\section*{Problem 1 - 281031}

For every natural number \(n\), its representation in base 2 (representation as a binary number), its representation in base 3, and so on, up to its representation in base 10 (representation as a decimal number) are considered.

If there are natural numbers \(n>1\) for which the last digit (units digit) in each of these representations (with bases \(2, 3, 4, \ldots, 10\)) is a 1, determine the smallest such natural number \(n\).
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For the number \(n\) to have the digit 1 as the last digit in positional notation systems with bases from 2 to 10, \(n\) must leave a remainder of 1 when divided by \(2, 3, 4, \ldots, 10\).

Thus, \(n-1>0\) must be divisible by \(2, 3, 4, \ldots, 10\). By definition, \(n-1\) is the least common multiple of \(2, 3, \ldots, 10\), so \(n-1=8 \cdot 9 \cdot 5 \cdot 7=2520\). The desired number is \(n=2521\).