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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. 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. 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 the that  the first clock does not have. For example suppose we want to solve the following equation \[  9x=0 \mod 12.  \]  There are $3$ numbers in the set of hours that give as a solution and these are $0$, $4$ and $8$. That is not so nice. Now suppose we want to solve the following equation  \[  9x=1 \mod 12.  \]  If you consider each multiple of 9 modulo 12 is in the set of hours $\{0,3,6,9\}$ hence the equation has no solution. In this case we say that the equation do not satisfy the "cancellation" property. In fact in a "standard" equation similar to the first one we can get multiply $9$ by its inverse, getting rid of the $9$, and obtaining the unique solution $0$.   By a similar computation  While in the second we and