The integer 2019 can be formed by placing two consecutive two-digit positive integers, 19 and 20, in decreasing order. What is the sum of all four-digit positive integers greater than 2019 that can be formed in this way?
(A) 476681
(B) 476861
(C) 478661
(D) 468671
(E) 468761
Solution 1

The smallest integer greater than 2019 that can be formed in this way is formed using the next two largest consecutive integers 20 and 21, giving the four-digit integer 2120.

The largest such integer is 9998 .

The list of such integers is

$$
2120,2221,2322, \ldots, 9796,9897,9998
$$

Each pair of consecutive numbers in this list differs by 101 since the number of hundreds increases by 1 and the number of ones increases by 1 between each pair.

Since the numbers in the list are equally spaced, then their sum will equal the number of numbers in the list times the average number in the list.

The average number in the list is $\frac{2120+9998}{2}=\frac{12118}{2}=6059$.

Since each number in the list is 101 greater than the previous number, then the number of increments of 101 from the first number to the last is $\frac{9998-2120}{101}=\frac{7878}{101}=78$.

Since the number of increments is 78 , then the number of numbers is 79 .

This means that the sum of the numbers in the list is $79 \times 6059=478661$.

## Solution 2

As in Solution 1, the list of such integers is $2120,2221,2322, \ldots, 9796,9897,9998$.

If the sum of these integers is $S$, then

$$
\begin{aligned}
S= & 2120+2221+2322+\ldots+9796+9897+9998 \\
= & (2100+2200+2300+\cdots+9700+9800+9900) \\
& \quad+(20+21+22+\cdots+96+97+98) \\
= & 100(21+22+23+\cdots+96+97+98+99) \\
& \quad+(20+21+22+\cdots+96+97+98) \\
= & 100(21+22+23+\cdots+96+97+98+99) \\
& \quad+(21+22+\cdots+96+97+98+99)+20-99 \\
= & 101(21+22+23+\cdots+96+97+98+99)-79
\end{aligned}
$$

There are 79 numbers in the list of consecutive integers from 21 to 99 , inclusive, and the middle number in this list is 60 .

Therefore, $S=101 \times 79 \times 60-79=478661$.

ANSWER: (C)