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A question on my homework asks me to give the amount of significant figures of $2900±100$. Would this be one, two, or both?

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Well, if it had been $2900\pm 50$ then there would clearly have been 2 significant figures, namely the 2 and 9. But how do we account for the fact that the error interval is wider here?

A principled answer would be that there are now $\log_{10}(\frac{2900}{2\cdot 100}) \approx 1.16$ digits of precision (the denominator here is the width of the interval specified, thus the factor of 2). However, in practice the "number of significant figures" are usually supposed to be an integer, and the convention is such that "10.01" and "99.99" both count as "four significant figures", so we should round the $1.16$ upwards to $2$ to get the (probably) expected answer.

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