CAN U STOP BRAGGING?! IMMA KILL U THE NEXT TIME YOU BRAG ABOUT YOUR LATEX

[size=200]PART A[/size]



$$n(2n+1)(7n+1)$$

We can approach this problem by seeing simplifying the $n$'s in the terms.

$$7n \pmod{6}\Longrightarrow n$$

So we can turn $n(2n+1)(7n+1)$ into $n(2n+1)(n+1)$


Let's split this: $n(2n+1)(n+1)\Longrightarrow n,2n+1,n+1$

Let's write these terms as $n\equiv 0 \pmod6$ first:

This one's easy. We already have a single term of $6$ meaning that it doesn't matter what the other terms are, the ending product will always be divisible by six.


Let's now write these terms as $n\equiv 1\pmod6$:

$$n\Longrightarrow 1\pmod 6$$
$$2n+1\Longrightarrow 1\cdot 2+1\pmod 6$$
$$3\pmod 6$$
$$n+1\Longrightarrow 1+1\pmod6$$
$$2\pmod 6$$

Taking these terms and multiplying these out:

$$(1\cdot 3\cdot 2) \pmod 6$$
$$\equiv 6 \pmod 6$$

And $6\pmod 6$ will tell us that this case is possible.


Let's do the same except we use terms as $n\equiv 2\pmod6$:

$$n\Longrightarrow 2\pmod 6$$
$$2n+1\Longrightarrow 2\cdot 2+1\pmod 6$$
$$5\pmod 6$$
$$n+1\Longrightarrow 2+1\pmod6$$
$$3\pmod 6$$

We can take these terms and multiply them out to get a result of:
$$2\cdot 3\cdot 5\pmod 6$$
$$30\pmod 6$$

Which is also divisible by six meaning that this case works as well.


Let's change up the case to $n\equiv 3\pmod6$:

$$n\Longrightarrow 3\pmod 6$$
$$2n+1\Longrightarrow 3\cdot 2+1\pmod 6$$
$$7\pmod 6$$
$$n+1\Longrightarrow 3+1\pmod6$$
$$4\pmod 6$$

Multiplying these terms out will give us:
$$3\cdot 7\cdot 4\pmod 6$$
$$84\pmod 6$$

Which is also divisible by six.


Let's change up the case to $n\equiv 4\pmod6$:

$$n\Longrightarrow 4\pmod 6$$
$$2n+1\Longrightarrow 4\cdot 2+1\pmod 6$$
$$9\pmod 6$$
$$n+1\Longrightarrow 4+1\pmod6$$
$$5\pmod 6$$

Multiplying these terms out will give us:
$$4\cdot 9\cdot 5\pmod 6$$
$$180\pmod 6$$


And $180$ is divisible by six.



The last but not least case: $n\equiv 5\pmod6$


$$n\Longrightarrow 5\pmod 6$$
$$2n+1\Longrightarrow 5\cdot 2+1\pmod 6$$
$$11\pmod 6$$
$$n+1\Longrightarrow 5+1\pmod6$$
$$6\pmod 6$$

Multiplying these terms out is unnecessary because the last term simply includes a six.



We skip $6\pmod 6$ and above because it just repeats the process of $0\pmod6$ and above.



[size=200]PART B[/size]
$$\pmod 12$$
Since we already proved $\pmod 6$ we know it's divisible by $3$, so now we just need to check whether it's divisible by $4$.


We can tackle this problem the same we tackled part A. We can list cases and check whether all of them are valid or not.


We can use the same terms: $n,2n+1,n+1$

Starting off: $n\equiv 0\pmod4$

$$n\Longrightarrow 4\pmod 4$$

We don't need to find the other terms because we already see one term of $4$ meaning that no matter what the other two terms multiply out to, the product will [b]always[/b] be a multiple of $4$.



Let's change the terms into $n\equiv 1\pmod4$:

$$n\Longrightarrow 1\pmod4$$
$$2n+1\Longrightarrow 2\cdot 1+1\pmod 4$$
$$3\pmod4$$
$$n+1\Longrightarrow 1+1\pmod4$$
$$2\pmod4$$

Multiplying these terms out:
$$1\cdot 3\cdot 2\pmod 4$$
$$6\pmod 4$$


And since $6$ isn't a multiple of $4$, $n\equiv 1\pmod4$ does not satisfy the conditions




Changing the terms into $n\equiv 2\pmod4$:

$$n\Longrightarrow 2\pmod4$$
$$2n+1\Longrightarrow 2\cdot 2+1\pmod 4$$
$$5\pmod4$$
$$n+1\Longrightarrow 2+1\pmod4$$
$$3\pmod4$$

Multiplying these terms out:
$$2\cdot 5\cdot 3\pmod 4$$
$$30\pmod 4$$


And since $30$ isn't a multiple of $4$, $n\equiv 1\pmod4$ does not satisfy the conditions



The last term is $n\equiv 3\pmod4$, so let's go plug that in.

$$n\Longrightarrow 3\pmod4$$
$$2n+1\Longrightarrow 2\cdot 3+1\pmod 4$$
$$7\pmod4$$
$$n+1\Longrightarrow 3+1\pmod4$$
$$4\pmod4$$

We do not need to multiply the three terms out because of the last term being $4$. This works because $4x$ will always be divisible and in this case, $x$ represents the other two numbers.

The only cases that worked was the first and the last.

So in this case, the only integers $n$ such that $n(2n+1)(7n+1)$, is $\bold{\boxed{3,7,11,15\cdots; \text{or multiples of:}  4}}$