Absolute greater than 10 10

Number Theory Level 3

If n ∈ Z n\in\mathbb Z

∣ 80 n + 1 ∣ > 10 \large \left| \dfrac{80}{n+1} \right|>10

n n satisfy the above inequality. Find the sum of all ∣ n ∣ |n| .

Inspiration


The answer is 56.

Md Mehedi Hasan by Md Mehedi Hasan , 22, Bangladesh

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1 solution

Md Mehedi Hasan
Dec 9, 2017

∣ 80 n + 1 ∣ > 10 6400 ( n + 1 ) 2 > 100 squaring both side ( n + 1 ) 2 6400 < 1 100 ( n + 1 ) 2 < 64 n 2 + 2 n − 63 < 0 ( n + 9 ) ( n − 7 ) < 0 \left|\dfrac{80}{n+1} \right|>10\\ \dfrac{6400}{(n+1)^2}>100\quad{\boxed{\color{#3D99F6}\text{squaring both side}}}\\ \dfrac{(n+1)^2}{6400}<\dfrac{1}{100}\\ (n+1)^2<64\\ n^2+2n-63<0\\ (n+9)(n-7)<0

The solution is { n ∈ Z : − 9 < n < 7 , n ≠ − 1 } \{n\in\mathbb Z:-9<n<7,n\neq-1\}

So, sum= ( ∑ n = − 8 6 ∣ n ∣ ) − ∣ − 1 ∣ = 57 − 1 = 56 \left(\sum_{n=-8}^{6}{|n|}\right)-|-1|\\=57-1\\=\boxed{56}

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