 Teacher resources and professional development across the curriculum

Teacher professional development and classroom resources across the curriculum            Solutions for Session 3, Part A

See solutions for Problems: A1 | A2 | A3    Problem A1

Counting in base two would look like this:

 1 10 11 100 101 110 111 1000 1001

A pattern you can observe is that each time you've reached the highest possible digit in a particular place value (which in this system is 1), you move to the next place value. This is the same pattern that occurs in base ten (as well as any other base), the only difference being that the highest possible digit in the base ten system is 9.   Problem A2

 a. The answer is 100110. The highest power of 2 we can make with 38 is 32, so our columns should read 32, 16, 8, 4, 2, and 1. We can use the 32 (38 - 32 = 6), so we record a 1 in that column. We can't use a 16 or an 8, so we record zeros in those columns. The next-highest power we can use is 4 (6 - 4 = 2), so we record a 1 in that column as well as the next one. Therefore, 38ten = 100110two. b. The answer is 111111. The highest power we can make with 63 is 32, so our columns should read 32, 16, 8, 4, 2, and 1, as before. We can use the 32 (63 - 32 = 31), so we record a 1 in that column. We can use the next-highest power, 16 (31 - 16 = 15), then the next-highest (15 - 8 = 7), then the next-highest (7 - 4 = 3), then the next-highest (3 - 2 = 1), and finally the last power (1 - 1 = 0). All six columns are filled with 1s, so 63ten = 111111two.   Problem A3

 a. This value is equal to (1 • 23) + (1 • 22) + (0 • 21) + (1 • 20) = 13. b. This value is equal to (1 • 24) + (1 • 23) + (1 • 22) + (1 • 21) + (1 • 20) = 31.     