We can simply add the units digits of the addends:
However, we could find the units digit with far less calculation. We could find their sum, which is, and note that the units digit is. Suppose we want to find the units digit of the following sum: This can be more easily understood with a few examples. If all we need to know about are remainders when integers are divided by, then we can work directly with those remainders in modulo. We don't always need to perform tedious computations to discover solutions to interesting problems. We know is a multiple of since is a multiple of. Īnd is the modulo residue of Another Solution: Otherwise,, which means that and are not congruent modulo. In general, two integers and are congruent modulo when is a multiple of. In modulo 5, two integers are congruent when their difference is a multiple of 5. The (mod 5) part just tells us that we are working with the integers modulo 5. We write this using the symbol : In other words, this means in base 5, these integers have the same residue modulo 5: For instance, we say that 7 and 2 are congruent modulo 5. There is a mathematical way of saying that all of the integers are the same as one of the modulo 5 residues. We say that is the modulo- residue of when, and. While this may not seem all that useful at first, counting in this way can help us solve an enormous array of number theory problems much more easily! This just relates each integer to its remainder from the Division Theorem. In general, for a natural number that is greater than 1, the modulo residues are the integers that are whole numbers less than : Because all integers can be expressed as, ,, , or in modulo 5, we give these integers their own name: the residue classes modulo 5. So, the integers from to, when written in modulo 5, are The same is true in any other modulus (modular arithmetic system). This is the way in which we count in modulo 12. Starting at noon, the hour hand points in order to the following: Let's use a clock as an example, except let's replace the at the top of the clock with a. 5.3.3 Solution using modular arithmetic.
5.1.4 Solution using modular arithmetic.5.1.3 Why we only need to use remainders.