Declining to insanity in 2017!

Algebra Level 5

$\large{405f(x) = \sqrt[x]{\sum_{k=1}^{2017}\sum_{i=1}^{3}{k^i}}}$ Find the smallest positive integer ${C}$ such that $\left \lfloor {f\left(\frac{10}{7}\right)}\right \rceil$ $+C$ is a palindrome .

Notation: $\lfloor \cdot \rceil$ denotes the nearest integer function .

The answer is 800.

1 solution

Akeel Howell
Jan 13, 2017

$\large{\sum_{k=1}^{2017}\sum_{i=1}^{3}{k^i} = \sum_{k=1}^{2017}{k^3}+\sum_{k=1}^{2017}{k^2}+\sum_{k=1}^{2017}{k}} \\ = \dfrac{(2017)^2(2018)^2}{4}+\dfrac{2017(2018)(4035)}{6}+\dfrac{2017(2018)}{2} = 4144587049347 \\ \therefore \large{405f\left(\dfrac{10}{7}\right) = \sqrt[\left(\dfrac{10}{7}\right)]{\sum_{k=1}^{2017}\sum_{i=1}^{3}{k^i}}} = 679574205.303640... \\ \Longrightarrow \lfloor f\left(\dfrac{10}{7}\right)\rceil = 1677961$ The smallest palindromic number greater than $1677961$ is $1678761$ and thus, $\large C = 1678761-1677961=\boxed{800.}$

I got the same sum as you have. However taking the appropriate roots, I get an irrational number which is 679574205.30364007312396166049323. The sum 4144587049347 is not a perfect 10th root of any number. The question could be modified to account for this.

Ajinkya Shivashankar - 4 years, 4 months ago

Thanks for bringing this to my attention. I modified the question so that we have $c+\lfloor f(\frac{10}{7})\rceil$ .

Akeel Howell - 4 years, 4 months ago

Declining to insanity in 2017!

The answer is 800.

1 solution

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