60. For what values of $n$ is the number $3\left(n^{2}+n\right)+7$ divisible by 5?
60. First method. Since $3\left(n^{2}+n\right)+7=3\left(n^{2}+n\right)+2+5$, it is sufficient to find those values of $n$ for which $3\left(n^{2}+n\right)+2$ is divisible by $5:$

$$
3\left(n^{2}+n\right)+2=5 t, \quad n=\frac{-3 \pm \sqrt{15(4 t-1)}}{6}
$$

and if $4 t-1=15 t_{1}^{2}$, then $n=\frac{-1 \pm 5 t_{1}}{2}$; therefore, $t_{1}$ must be odd, i.e., $t_{1}=2 t_{2}+1$, then $n_{1}=2+5 t_{2}, n_{2}=-5 t_{2}-3$, these formulas can be combined: $n=5 t+2$.

Second method.

Hint. Consider the cases: $n=5 m, n=5 m \pm 1, n=5 m \pm 2$.