The positive integers are written into rows so that row n includes every integer m with the properties: a) $m$ is a multiple of $n$; b) $m \leq n^2$ ; c) $m$ is not in the earlier row.
Determine the largest possible integer $n$ with the property that row $n$ does not include $n^2-10n$.
I have started by listing rows and finding some similarities. I have found that the smallest such integer $n=11$. But how to find the largest one?
On
For $n=30$ we have $n^2-10n=600$.
Note that $600$ is not in rows $1$ through $24$ since $600 > 24^2$.
But $600$ is in row $25$ since $600=24{\,\cdot\,}25$.
Hence row $30$ does not contain $600$.
Claim:$\;$For all $n > 30$, row $n$ contains $n^2-10n$.
Proof:
Suppose otherwise.
Let $n > 30$ be such that row $n$ does not contain $n^2-10n$.
Since $n^2-10n$ is a multiple of $n$ and$\;0 < n^2-10n < n^2$, it follows that $n^2-10n$ must be contained in row $m$ for some $m < n$.
Thus we must have $n^2-10n\le m^2$ and $m{\,\mid\,}(n^2-10n)$.
Noting that $$ n^2-10n=(n-6)^2+(2n-36) > (n-6)^2 $$ it follows that $m=n-r$ for some positive integer $r\le 5$.
Then from the identity $$ n^2-10n=(n-r)(n+r-10)-(10r-r^2) $$ it follows that $(n-r){\,\mid\,}(10r-r^2)$.
Then we get $n-r\le 10r-r^2$, so $n\le 11r-r^2$, hence $n\le 30$. contradiction.
This completes the proof of the claim.
Since $n^2 - 10n \le 0$ for $1 \le n \le 10$, none of those rows include $n^2 - 10n$. For $n \ge 11$, since $n(n - 10)$ is a positive integral multiple of $n$ and it's $\le n^2$, it meets the first $2$ properties. Thus, row $n$ would only not include $n^2 - 10n$ if it didn't meet the third property, i.e., that value was already included in an earlier row $n - k$ for some integer $k \gt 0$.
Note that $n - k$ must be a factor of $n(n - 10)$ where $n(n - 10) \le (n - k)^2$. Since $n(n - 10) \gt (n - 10)^2$, this means $1 \le k \le 9$. Considering each $k$ in turn, note some factor of $n - k$ must divide into $n$, and the remaining factor into $n - 10$. For $k = 1$, since $\gcd(n, n - 1) = 1$, this requires $n - 1 \mid n - 10$, which is impossible. For $k = 2$, with $\gcd(n, n - 2) \le 2$ and $\gcd(n - 2, n - 10) \le 8$, then the largest possible value of $n - 2 = 2 \times 8 = 16$ gives $n = 18$ and $n(n - 10) = 144 = 16\times 9$. In this case, since $9 \le 16$, then $n(n - 10)$ would have already been included in row $16$.
Similarly, for $k = 3$, we get a maximum value of $n - 3 = 3 \times 7 = 21$, so $n = 24$ and $n(n - 10) = 336 = 21\times 16$. With $k = 4$, we have $n - 4$ being at the most $4 \times 6 = 24$, so $n = 28$ and $n(n - 10) = 504 = 24\times 21$. Next, $k = 5$ has a max. $n - 5 = 5 \times 5 = 25$, so $n = 30$ and $n(n - 10) = 600 = 25\times 24$.
With $k = 6$'s max. of $n - 6 = 6 \times 4 = 24$, then $n = 30$ again. However, here $25 \gt 24$, so it doesn't meet the second condition (i.e., $m \le n^2$). For $k = 7$, $8$ and $9$, the max. values of $n - k$ give decreasing values of $n$, i.e., all less than $30$ (also, as with $n = 6$, all of $(n - k)^2$ are less than $n(n - 10)$).
Thus, the largest value of $n$ where row $n$ doesn't include $n(n - 10)$ is $n = 30$, with $600 = 25 \times 24$ already being included in row $25$.