(solved by Sébastien Miquel). The natural number $A$ has the following property: the number $1+2+\cdots+A$ is written (in base 10) as the number $A$ followed by three other digits. Find $A$.
. The property of the statement translates to the inequality:

$$
0 \leqslant \frac{A(A+1)}{2}-1000 A=A\left(\frac{A+1}{2}-1000\right) \leqslant 999 .
$$[^5]

If $A<1999$, then the factor $\frac{A+1}{2}-1000$ is negative and the inequality is not satisfied. If $A \geqslant 2000$, then the previous factor is greater than $\frac{1}{2}$ and the product $A\left(\frac{A+1}{2}-1000\right)$ exceeds 1000. The only solution is therefore $A=1999$, which can be verified if desired that it indeed satisfies the condition.