16. A flea is initially at the origin of the Cartesian plane and can move to integer coordinate points by choosing one of these three moves each time:

- from the point $(x, y)$ it jumps to the point $(x+2, y+4)$;
- from the point $(x, y)$ it jumps to the point $(x, y+5)$;
- from the point $(x, y)$ it jumps to the point $(x-2, y-9)$.

How many paths, realizable by the flea with its moves, lead from the origin $(0,0)$ to the point $(0,2016)$?
(A) none
(B) precisely 1
(C) a number between 10 and 30
(D) a number between 30 and 60
(E) infinitely many

## Unione Matematica Italiana PROGETTO OLIMPIADI DI MATEMATICA

Ministry of Education, University and Research

Scuola Normale Superiore

## T2 Archimedes Games - Biennio Competition

## November 23, 2016

- The test consists of 16 problems. Each question is followed by five answers indicated by the letters $(\mathbf{A}),(\mathbf{B}),(\mathbf{C}),(\mathbf{D}),(\mathbf{E})$: only one of these answers is correct, the other 4 are incorrect.
- Each correct answer is worth 5 points, each incorrect answer is worth 0 points, and each problem left unanswered is worth 1 point.
- For each of the problems, you must write the letter corresponding to the answer you believe is correct in the grid provided below. No erasures or corrections are allowed on the grid. The use of any type of calculator or communication device is not permitted.

## The time you have to complete the test is 110 minutes

 Good work and have fun! NAME

LAST NAME

class:
![](https://cdn.mathpix.com/cropped/2024_04_17_ccb981cb4d80b90fdff2g-3.jpg?height=94&width=1226&top_left_y=1160&top_left_x=144)
Problem 16. The answer is (A).

If the flea makes $a$ moves of the first type, $b$ of the second, and $c$ of the third, the total displacement in the horizontal direction will be $2a - 2c$. We want to move from the point $(0,0)$ to the point $(0,2016)$, and therefore $2a - 2c$ must be equal to 0, from which $a = c$. The number of moves of the first and third type in our path must be the same. The total displacement in the vertical direction will then be

$$
4a + 5b - 9c = 4a - 5b - 9a = 5(b - a)
$$

which is necessarily a multiple of 5. It will therefore be impossible to start from the point $(0,0)$ and reach $(0,2016)$, since the required vertical displacement is 2016, which is not a multiple of 5.

(Problem proposed by C. Casamento Tumeo)