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This is probably easy, but I was just wondering if there is a nice and easy formula for the topological Euler characteristic of a K3 surface $X$ with say $k$ nodes. If there is no general formula, is it known what the answer is for $k=1$?

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up vote 5 down vote accepted

The nodes do not modify the birational invariants of a surface. So if we blow-up the $k$ nodes of $X$ we obtain a smooth K3 surface $S$, containing $k$ $(-2)$-curves, whose topological Euler number is $24$. Coming back to $X$, we substitute each $(-2)$-curve (which is topologically a sphere, so has Euler number $2$) with a point. So the Euler number of $X$ is $24-k$.

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