p1. Given a set of $n$ the first natural number. If one of the numbers is removed, then the average number remaining is $21\frac14$ . Specify the number which is deleted. p2. Ipin and Upin play a game of Tic Tac Toe with a board measuring $3 \times 3$. Ipin gets first turn by playing $X$. Upin plays $O$. They must fill in the $X$ or $O$ mark on the board chess in turn. The winner of this game was the first person to successfully compose a sign horizontally, vertically, or diagonally. Determine as many final positions as possible, if Ipin wins in the $4$th step. For example, one of the positions the end is like the picture on the side. p3. Numbers $ 1$ to $10$ are arranged in pentagons so that the sum of three numbers on each side is the same. For example, in the picture next to the number the three numbers are $16$. For all possible arrangements, determine the largest and smallest values of the sum of the three numbers. p4. Define $$S(n)=\sum_{k=1}^{n}(-1)^{k+1}\,\, , \,\, k=(-1)^{1+1}1+(-1)^{2+1}2+...+(-1)^{n+1}n$$Investigate whether there are positive integers $m$ and $n$ that satisfy $S(m) + S(n) + S(m + n) = 2011$ p5. Consider the cube $ABCD.EFGH$ with side length $2$ units. Point $A, B, C$, and $D$ lie in the lower side plane. Point $I$ is intersection point of the diagonal lines on the plane of the upper side. Next, make a pyramid $I.ABCD$. If the pyramid $I.ABCD$ is cut by a diagonal plane connecting the points $A, B, G$, and $H$, determine the volume of the truncated pyramid low part.
Problem
Source:
Tags: algebra, geometry, combinatorics, number theory, indonesia juniors
04.11.2021 00:51
Quote: P3. Numbers $ 1$ to $10$ are arranged in pentagons so that the sum of three numbers on each side is the same. For example, in the picture next to the number the three numbers are $16$. For all possible arrangements, determine the largest and smallest values of the sum of the three numbers.
04.11.2021 01:47
parmenides51 wrote: p4. Define $$S(n)=\sum_{k=1}^{n}(-1)^{k+1}\,\, , \,\, k=(-1)^{1+1}1+(-1)^{2+1}2+...+(-1)^{n+1}n$$Investigate whether there are positive integers $m$ and $n$ that satisfy $S(m) + S(n) + S(m + n) = 2011$
27.05.2023 20:23
P4 Let the term form that expansion is $a_i$. Notice that $a_{2k-1}+a_{2k}=-1,\forall k\in \mathbb{N}$. thus we get these equaity; $S(n)=-\frac{n}{2}, \forall n \in$ even $S(n)=-\frac{(n-1)}{2}+n, \forall n \in$ odd Next we want to know if there any $m$ and $n$ that satisfy $S(m)+S(n)+S(m+n)=2011$ If both $m,n$ are even $\Rightarrow m+n$ even, we get $$S(m)+S(n)+S(m+n)=-\frac{n}{2}-\frac{m}{2}-\frac{(m+n)}{2}$$Because $m,n\in \mathbb{N}$ then $S(m)+S(n)+S(m+n)$ will be negative If both $m,n$ are odd $\Rightarrow m+n$ even, we get $$S(m)+S(n)+S(m+n)=-\frac{(m-1)}{2}+m-\frac{(n-1)}{2}+n-\frac{(m+n)}{2}=1$$If one of $m,n$ is odd and the rest is even, WLOG $m$ odd and $n$ even $\Rightarrow m+n$ even, we get $$S(m)+S(n)+S(m+n)=-\frac{(m-1)}{2}+m-\frac{n}{2}-\frac{((m+n)-1)}{2}+(m+n)=m+1$$Therefore $S(m)+S(n)+S(m+n)=2011$ can be satisfy if $m+1=2011\iff m=2010$, but must $m$ be odd and $n$ even. Contradiction. $\therefore$ There is no $m,n$ that satisfy
28.05.2023 04:35
28.05.2023 05:13
not too lazy to LaTeX this time lol
28.05.2023 20:14
Hello_Kitty wrote: P1: We get $q=3$ with..........: Isn't it 21 1/4 not 21/4