Determine the integer equal to

$$
2020+1919+1818+1717+1616+1515+1414+1313+1212+1111+1010 .
$$
Solution 1

Let $S$ be the sum in the problem. We can group terms and rewrite the sum as

$$
\begin{aligned}
S & =(1010+2020)+(1111+1919)+(1212+1818)+(1313+1717)+(1414+1616)+1515 \\
& =3030+3030+3030+3030+3030+1515 \\
& =5 \times 3030+1515 \\
& =16665 .
\end{aligned}
$$

## Solution 2

The numbers being added are the multiples of 101 from $10 \times 101$ through $20 \times 101$.

Thus, if we set $S$ to be the sum in the problem, we can factor out 101 to get

$$
S=101(10+11+12+13+14+15+16+17+18+19+20) \text {. }
$$

It can be checked by hand that $10+11+\cdots+19+20=165$. Alternatively, using the fact that

$$
1+2+3+4+\cdots+(n-1)+n=\frac{n(n+1)}{2}
$$

for any positive integer $n$, the sum can be computed as

$$
\begin{aligned}
10+11+\cdots+19+20 & =(1+2+3+\cdots+19+20)-(1+2+\cdots+8+9) \\
& =\frac{20(21)}{2}-\frac{9(10)}{2} \\
& =10 \times 21-9 \times 5 \\
& =210-45 \\
& =165 .
\end{aligned}
$$

Therefore, $S=101 \times 165=(100+1) \times 165=16500+165=16665$.