Absolute Madness

Algebra Level 5

∑ k = 1 n ∣ x k ∣ = 19 + ∣ ∑ k = 1 n x k ∣ . \displaystyle \sum_{k=1}^{n} |x_{k}|=19+\left|\sum_{k=1}^{n} x_{k}\right|\,.

If x i x_i is real, i ∈ N i\in\mathbb{N} ,such that ∣ x i ∣ < 1 |x_{i}|<1 .

What is the smallest possible value of n n such that it satisfy the equation above?


The answer is 20.

Rohit Udaiwal by Rohit Udaiwal , 20, India

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1 solution

Sal Gard
May 11, 2016

The absolute value of the sum of the x's is 19 less than the sum of the absolute values. Now let the positive numbers' sum be x and the negative numbers' y. x+y=n and x-y=n-19, implying y=9.5. Since y=9.5, z+9.5=19+abs(z-9.5). To minimize z, let abs(z-9.5)=9.5-z. Now we have z=9.5. The nearest whole value that works is 10 and the nearest value to y=9.5 is 20, so 10+10=20.

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