p1. In how many ways can $2019$ be written as the sum of consecutive positive integers? p2. On a table there are $2019$ chips. Two players take turns drawing chips. The first player to play can draw any odd number of chips between $ 1$ and $99$. The other player can draw any even number of chips between$ 2$ and $100$. The player who can not continue playing loses. Determine if any of the players has a winning strategy. p3. Consider a rectangle $ABCD$ with $| AB | > | BC |$ and let $E$ be the midpoint of $CD$ side. $F$ is chosen in $CD$ such that $| CF | = | BC |$. Suppose $AC \perp BE$. Prove that $| AB | = | BF |$. p4. Each face of a cube of dimensions $1000\times 1000\times 1000$ is divided into $1000^2$ unit squares of $1\times 1$. Determine the largest number of little squares units that can be marked on the cube in such a way that no pair of them share a side in common.
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Tags: algebra, geometry, combinatorics, number theory, chilean NMO
BackToSchool
21.10.2021 17:03
Find prime factorization of $2019$ as $3 \times 673$.
The only way to write $2019$ as the sum of consecutive positive integers shall be $672+673+674$.
BackToSchool
21.10.2021 17:21
Note that $$2019 = 101 \times 19 + 100$$The first player has winning strategy.
The first player should draw 99 chips on his/her first draw. In his/her following turns, he/she always make sure that total $101$ chips be draw in each round "second play + first player", thus leaving $1$ chip at the end for which the second player can do nothing.
imas_713_a
02.03.2022 09:46
BackToSchool wrote:
Find prime factorization of $2019$ as $3 \times 673$.
The only way to write $2019$ as the sum of consecutive positive integers shall be $672+673+674$.
I think you have missed some cases. Like, $1009+1010$ We can write the sum of 1 to N as $$\frac{n^{2}+n}{2}$$ Now, the sum of some consecutive integer will always be in the form of, $$\frac{n^{2}+n}{2} - \frac{k^{2}+k}{2}\\ \Rightarrow \frac{(n-k)(n+k+1)}{2} \\ $$Now its easy to see cases by applying $$ \frac{(n-k)(n+k+1)}{2} =2019 $$And see the factorisation of $4038$