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\[  26=2 \times 12 + 2.   \]  Obviously the remainder depends on how many are the hours the clock. Hence every number has a unique representation after fixing the number of hours. Because of this reduction we always assume that the labelling of the hours is exactly  \[  0,1,2,\ldots, n-1  \]  where $n$ is the number of the hours of the clock.  Over these clocks we can define a very nice arithmetic that is called the clock arithmetic, or modular arithmetic. That is we can enrich the set of "hours" with addition and multiplication in a natural way. That is we can add or multiply the numbers in the standard way and then apply the division with respect to the fixed hours and find the remainder. remainder, hence the hour. As for the case $\Fb$ we represent the tables of the two operations. First of all we consider the sum.  In both clocks, $\ZZ_{12}$ and $\ZZ_7$ the sum is nothing but a rotation of the first row by one element. Now focus on the multiplication  The two clocks behaves in different way. In $\ZZ_7$ the multiplication produce a much random table, but, in each row we find al the elements of $\ZZ_7$. If we study the table of $\ZZ_{12}$ there are some rows that are complete some other not.   We observe that  For example $11+15=26$ that in the clock of $12$ hours is $2$. Also the multiplication behaves nicely in the two clocks using the same method.   Unfortunately the second clock has a pathological behavior that the first clock does not have. For example suppose we want to solve the following equation