5. Solve the cryptarithm (the same letters correspond to the same digits, different letters correspond to different digits)

$$
\mathrm{MSU}+\mathrm{MSU}+\mathrm{MSU}+\mathrm{MSU}+\mathrm{OLYMP}+\mathrm{OLYMP}=\mathrm{MOSCOW}
$$
Answer: $C=5, L=7, M=1, O=9, P=2, S=4, U=3, W=6, Y=0,143+143+$ $143+143+97012+97012=194596$.

Solution: Note that since $M S U \leq 987, O L Y M P \leq 98765$, then

$M S U+M S U+M S U+M S U+O L Y M P+O L Y M P \leq 987 \times 4+98765 \times 4=201478$.

Thus, $M$ can only be 1 or 2. $M=2$ does not work, because in this case

$M S U+M S U+M S U+M S U+O L Y M P+O L Y M P \leq 298 \times 4+98765 \times 4=198722$.

Therefore, $M=1$. Note that when adding $O$ and $O$, the result is $1 O$, with a carry of no more than 2 from the previous digits. Therefore, $O$ can only be 8 or 9. Suppose $O=8$, then

$M S U+M S U+M S U+M S U+O L Y M P+O L Y M P \leq 198 \times 4+89716 \times 4=180314$.

But then $S$ must be 0, so

$M S U+M S U+M S U+M S U+O L Y M P+O L Y M P \leq 109 \times 4+89716 \times 4=179868$, which leads to a contradiction. Therefore, $O=9$.