Neither quadratic, nor cubic, it's 2011-ic!
p ( x ) = x 2 0 1 1 + n = 1 ∑ 2 0 1 0 a n x 2 0 1 1 − n + 1
Let x 1 , x 2 , x 3 , x 4 , ⋯ , x 2 0 1 1 be the roots of the monic polynomial above and let y 1 , y 2 , y 3 , y 4 , ⋯ , y 2 0 1 1 be defined as:
y 1 y 2 y 3 ⋮ y 2 0 1 0 y 2 0 1 1 = x 1 x 2 x 3 . . . x 1 0 = x 2 x 3 x 4 . . . x 1 1 = x 3 x 4 x 5 . . . x 1 2 = ⋮ = x 2 0 1 0 x 2 0 1 1 x 1 . . . x 8 = x 2 0 1 1 x 1 x 2 . . . x 9
If the value of ( y 1 y 2 y 3 . . . y 2 0 1 1 ) 2 0 1 1 2 0 1 0 can be expressed as lo g 1 0 a . Find a .
The answer is 10.
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1 solution
It is an overrated question
But still it was a good one.
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I know this is a simple problem. At first glance, I too thought of it as a problem that was gonna take a lot of paperwork but when I found the way to solve this, I was laughing at myself that I was thinking so much for a simple problem only because it was constructed so well.
Relevant wiki: Vieta's Formula - Higher Degrees
From p ( x ) , we get,
i = 1 ∏ 2 0 1 1 x i = − 1 Using Vieta’s formulas
We observe that,
( y 1 ⋅ y 2 ⋅ y 3 . . . y 2 0 1 1 ) = x 1 ⋅ x 2 ⋅ x 3 . . . x 2 0 1 1 1 0 = ( i = 1 ∏ 2 0 1 1 x i ) 1 0 = ( − 1 ) 1 0 = 1
Hence,
( y 1 ⋅ y 2 ⋅ y 3 . . . y 2 0 1 1 ) 2 0 1 1 2 0 1 0 = ( 1 ) 2 0 1 1 2 0 1 0 = 1
And,
lo g 1 0 a = 1 ⟹ a = 1 0