Looking For the New Year $2020$ .

We note that the last number of row $n$ is the $n$ th triangular number $T_n = \dfrac {n(n+1)}2$ . Assumming $2020 le T_n$ , then we have:

$\begin{aligned} \frac {n(n+1)}2 & \ge 2020 \\ n(n+1) & \ge 4040 \\ \implies n & = \left \lceil \sqrt{4040} \right \rceil & \small \blue{\text{where }\lceil \cdot \rceil \text{ denotes the ceiling function.}} \\ & = \boxed{64} \end{aligned}$

Checking: $T_{63} = 2016$ and $T_{64} = 2080$ so $2020$ is on the $\boxed{64}$ th row.

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Reference: Ceiling function .

The $i$ 'th row starts with $\dfrac{i^2-i+2}{2}$ . So $\dfrac{i^2-i+2}{2}\leq {2020}$ or $i\leq {64.02}$ . This row ends with $\dfrac{i^2+i}{2}$ . So $\dfrac{i^2+i}{2}\geq {2020}$ or $i\geq {63.06}$ . Therefore $i=\boxed {64}$

The right end of row n is the number n(n+1)/2. The sum of all naturals up to n. To solve: n (n+1)/2=2020 or n^2+n-4040=0 n=63 and a bit (no more precision is needed) 2020 is past row 63 but not as far as the end of row 64. Therefore it is in row 64.