# Quiz 2 - Solutions

### Part (a)

47 divided by 7 leaves 6 remainder 5.
In other words, 47 = 6x7 + 5. Therefore a=5 works.

### Part (b)

There were several ways to approach this. One is to compute all the values:
• f(0)=3
• f(1)=4
• f(2)=5
• f(3)=6
• f(4)=0 [as 4+3=7 which is congruent to 0 mod 7]
• f(5)=1 [as 5+3=8 which is congruent to 1 mod 7]
• f(6)=2 [as 6+3=9 which is congruent to 2 mod 7]
Since all the numbers A appear once (and only once) in the right hand column, f is bijective. [You could have also shown this by drawing two columns of 0's through 6's, and then linking up values and their images. The crucial point is that you have to connect this with being bijective, by explicitly stating that each number in the second column corresponds to EXACTLY one number in the first column.]

Another way to establish that f is bijective is to show that it has an inverse. In this case, the inverse is the function g(y) which equals the remainder of y-3 upon division by 7. [Because if y is congruent to x+3, then x is congruent to y-3. Note that x is NOT equal to y-3; you have to take remainders, or else you won't have numbers between 0 and 6.]