We are in the right year for this!

Since $\left \lceil \sqrt{202} \right \rceil = 15$ , we have $\left \lceil \sqrt{k} \right \rceil \in \{1,2,3,...15\}$ . We note that $\left \lceil \sqrt{k} \right \rceil = n$ for $(n-1)^2 < k \le n^2$ and the number of $k=n$ is $n^2 -(n-1)^2 = 2n-1$ for $n=14$ and the number of $\left \lceil \sqrt{k} \right \rceil = 15$ is $202 - 14^2 = 202-196 = 6$ .

$\begin{aligned} \sum_{k=1} ^{202} \left \lceil \sqrt{k} \right \rceil & = \sum_{n=1} ^{14} n(2n-1) + 6(15) \\ & = \sum_{n=1} ^{14} (2n^2-n) + 90 \\ & = 2 \sum_{n=1} ^{14} n^2 - \sum_{n=1} ^{14} n + 90 \\ & = 2 \left( \frac{14(15)(29}{6} \right) - \frac{14(15)}{2} + 90 \\ & = \boxed{2015} \end{aligned}$

Although I solved the question but when I got 2015 I realized that the title gives the hint for the answer.

First, determine what ceil(sqrt(202)) is to know where to stop. ceil(sqrt(202)) = 15. So we're going to sum up the numbers between 1 and 15 a varying number of times. Be careful to stop before reaching 15^2 = 225 because that's too high (we're only going up to 202).
ceil(sqrt(1)) = 1
ceil(sqrt(2)) = 2
ceil(sqrt(3)) = 2
ceil(sqrt(4)) = 2
ceil(sqrt(5)) = 3
ceil(sqrt(6)) = 3
ceil(sqrt(7)) = 3
ceil(sqrt(8)) = 3
ceil(sqrt(9)) = 3
... There's a pattern. They each appear an odd number of times. First 1, then 3, then 5, etc. We go up to 14 (which should appear 27 times), then we use 202-14^2 = 6 to determine that we sum up 15 only 6 times. So doing 1(1) + 3(2) + 5(3) + 7(4) + 9(5)..........27(14) + 6(15) = 2015 gives us our answer.
The answer is 2015.

i took a long road.. - 1-2^2 u have 2 nos ..u can follow a pattern between consecutive squares.. multiply ad add+90(remember we still have 6 nos after 196

just guessed using the title. got it correct, however this won't work in a year from now

He say that we are in the right year for this and it is 2015... didnt even have to look at the problem...

I guessed 2015 cuz thats the current year!

i took the long road: making a table.

for k = 1 only, ceil(sqrt(k)) = 1 for k = 2 to 4, ceil(sqrt(k)) = 2 for k = 5 to 9, ceil(sqrt(k)) = 3 all the way to for k = 170 to 196, ceil(sqrt(k)) = 14 for k = 197 to 202, ceil(sqrt(k)) = 15

then for each row, multiply each unique ceil(sqrt(k)) by the number of k's belonging to it

so for k = 1, product = 1 for k = 2 to 4, product = 6 for k = 5 to 9, product = 3*(9-5+1) = 15 all the way to for k = 170 to 196, product = 378 for k = 197 to 202, product = 90

then add all these products to get 2015