A. I. F. Shark
To a natural number $A$, three digits were appended on the right. The resulting number turned out to be equal to the sum of all natural numbers from 1 to $A$. Find $A$.

#
Let the appended digits form the number $B, 0 \leq B \leq 999$. Then the resulting number is, on the one hand, $1000 A+B$, and on the other hand, $-1+2+\ldots+A=\frac{1}{2} A(A+1)$. The equation $1000 A+B=\frac{1}{2} A(A+1)$ transforms into $A(A-1999)=2 B$, from which $0 \leq A(A-1999) \leq 1998$. Since the left inequality here is possible only for $A \geq 1999$, and the right one for $A<2000$, then $A=1999$.

## Answer

1999.00