Gaba · 2 years, 7 months ago

Log in to reply

Is the answer 91? Abhineet Nayyar · 2 years, 7 months ago

Log in to reply

@Abhineet Nayyar Yes , it is 91 :D Krishna Ar · 2 years, 7 months ago

Log in to reply

@Krishna Ar Yayy!! lol:D:D Abhineet Nayyar · 2 years, 7 months ago

Log in to reply

91 Gaurav Singh · 1 year, 9 months ago

Log in to reply

91 Akshat Sharda · 1 year, 9 months ago

Log in to reply

91 Sahil Nare · 2 years ago

Log in to reply

The discriminant of the above equation should be a perfect square for the equation to have integral roots. So the smallest value of n for which it is a perfect square is 91. Sanchit Ahuja · 2 years, 7 months ago

Log in to reply

Explain? Shiv Kumar · 2 years, 7 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...