187 – Quiz 8

Here is quiz 8.

Problem 1 asks to find integers such that  

One way of doing this is to follow the Euclidean algorithm to compute the g.c.d. of 13 and 8:

Now we work from these equations, starting with the last one and going up, expressing the remainders as linear combinations of the other two numbers:

Since we have

Since we have

Since we have

We have found that satisfy  

There are, of course, other ways we could have found these numbers, including trial and error: Simply list the multiples of 13 in order: 13, 26, 39, …, and note that 39 is 1 shy of a multiple of 8: giving the same solution as before.

Problem 2 asks for integers neither of which is 0, and such that

Perhaps the easiest solution is to make

Problem 3 asks for integers different from those in problem 1, such that

One way of finding these numbers is by adding the solutions to problems 1 and 2:

so

We could have also subtracted the solution to problem 2 from the solution to problem 1, to obtain

Or we could have squared the solution to problem 1: or

or

Problem 4 asks  for the number of injective functions from into

Let be 1-1. There are 5 possibilities for the value Whatever it is, can take any value in other than so there are 4 possibilities for The value can be anything in other than or so there are 3 possibilities for Finally, can take any of the remaining 2 values. This gives a total of injective functions.

Note that this is the same as the number of lists of length 4 without repetitions with values taken from the set

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