4. We denote (for the sake of simplifying the writing) the branch with number $n$ as $c_{n}$. Then, on $c_{1}$ the cricket jumps 0 times, on $c_{2}$ the cricket jumps 2 times, on $c_{n}$ the cricket jumps $n$ times. In total, before flying off $c_{n}$, the cricket has jumped $1+2+3+\ldots+n=\frac{n(n+1)}{2}$ times.

From $\frac{n(n+1)}{2} \leq 100<\frac{n(n+1)}{2}+n+1$ we deduce, possibly by trial, $n=13$. (That is, the last jump from branch $c_{13}$ is the 91st, so the cricket jumps the 100th time when it is on branch $c_{14}$).

NOTE

- Any correct solution is graded accordingly.
4. In the morning, a sparrow flies to a branch of a tree and chirps once, then flies to a second branch and chirps twice, then to a third branch and chirps three times, and so on (on the 20th branch it chirps 20 times...). On which branch is the sparrow when it chirps for the 100th time in the morning?

## NOTE

- All subjects are mandatory;
- Each subject is graded out of 7 points;
- No points are awarded by default;
- The actual working time is 2 hours from the moment the subject is received.


## Mathematics Olympiad  Local phase - February 16, 2014

## Grade 6 - grading rubric