Sum to 2016

Algebra Level 4

$\left \lfloor{\sqrt{6}}\right \rfloor +\left \lfloor{\sqrt{7}}\right \rfloor+\left \lfloor{\sqrt{8}}\right \rfloor+\ldots+\left \lfloor{\sqrt{n}}\right \rfloor=2016$

Find the value of the integer $n$ , satisfying the above equation.

Inspiration

The answer is 216.

1 solution

Atul Shivam
Jan 21, 2016

hint: upon simplifying it will give $2+2+2+2+3+3+3+3+3+3+3+4+4+...=2016$ now use method of difference to get $n$ equal to $\boxed{216}$

Sum to 2016

The answer is 216.

1 solution

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