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To find the sum of even divisors, note that an even divisor must include at least one factor of 2, so \( a \geq 1 \). Thus:
\( a = 1 \) or \( a = 2 \)
\( b = 0 \) or \( b = 1 \)
\( c = 0 \) or \( c = 1 \)
We compute the sum:
\sum_{\substack{a=1\\b=0\\c=0}}^{2,1,1} 2^a \cdot 3^b \cdot 5^c
= \left( \sum_{a=1}^{2} 2^a \right) \cdot \left( \sum_{b=0}^{1} 3^b \right) \cdot \left( \sum_{c=0}^{1} 5^c \right)
\( \sum_{a=1}^{2} 2^a = 2 + 4 = 6 \)
\( \sum_{b=0}^{1} 3^b = 1 + 3 = 4 \)
\( \sum_{c=0}^{1} 5^c = 1 + 5 = 6 \)