Solution:
We claim that the only prime number of which the sequence must contain a multiple is $p=2$. To prove this, we begin by noting that
$$a_{n+1}=\frac{a_{n}\left(a_{n}+1\right)}{2}-10=\frac{\left(a_{n}-4\right)\left(a_{n}+5\right)}{2}$$
Let $p>2$ be an odd prime, and choose $a_{1} \equiv -4 \pmod{p}$, so $2 a_{2} \equiv (-4-4)(-4+5) \equiv -8 \pmod{p}$, whence $a_{2} \equiv -4 \pmod{p}$, since $p$ is odd. By induction, $a_{n} \equiv -4 \pmod{p} \not\equiv 0 \pmod{p}$ for all $n$, and so the sequence need not contain a multiple of $p$.
We are left to show that the sequence must contain an even number. Suppose to the contrary that $a_{n}$ is odd for $n=1,2, \ldots$ We observe that
$$a_{n+1}-a_{n}=\frac{a_{n}\left(a_{n}+1\right)}{2}-\frac{a_{n-1}\left(a_{n-1}+1\right)}{2}=\frac{a_{n}-a_{n-1}}{2}\left(a_{n}+a_{n-1}+1\right)$$
By assumption, $a_{n}+a_{n-1}+1$ is odd for $n=1,2, \ldots$, so this shows that $v_{2}\left(a_{n+1}-a_{n}\right)=v_{2}\left(a_{n}-a_{n-1}\right)-1$, and so there exists $N$ such that $v_{2}\left(a_{N+1}-a_{N}\right)=0$. This is a contradiction, because $a_{n+1}-a_{n}$ is even for $n=1,2, \ldots$ by assumption, and thus completes the proof.

Solution:
For odd $p$, proceed as in solution 1. Now let $p=2$, and suppose that every term of the sequence is odd. We claim that it follows that $a_{n} \equiv 5 \pmod{2^{k}}$ for every integer $n \geq 1$ and every integer $k \geq 1$. We proceed per induction on $k$. For $k=1$ this simply states that $a_{n}$ is odd for all integers $n \geq 1$, as assumed. Now suppose it is true for $k=r$. Let $k=r+1$. Take any integer $n \geq 1$. Note that, by the induction hypothesis, $a_{n} \equiv 5 \pmod{2^{r}}$. Therefore there exists an integer $s$ such that $a_{n}=2^{r} s+5$. Now note that
$$a_{n+1}=\frac{\left(a_{n}-4\right)\left(a_{n}+5\right)}{2}=\frac{\left(2^{r} s+1\right)\left(2^{r} s+10\right)}{2}=\left(2^{r} s+1\right)\left(2^{r-1} s+5\right) \equiv 2^{r-1} s+5 \pmod{2^{r}}$$
By the induction hypothesis, $a_{n+1} \equiv 5 \pmod{2^{r}}$. Therefore $s$ is even, such that $a_{n}$ is of the form $2^{r+1} s+5$ for any integer $n \geq 1$, which concludes the induction. From this property, it follows that $a_{1}-5$ is divisible by $2^{k}$ for every integer $k \geq 1$, which is only possible if $a_{1}=5$. But $a_{1}>5$, so this is a contradiction.