I like to look for mistakes in other people's works. So, I don't have time to find my own mistakes.
I was reading a wonderful book yesterday:
My Numbers, My Friends (Popular Lectures on Number Theory) by the author Paulo Ribenboim.
$$...\;if\;p\;is\;a\;prime\;dividing\;both\;P,\;Q,\;then...$$
$\blacksquare$
Those who wish can read as much as they want by clicking on the image of the book cover.
Click on the image to download.
For a wider collection click here. The password for opening encrypted files is ogeometrie
Click on the image to download.
For a wider collection click here. The password for opening encrypted files is ogeometrie
Nu este sigur daca voi reusi actualizarea ERATA la fiecare numar al revistei.
E.g $\{1,2,3,4,5\}=\{1,4\}\cup \{2,3\}\cup \{5\}$ ;
$\{1,2,3,4,5,6\}=\{1,6\}\cup \{2,5\} \cup \{3,4\}.$
The following is one of the old contest problems (22nd All Soviet Union Math Contest, 1988).
Let $m,\; n,\; k$ be positive integers with $m \geqslant n$ and $1+2+\cdots +n=m\cdot k$. Prove that the numbers $1,\;2,\;\cdots,n$ can be divided into $k$ groups in such a way that the sum of the numbers in each group equals $m$.
CiP example: $1+2+3+4+5+6+7+8+9=45$;
from $45=9\cdot 5$ the partition of $k=5$ groups is obtained, each with a sum of $m=9$
$$\{1,\;8\}\cup \{2,\;7\}\cup\{3,\;6\}\cup \{4,\;5\}\cup\{9\};$$
from $45=15 \cdot 3$ the partition of $k=3$ goups is obtained, each with a sum of $m=15$.
$$\{1,\;2,\;3,\;9\} \cup \{4,\;5,\;6\} \cup \{7,\;8\}.$$
CiP exampl$e^{bis}$: $1+2+3+4+5+6+7+8=36=12 \cdot 3$ and we have the partition of $k=3$ subsets, with the sum of $m=12$
$$\{1,\;2,\;3,\;6\} \cup \{4,\;8\} \cup \{5,\;7\}.$$
It would seem that the answer to the original question is $n=3\cdot p-1$ or $n=3\cdot p$. That is, in the case of $n=3\cdot p+1$, there is no such partition. (Indeed, if $n=3\cdot p+1$, then, because $$1+2+ \cdots +n=\frac{n(n+1)}{2}=\frac{(3p+1)(3p+2)}{2}=9\cdot \frac{p(p+1)}{2}+1,$$ this sum should be a multiple of 3, which is not true.)
If $n=3p$ then $1+2+ \cdots +n=\frac{3p(3p+1)}{2}=3 \cdot \frac{p(3p+1)}{2}=3\cdot \frac{p(p+1+2p)}{2}=3\cdot m$,
and $p$ or $p+1$ is divisible by $2$ so $m$ is integer.
If $n=3p-1$ then $\frac{n(n+1)}{2}=\frac{(3p-1) p}{2}=\frac {(2p+p-1)\cdot p}{2}=3 \cdot m$
with $m$ integer by the same arguments.
So in these cases we fall over the general problem given to the 22nd All Soviet Union Math Contest, 1988.
Another example in the case of $n=15$ is here : math.stackexchange.com/questions/31929/partition-of-equal-size-and-equal-sum
The proof is given by induction, resulting in a way of constructing the partitions.
The general problem has an object if $n \geqslant 3$.
For $n=3$ we have only one partition $\{1,\;2\} \cup \{3\}.$
For $n=4$ such partitions do not exsist.
For $n=5,\;6$ are just the examples presented previously.
For $n=7$ such partitions do not exist.
For $n=8$, because $1+2+\cdots +8=36=9 \cdot 4=12 \cdot 3=18 \cdot 2$ we have in addition to the presented previously and the partitions
$$\{1,\;2,\;3,\;4,\;8\}\cup \{5,\;6,\;7\},\;\{1,\;3,\;6,\;8\}\cup \{2,\;4,\;5,\;7\}$$
$$\{1,\;4,\;5,\;8\}\cup \{2,\;3,\;6,\;7\}\;,\;\{1,\;4,\;6,\;7\}\cup \{2,\;3,\;5,\;8\}$$
$$\{1,\;2,\;4,\;5,\;6\}\cup \{3,\;7,\;8\}\;,\;\{1,\;2,\;3,\;5,\;7\} \cup \{4,\;6,\;8\}$$
$$\{1,\;2,\;7,\;8\}\cup \{3,\;4,\;5,\;6\}.$$
For $n=9$, because $1+2+\cdots+9=45=9\cdot 5=15 \cdot 3$ are just the examples presented previously.
Click on the image to download.
For a wider collection click here. The password for opening encrypted files is ogeometrie
Click on the image to download.
For a wider collection click here. The password for opening encrypted files is ogeometrie
