Let's calculate the following powers: a) $101^{7}$ b) 0.9998 (to 7 decimal places).
a)

\[
\begin{gathered}
101^{7}=\left(10^{2}+1\right)^{7}=\binom{7}{0} 10^{14}+\binom{7}{1} 10^{12}+\binom{7}{2} 10^{10}+\binom{7}{3} 10^{8}+ \\
+\binom{7}{4} 10^{6}+\binom{7}{5} 10^{4}+\binom{7}{6} 10^{2}+\binom{7}{7}
\end{gathered}
\]

Writing the value of each term below the others:

\[
\begin{aligned}
& 100000000000000 \\
& 7000000000000 \\
& 210000000000 \\
& 3500000000 \\
& 35000000 \\
& 210000 \\
& 700 \\
& \begin{array}{r}
1 \\
\hline 107213535210701
\end{array}
\end{aligned}
\]

Szabó Endre (Gyöngyös, Vak Bottyán g. I. o. t.)

b) In some copies, the exponent was washed off or not visible at all, which is why we removed this example from the competition.

\[
0.9998^{5}=\left(1-2 \cdot 10^{-4}\right)^{5}
\]

The 4th term: \(-\binom{5}{3} 2^{3} \cdot 10^{-12}=-80 \cdot 10^{-12}=-8 \cdot 10^{-11}\).

Considering only the first 3 terms, the error (due to the alternating signs and decreasing absolute values of the terms) is smaller than the 4th term, which does not affect the first 7 decimal places.

Thus:

\[
\begin{gathered}
\left(1-2 \cdot 10^{-4}\right)^{5}=\binom{5}{0}-\binom{5}{1} 2 \cdot 10^{-4}+\binom{5}{2} 2^{2} \cdot 10^{-8}=1-10 \cdot 10^{-4}+40 \cdot 10^{-8}= \\
=1-10^{-3}+4 \cdot 10^{-7}=1.0000004-0.001=0.9990004
\end{gathered}
\]