Test 3 Section 2 Part 5


Problems 18-20 on starting on page 522
Test 3 Section 2 Part 5

3 Replies to “Test 3 Section 2 Part 5”

  1. For #20, a faster way of doing it is subtracting 3 from 15 (since you know the answer has a remainder of 3), and finding all the factors of that number. You can do this b/c k will go into the part of the answer that doesn’t have a remainder perfectly (i.e. w/o a remainder). So 15 – 3 = 12, and you would find all the factors of 12, which are 4,3,6, and 2. Except k can’t equal 3 b/c it wouldn’t make sense to have a remainder of 3 if k=3. So k would be true for 3 values instead of 4.

Leave a Reply

Your email address will not be published. Required fields are marked *