⎩ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎧ a ≡ 1 0 1 ( m o d 2 0 1 7 ) b ≡ 1 0 6 ( m o d 2 0 1 7 ) c ≡ 6 7 2 ( m o d 2 0 1 7 ) x ≡ r ( m o d 2 0 1 7 ) x 3 + 2 0 a x 2 − 1 9 b x − 3 c = 0
Given that r ( 0 ≤ r ≤ 2 0 1 6 ) satisfies the system of equations above, find r .
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Same solution...lovely problem and nice solution...upvoted and liked!!!!!
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Since x 3 + 2 0 a x 2 − 1 9 b x − 3 c = 0 we must have x 3 + 2 0 × 1 0 1 x 2 − 1 9 × 1 0 6 x − 3 × 6 7 2 x 3 + 2 0 2 0 x 2 − 2 0 1 4 x − 2 0 1 6 ( x + 1 ) 3 = x 3 + 3 x 2 + 3 x + 1 ≡ ≡ ≡ 0 0 0 modulo 2 0 1 7 , and hence we must have x ≡ − 1 ( m o d 2 0 1 7 ) . This means that we must have r = 2 0 1 6 , provided that a solution can be found.
Let us look for a solution with x = − 1 . For this to work, we require that 2 0 a + 1 9 b = 3 c + 1 , and choosing a = 2 1 1 8 , b = 1 0 6 , c = 1 4 7 9 1 does the trick. Many other solutions are possible.