p1. Is there any natural number n such that $n^2 + 5n + 1$ is divisible by $49$ ? Explain. p2. It is known that the parabola $y = ax^2 + bx + c$ passes through the points $(-3,4)$ and $(3,16)$, and does not cut the $x$-axis. Find all possible abscissa values for the vertex point of the parabola. p3. It is known that $T.ABC$ is a regular triangular pyramid with side lengths of $2$ cm. The points $P, Q, R$, and $S$ are the centroids of triangles $ABC$, $TAB$, $TBC$ and $TCA$, respectively . Determine the volume of the triangular pyramid $P.QRS$ . p4. At an event invited $13$ special guests consisting of $ 8$ people men and $5$ women. Especially for all those special guests provided $13$ seats in a special row. If it is not expected two women sitting next to each other, determine the number of sitting positions possible for all those special guests. p5. A table of size $n$ rows and $n$ columns will be filled with numbers $ 1$ or $-1$ so that the product of all the numbers in each row and the product of all the numbers in each column is $-1$. How many different ways to fill the table?
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Tags: geometry, algebra, combinatorics, number theory, indonesia juniors
DottedCaculator
04.11.2021 23:26
$n^2+5n+1=(n-1)^2+7n$. This means $n\equiv1\pmod7$, which is impossible since $49\nmid7n$
biscuit02
05.11.2021 10:02
$y = ax^2 + bx + c$
$4=9a-3b+c$
$16=9a+3b+c$
Solving gives $b=2$ and $9a+c=10$
Since parabola does not cut x-axis, $b^2-4ac<0\implies a>\frac{1}{c}\implies 9a+c>\frac{9}{c}+c$
$c^2-10c+9<0$
$1<c<9$
x-coordinate of vertex $= -\frac{b}{2a}= -\frac{1}{a}$
Also $9a+c=10 \implies -\frac{1}{a}=\frac{9}{c-10}$
Possible x-coordinate of vertex $=\boxed{-\frac{9}{8},-\frac{9}{7} ,-\frac{3}{2} ,-\frac{9}{5} ,-\frac{9}{4} ,-3 ,-\frac{9}{2}}$
Quote: p3. It is known that $T.ABC$ is a regular triangular pyramid with side lengths of $2$ cm. The points $P, Q, R$, and $S$ are the centroid of triangle $ABC$, respectively. triangle $TAB$, triangle $TBC$, and triangle $TCA$. Determine the volume of the pyramid $P.QRS$ . Is there missing information ?
Arrange the 8 men in $8!$ ways
Choose and arrange 5 positions out of the 9 possible positions for women in $9P5$ ways
Total = $\boxed{8! \times 9P5}$
For $1\times 1$ table, number of ways $= 1 = 1!$
For $2\times 2$ table, number of ways $= 2 = 2!$
For $3\times 3$ table, number of ways $= 6 = 3!$
For $n\times n$ table, number of ways $= \boxed{n!}$
parmenides51
05.11.2021 22:29
@above, wording has been updated, I hope it's ok now
ind1v
06.11.2021 08:07
$\text{Suppose that there exists a natural number n that satisfies the condition}$
$n^2 + 5n + 1 = n^2 + 5n - 6 + 7 = (n + 6)(n - 1) + 7 \equiv 0 \pmod{7}$
$n + 6 \equiv n - 1 \pmod{7} \rightarrow n \equiv 1 \pmod{7} \rightarrow (n + 6)(n - 1) \equiv 0 \pmod{49}$
$\rightarrow 7 \equiv 0 \pmod{49} \text{ which is clearly not correct}$