Floor sum.....

IIT JEE 2015 Discussion Forum › Algebra questions › Floor sum.....

\hspace{-16}\bf{(1)}$\;\; Find value of $\bf{x}$ in $\bf{\lfloor 19x+97 \rfloor = 19+97x}$\\\\ $\bf{(2)}$\; \; Calculate Sum of $\bf{\sum_{k=1}^{2012}\lfloor \sqrt{k} \rfloor =}$\\\\ Where $\bf{\lfloor x \rfloor =}$ Floor Function.

#Mathematics #Algebra

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3 Answers

1057

Ketan Chandak ·2012-05-20 20:35:17

for the second question....use the result that difference between 2 consecutive squares are in AP if u sum them....
since...
(n+1)²-n²=2n+1
so in the given question n will occur 2n+1 times...
this way this question can be summed using general summation formulas...

i guess the answer is 59202.....

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1057

Ketan Chandak ·2012-05-20 21:19:37

for the first one...
[19x]=97x-78
now maybe put x as k/97 where k is a integer and then solve it....

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·2012-05-20 21:34:40

1) x=1 is the only solution

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