p1. Katia and Mariela play the following game: In each of their three turns, Katia replaces one of the stars in the expression $\star \star\star\star\star\star\star\star\star$ for some a digit of between $1$, $2$, $3$, $4$, $5$, $6$, $7$, $8$, $9$ that has not been used before in the game. In shifts of Mariela, she replaces two of the stars with two different digits that have not been used. Katia starts playing and plays alternately. Mariela wins if the resulting number at the end of the game is divisible by $27$. Does Mariela have any to ensure the triumph? p2. Four line segments divide a rectangle into $ 8$ regions, such as illustrated in the figure. For the three marked regions, the area is as follows: $3$, $5$ and $ 8$. How much is the area of the gray quadrilateral? p3. In the interior of a square with side $ 1$, $2016$ points are marked. Prove that it is possible, regardless of the position of the points, to join them all using a path continuous whose length is not greater than $146$. p4. Find all pairs of prime numbers $(p, q)$ for which the following equality be true $$7pq^2 + p = q^3 + 43p^3 + 1$$ PS. Senior p1 was also Juniors p2.
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Tags: algebra, geometry, number theory, combinatorics, chilean NMO
28.06.2023 00:00
p4 If both $p,q$ are odd than the $LHS$ is even and the $RHS$ is odd which we cannot have. So atleast one of $p,q$ must be 2. Case 1:$p=2$ Then the equation becomes: $14q^2+2=q^3+344+1=q^3+345$ $14q^2-q^3=343$ $q^2(14-q)=343=49\cdot 7=7^2(14-7)$ So, $q=7$ Case 2:$q=2$ Then the equation becomes: $28p+p=8+43p^3+1=43p^3+9$ $A=29p=43p^3+9=B$ In this case since $p$ is odd A is odd but B is even which is a contradiction. So the only solution is $(p,q)=(2,7)$ (Same idea as #2 and #3)
09.02.2024 06:04
JanHaj wrote: p4 If both $p,q$ are odd than the $LHS$ is even and the $RHS$ is odd which we cannot have. So atleast one of $p,q$ must be 2. Case 1:$p=2$ Then the equation becomes: $14q^2+2=q^3+344+1=q^3+345$ $14q^2-q^3=343$ $q^2(14-q)=343=49\cdot 7=7^2(14-7)$ So, $q=7$ Case 2:$q=2$ Then the equation becomes: $28p+p=8+43p^3+1=43p^3+9$ $29p=43p^3+9$ $29p-43p^3=p(29-43p^2)=9$ Now it is quite clear that we should have $29-43p^2>1$ but because $p>1$(since $p$ is prime), instead we get: $29-43p^2>29-43\cdot (1)^2=-14>1$ which is absurd. Therefore the only solution is $(p,q)=(7,2)$ little mistake, the solution is $(p,q)=(2,7)$