Find the number of integers between 1 and 25 so that the expression is divided by 6

Question

Find the number of integers so that the $n^2+3n+2$ is divided by 6 when $1 \le n \le 25$.

I start by $6\mid n^2+3n+2 6|(n+1)(n+2) \implies 6k=(n+1)(n+2) $ when $ k \in Z$ we can see that for $n=$ $1,2,4,5,7,8,10,11....$ will work, but I am not sure how to continue from here .

(some HINT can be very appreciated ).

Additionally I would love see a combinatorics approach for this .

Thank you guys.

Answer 1

Being divisible by $6$ means being even and divisible by $3$.

• Modulo $2$, $n^2+3n+2\equiv n^2+n\equiv n+n \text{ (by Fermat) }\equiv 0$. So it is always even.
• Modulo $3$, $n^2+3n+2\equiv n^2+2$, and it is congruent to $0$ if and only if $n^2\equiv 1\mod 3$, which happens only if $n\equiv \pm 1$, in other words, if and only if $n$ is not divisible by $3$.

So we simply can take the list of integers between $1$ and $25$ and remove the $8$ multiples of $3$ from this list. There remains $17$ integers.

Answer 2

Hint: Either one of the following cases happen:

(1) $6$ divides one of $n+1$ and $n+2$.

(2) one of $n+1$ and $n+2$ is divisible by $3$ and the another is divisible by $2$.

Answer 3

For $n=6$ we have $$7\times 8=56$$ and $56$ isn't divisible by $6$. No induction works here!

Answer 4

As $6=2\cdot3$

$n^2+3n+2=(n+1)(n+2)$ is always divisible by $2$ being a product of two consecutive integers

Again, $n^2+3n+2\equiv n^2-1\pmod3$

So, $3\nmid n$

Now the Cardinality of the complement set i.e., $\{n,1\le n\le25;3\mid n\}$ is $\left\lfloor\dfrac{25}3\right\rfloor=8$