6. (20 points) Let a sequence of non-negative integers be given

$$
k, k+1, k+2, \ldots, k+n
$$

Find the smallest $k$, for which the sum of all numbers in the sequence is equal to 100.

#
# Answer: 9.

Solution. The given sequence of numbers is an arithmetic progression with $n+1$ terms. Its sum is $\frac{(n+1)(2 k+n)}{2}$. Therefore, the condition can be rewritten as $(n+1)(2 k+n)=200$, where $n$ is a non-negative integer. It is clear that $k$ decreases as $n$ increases. Thus, we need to find the largest $n$ for which $k$ is an integer. Note that

$$
200=(n+1)(2 k+n)>n^{2} \Longrightarrow n<\sqrt{200} \Longrightarrow n \leqslant 14
$$

Moreover, $200=2^{3} \cdot 5^{2}$. Let's consider two cases.

1) $n$ is even. The largest $n \leqslant 14$ for which 200 is divisible by $n+1$ is 4, from which $k=\frac{40-4}{2}=18$.
2) $n$ is odd. Then $n$ can only be $1, 3, 7$ (otherwise 200 is not divisible by $n+1$). For $n=7$, we get $k=9$. This will be the smallest.