\section*{Problem 3 - 021133}

In how many different ways can the number 99 be expressed as the sum of three distinct prime numbers?

(Two cases are considered the same if the same addends appear, merely in a different order.)
}

The following ordered pairs of prime numbers \(a, b\) satisfy the equation \(a+b=99-c\) for a given \(c>b>a, c>33\):

\begin{tabular}{c|c|c|c|c}
\(c\) & \(99-c\) & \((a, b)\) with \(a+b=99-c\) & \((a, b)\) with \(c>b>a\) & Number \\
\hline 97 & 2 & \(\varnothing\) & \(\varnothing\) & 0 \\
89 & 10 & \((3,7),(5,5)\) & \((3,7)\) & 1 \\
83 & 16 & \((3,13),(5,11)\) & \((3,13),(5,11)\) & 2 \\
79 & 20 & \((3,17),(7,13)\) & \((3,17),(7,13)\) & 2 \\
73 & 26 & \((3,23),(7,19),(13,13)\) & \((3,23),(7,19)\) & 2 \\
71 & 28 & \((5,23),(11,17)\) & \((5,23),(11,17)\) & 2 \\
67 & 32 & \((3,29),(13,19)\) & \((3,29),(13,19)\) & 2 \\
61 & 38 & \((7,31),(19,19)\) & \((7,31)\) & 1 \\
59 & 40 & \((3,37),(11,29),(17,23)\) & \((3,37),(11,29),(17,23)\) & 3 \\
53 & 46 & \((3,43),(5,41),(17,29),(23,23)\) & \((3,43),(5,41),(17,29)\) & 3 \\
47 & 52 & \((5,47),(11,41),(23,29)\) & \((11,41),(23,29)\) & 2 \\
43 & 56 & \((3,53),(13,43),(19,37)\) & \((19,37)\) & 1 \\
41 & 58 & \((5,53),(11,47),(17,41),(29,29)\) & \(\varnothing\) & 0 \\
37 & 62 & \((3,59),(19,43)\) & \(\varnothing\) & 0
\end{tabular}

For \(c \leq 33\), \(a+b=99-c \geq 66\), i.e., \(b>33 \geq c\) because \(a < b < c\) must hold, and any triples \((a, b, c)\) could be reordered so that \(c>b>a\) holds.

Thus, 99 can be expressed as the sum of three different prime numbers in \(1+2+2+2+2+2+1+3+3+2+1=21\) different ways.