A number is called Norwegian if it has three distinct positive divisors whose sum is equal to 2022. Determine the smallest Norwegian number. (Note: The total number of positive divisors of a Norwegian number is allowed to be larger than 3.) (Cyprus) Answer: 1344
Observe that 1344 is a Norwegian number as 6, 672 and 1344 are three distinct divisors of 1344 and \(6 + 672 + 1344 = 2022\). It remains to show that this is the smallest such number. Assume for contradiction that \(N \frac{2022}{1344} = \frac{337}{224} = \frac{3}{2} + \frac{1}{224}\). If \(a > 1\) then 
\[ \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \leqslant \frac{1}{2} + \frac{1}{3} + \frac{1}{4} = \frac{13}{12} \]
\[ \frac{3}{2} + \frac{1}{224} - 1 - \frac{1}{3} > \frac{1}{6} \]
so \(c = 4\) or \(c = 5\). If \(c = 4\) then 
\[ 2022 = N \left(1 + \frac{1}{3} + \frac{1}{4}\right) = \frac{19}{12} N \]
but this is impossible as \(19 \nmid 2022\). If \(c = 5\) then 
\[ 2022 = N \left(1 + \frac{1}{3} + \frac{1}{5}\right) = \frac{23}{15} N \]
which again is impossible, as \(23 \nmid 2022\). Case \(b = 2\). Note that \(c \frac{3}{2} + \frac{1}{224} - 1 - \frac{1}{2} = \frac{1}{224}\). It holds that 
\[ 2022 = N \left(1 + \frac{1}{2} + \frac{1}{c}\right) = \frac{3c + 2}{2c} N \Rightarrow (3c + 2) N = 4044c \]
Since \((c, 3c - 2) = (c, 2) \in \{1, 2\}\), then \(3c + 2 \mid 8088 = 2^3 \cdot 3 \cdot 337\) which implies that \(3c + 2 \mid 2^3 \cdot 337\). But since \(3c + 2 \geqslant 3 \cdot 3 + 2 > 8 = 2^3\) and \(3c + 2 \neq 337\), then it must hold that \(3c + 2 \geqslant 2 \cdot 337\), contradicting \(c < 224\).