31. What is the remainder when the number $2^{99}$ is divided by 7?

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31. Note that $2^{99}=8^{33}=(7+1)^{33}=(7+1) \ldots(7+1)$. Expand the brackets. The resulting terms will be divisible by 7, except for 1. Thus, $2^{99}$ can be written in the form: $7 x$ +1 . Answer: 1.