Jaime loves to add sequences of consecutive integers.

a) What result does Jaime find when he adds the first 2,019 positive integers?

b) Jaime adds 100 consecutive numbers and finds the sum to be 17,050. What is the smallest number in this sequence?

c) By adding 180 consecutive numbers, Jaime found the result to be 3,690. What is the smallest of them?

d) Jaime added 100 consecutive positive numbers, but made a mistake, replacing one of them with its square, thus obtaining 7,500. Which number was squared?
Solution

a) If the first is 1 and the last is 2,019, then the sum is $\frac{(1+2,019) \cdot 2,019}{2}=2,039,190$.

b) Let $n$ be the smallest of them, then the largest is ( $n+99$ ). Therefore, we have:

$$
\begin{aligned}
\frac{(n+n+99) \cdot 100}{2} & =17,050 \\
2 n+99 & =341 \\
2 n & =242 \\
n & =121
\end{aligned}
$$

c) Let $n$ be the smallest of them, then $(n+179)$ is the largest. Thus:

$$
\begin{aligned}
\frac{(n+n+179) \cdot 180}{2} & =3,690 \\
2 n+179 & =41 \\
2 n & =-138 \\
n & =-69
\end{aligned}
$$

d) Let $n$ be the smallest of them and $k$ be the number that was replaced by its square, then:

$$
\begin{aligned}
\frac{(n+n+99) \cdot 100}{2}-k+k^{2} & =7,500 \\
50(2 n+99)+k(k-1) & =7,500 \\
k(k-1) & =50(150-2 n-99) \\
k(k-1) & =50(51-2 n)
\end{aligned}
$$

For $2 \cdot 5^2 \cdot (51-2 n)$ to be the product of two consecutive numbers, we must have $n=1$ or $n=19$. Therefore, we have two possible scenarios, that is, the number added as a square could have been 50, if $n=1$, or 26, if $n=19$.