Modular arithmetic and algebra

Algebra Level 5

{ a 101 ( m o d 2017 ) b 106 ( m o d 2017 ) c 672 ( m o d 2017 ) x r ( m o d 2017 ) x 3 + 20 a x 2 19 b x 3 c = 0 \begin{cases} a \equiv 101 \pmod{2017} \\ b \equiv 106 \pmod{2017} \\ c \equiv 672 \pmod{2017} \\ x \equiv r \pmod{2017} \\ x^3+20ax^2-19bx-3c=0 \end{cases}

Given that r r ( 0 r 2016 0\leq r \leq 2016 ) satisfies the system of equations above, find r r .


The answer is 2016.

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1 solution

Mark Hennings
Sep 20, 2016

Since x 3 + 20 a x 2 19 b x 3 c = 0 x^3 + 20ax^2 - 19bx - 3c = 0 we must have x 3 + 20 × 101 x 2 19 × 106 x 3 × 672 0 x 3 + 2020 x 2 2014 x 2016 0 ( x + 1 ) 3 = x 3 + 3 x 2 + 3 x + 1 0 \begin{array}{rcl} x^3 + 20\times101x^2 - 19\times106x - 3\times672 & \equiv & 0 \\ x^3 + 2020x^2 -2014x - 2016 & \equiv & 0 \\ (x+1)^3 \; = \; x^3 + 3x^2 + 3x + 1 & \equiv & 0 \end{array} modulo 2017 2017 , and hence we must have x 1 ( m o d 2017 ) x \equiv -1 \pmod{2017} . This means that we must have r = 2016 r = \boxed{2016} , provided that a solution can be found.

Let us look for a solution with x = 1 x=-1 . For this to work, we require that 20 a + 19 b = 3 c + 1 20a + 19b = 3c+1 , and choosing a = 2118 a=2118 , b = 106 b=106 , c = 14791 c=14791 does the trick. Many other solutions are possible.

Same solution...lovely problem and nice solution...upvoted and liked!!!!!

rajdeep brahma - 3 years ago

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