\section*{Problem 3 - V01003}

Decompose 900 into two summands such that the sum of their reciprocals is equal to the reciprocal of 221!
}

The two sought addends we denote with \(a\) and \(b\). The two conditions are

\[
a+b=900, \quad(1) \quad \frac{1}{a}+\frac{1}{b}=\frac{1}{221}
\]

(2) can be rewritten as

\[
\frac{a+b}{a b}=\frac{900}{a b}=\frac{1}{221}
\]

so that \(a b=900 \cdot 221=198900\) holds. With \(a=900-b\) from (1) it follows that

\[
\begin{aligned}
a \cdot b=(900-b) \cdot b & =198900 \\
b^{2}-900 \cdot b & =-198900 \\
b^{2}-900 \cdot b+450^{2} & =450^{2}-198900 \\
(b-450)^{2} & =3600
\end{aligned}
\]
and thus the possibilities \(b_{1}=390\) and \(b_{2}=510\). Therefore, the two addends - taking into account the commutativity of addition - are accordingly 390 and 510.