3. The positive integers are written into rows so that Row $n$ includes every integer $m$ with the following properties:
(i) $m$ is a multiple of $n$,
(ii) $m \leq n^{2}$, and
(iii) $m$ is not in an earlier row.
The table below shows the first six rows.
(a) Determine the smallest integer in Row 10.
(b) Show that, for all positive integers $n \geq 3$, Row $n$ includes each of $n^{2}-n$ and $n^{2}-2 n$.
(c) Determine the largest positive integer $n$ with the property that Row $n$ does not include $n^{2}-10 n$.
3. (a) The positive multiples of 7 that are no larger than $7^{2}=49$ are
$7,14,21,28,35,42,49$
None of these integers is in an earlier row, so they are exactly the integers in Row 7 . The positive multiples of 8 that are no larger than $8^{2}=64$ are
$$
8,16,24,32,40,48,56,64
$$

Of these, 8,16 , and 24 are in earlier rows, so Row 8 is $32,40,48,56,64$.
The positive multiples of 9 that are no larger than $9^{2}=81$ are
$$
9,18,27,36,45,54,63,72,81
$$

Of these, 9,18 , and 36 are in earlier rows, so Row 9 is $27,45,54,63,72,81$.
The positive multiples of 10 that are no larger than $10^{2}=100$ are
$10,20,30,40,50,60,70,80,90,100$

Of these, 10, 20, 30, and 40 are in earlier rows, so the smallest integer in Row 10 is 50 .
(b) Consider an integer $n \geq 3$. Notice that $n, n-1$, and $n-2$ are all positive integers.
Factoring, the quantity $n^{2}-n$, we get $n^{2}-n=n(n-1)$.
Since $n>n-1, n(n-1)>(n-1)(n-1)=(n-1)^{2}$, which means that $n(n-1)$ is too
large to be in Row $n-1$. This also implies that $n(n-1)$ is larger than each of $1^{2}, 2^{2}, 3^{2}$, and so on up to $(n-2)^{2}$,
so we conclude that $n(n-1)$ is not in any of the first $n-1$ rows.
We also have that $n(n-1)$ is a multiple of $n$, and since $n$ is positive, $n^{2}-nn-2$ implies that $n(n-2)>(n-2)(n-2)=(n-2)^{2}$. This implies that $n^{2}-2 n$ is larger than $k^{2}$ for all positive integers $k$ with $k \leq n-2$. As well, since $n$ is positive, $n^{2}-2 n0$.
Since $n(n-10)=n^{2}-10 n$ and $n$ is positive, we must have $n^{2}-10 nn-10$ and $n-10$ is positive, we have that $n(n-10)>(n-10)(n-10)=(n-10)^{2}$. This shows that $n(n-10)$ is too large to be in Row $m$ for any $m \leq n-10$.
The previous two sentences combine to show that $n^{2}-10 n$ must be in one of Row $n-9$, must be in ond so on up to Row $n$. We are assuming that $n^{2}-10 n$ is not in Row $n$, so it This means we have 9 cases to consider. $n^{2}$. on to Row $n-1$. and in each $k$ from 1 through 9 , we will consider the case when $n^{2}-10 n$ is in Row $n-k$, and in each of these cases, we will show that $n \leq 30$. This will show that no matter which of the 9 rows $n^{2}-10 n$ is in, it cannot be larger than 30 .
Suppose $n^{2}-10 n$ is in Row $n-9$. Among other things, this means $n^{2}-10 n$ must be a multiple of $n-9$
The expression $n^{2}-10 n$ does not have an obvious factor of $n-9$. It will be useful to find an algebraic expression that is a multiple of $n-9$ and is "close" to $n^{2}-10 n$. After some trial and error, you might notice that $(n-9)(n-1)=n^{2}-10 n+9$, which can be rearranged to get $9=(n-9)(n-1)-\left(n^{2}-10 n\right)$.
We are assuming $n^{2}-10 n$ is a multiple of $n-9$, so the expression $(n-9)(n-1)-\left(n^{2}-10 n\right)$ is the difference of two multiples of $n-9$.
The difference between two multiples of $n-9$ is itself a multiple of 9 , which means $9=(n-9)(n-1)-\left(n^{2}-10 n\right)$ must be a multiple of $n-9$ If 9 is a multiple of $n-9$, then we must have that $9 \geq n-9$, which can be rearranged to get that $n \leq 18$
We have now shown that if $n^{2}-10 n$ is in Row $n-9$, then $n \leq 18$, and so $n \leq 30$.
This kind of reasoning can be applied to the remaining eight cases.
For example, suppose $n^{2}-10 n$ is in Row $n-8$.
Observe that $(n-8)(n-2)=n^{2}-10 n+16$, and so $16=(n-8)(n-2)-\left(n^{2}-10 n\right)$.
If $n^{2}-10 n$ in Row $n-8$, then $n^{2}-10 n$ is a multiple of $n-8$, which means the expression $16=(n-8)(n-2)-\left(n^{2}-10 n\right)$ is a multiple of $n-8$ If 16 is a
,
We will now quickly include the results of examining the remaining seven cases. The calculations are similar to those for Rows $n-9$ and $n-8$ and are not included. Therefore, $n-7 \leq 21$ or $n \leq 28$, and so $n \leq 30$.

Therefore, $n-2 \leq 16$ or $n \leq 18$, and so $n \leq 30$.

To recap, we assumed that $n^{2}-10 n$ is not in Row $n$, then deduced that it must be in Row $k$ for some $k$ from $n-9$ through $n-1$. We then considered each each of these 9 possibilities individually and showed that $n \leq 30$ in each case.
Therefore, if $n^{2}-10 n$ is not in Row $n$, then $n \leq 30$. Since $n=30$ has the property that $n^{2}-10 n$ is not in Row $n$, we can conclude that 30 is the largest integer with this property.