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# Engineering

Sum of all integral values of a belongs [1, 500] for which the equation as a [x]^3 + x - a = 0 has a solution ([.] denotes the greatest integer function) is ?

Answers (1)
S Sayak

[x]3 + x - a = 0

x = a - [x]3 

a is an integer

[x]3 is an integer

x is an integer

a = x + [x]3 

for x = 0, a = 0

for x = 1, a = 2

for x = 7, a = 350

for x = 8, a = 520

Therefore the required sum is

\\\sum_{1}^{7}(n+n^{3})\\ =\left ( \frac{(n)(n+1)}{2}+\frac{(n)^{2}(n+1)^{2}}{4} \right )\\ =\frac{7\times 8}{2}+\frac{7^{2}\times 8^{2}}{4}\\ =28+784\\ =812

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