Converting decimal to binary by repeated division by 2
Name:
Date:
Class:
A
Quick Questions
Shade in the A or B circle for each question.
1. When converting 13 to binary by repeated division, which remainders do you get, in the order you calculate them?
2. How do you read the remainders to write the binary number?
3. When do you stop dividing by 2?
4. Which remainders are possible when you divide by 2?
5. What is 6 in binary using repeated division by 2?
B
Match the Pairs
Draw a line connecting each item on the left to its match on the right.
| Left column | Right column | |
|---|---|---|
| 113 ÷ 2 = 6 remainder 1 | AThe first division step for 13: the quotient is 6 and the remainder 1 is noted. This remainder becomes the last binary digit (the least significant bit). | |
| 21 ÷ 2 = 0 remainder 1 | BThe remainders were read from first to last (top to bottom) instead of last to first. The correct answer is 1101. | |
| 3Read the remainders from last to first (bottom to top) to build the binary number. For 13 the remainders in order are 1, 0, 1, 1. | C1101 | |
| 4The place-value method gives the same answer as repeated division. 13 = 8 + 4 + 0 + 1, so use the places 8, 4, 2, 1. | D8 + 4 + 1 = 13 → 1101 | |
| 513 → remainders 1, 0, 1, 1 → answer 1011 | EThe division stopped when the quotient was 1. You must keep dividing until the quotient is 0 (1 ÷ 2 = 0 remainder 1), or the leftmost bit is missing. | |
| 613 ÷ 2 = 6, 6 ÷ 2 = 3, 3 ÷ 2 = 1 → stop | FThe final division step: the quotient reaches 0, so you stop. This last remainder is the first binary digit (the most significant bit). |
C
Definitions
repeated division by 2
D
Fill in the Blanks
Complete each line of code by filling in the missing parts.
1.
13 / _ = 6 remainder 1
2.
13 / 2 = 6 remainder 1
6 / 2 = 3 remainder _
3.
3 / 2 = 1 remainder 1
1 / 2 = _ remainder 1
quotient is 0, so stop
4.
Read the remainders from last to _____
5.
8 + 4 + 1 = 13 so 13 is ____
E
Spot the Error
Shade in the letter of the line that contains an error.
Question 1
A
13 / 2 = 6 remainder 1
B
6 / 2 = 3 remainder 1
C
3 / 2 = 1 remainder 1
D
1 / 2 = 0 remainder 1
E
Read bottom to top: 1101
Question 2
A
10 / 2 = 5 remainder 0
B
5 / 2 = 3 remainder 1
C
2 / 2 = 1 remainder 0
D
1 / 2 = 0 remainder 1
E
Read bottom to top: 1010
Question 3
A
6 / 2 = 3 remainder 2
B
3 / 2 = 1 remainder 1
C
1 / 2 = 0 remainder 1
D
Read bottom to top: 110
Question 4
A
9 / 2 = 4 remainder 1
B
4 / 2 = 2 remainder 0
C
2 / 2 = 1 remainder 0
D
1 / 2 = 0 remainder 0
E
Read bottom to top: 1001
Question 5
A
n ← 11
B
WHILE n > 1
C
r ← n MOD 2
D
n ← n DIV 2
E
END WHILE
Answer Key
Converting decimal to binary by repeated division by 2
Section A: Quick Questions
Q1
B — 1, 0, 1, 1
Q2
A — From last to first (bottom to top)
Q3
A — When the quotient reaches 0
Q4
B — Only 0 or 1
Q5
A — 110
Section B: Match the Pairs
Q1
13 ÷ 2 = 6 remainder 1 → The first division step for 13: the quotient is 6 and the remainder 1 is noted. This remainder becomes the last binary digit (the least significant bit).
Q2
1 ÷ 2 = 0 remainder 1 → The final division step: the quotient reaches 0, so you stop. This last remainder is the first binary digit (the most significant bit).
Q3
Read the remainders from last to first (bottom to top) to build the binary number. For 13 the remainders in order are 1, 0, 1, 1. → 1101
Q4
The place-value method gives the same answer as repeated division. 13 = 8 + 4 + 0 + 1, so use the places 8, 4, 2, 1. → 8 + 4 + 1 = 13 → 1101
Q5
13 → remainders 1, 0, 1, 1 → answer 1011 → The remainders were read from first to last (top to bottom) instead of last to first. The correct answer is 1101.
Q6
13 ÷ 2 = 6, 6 ÷ 2 = 3, 3 ÷ 2 = 1 → stop → The division stopped when the quotient was 1. You must keep dividing until the quotient is 0 (1 ÷ 2 = 0 remainder 1), or the leftmost bit is missing.
Section C: Definitions
Q1
repeated division by 2 — Divide a number by 2 again and again until the quotient is 0, noting each remainder. Read the remainders from last to first to get the binary number. Example: 13 is 1101.Section D: Fill in the Blanks
Q1
13 / 2 = 6 remainder 1Q2
13 / 2 = 6 remainder 1
6 / 2 = 3 remainder 0Q3
3 / 2 = 1 remainder 1
1 / 2 = 0 remainder 1
quotient is 0, so stopQ4
Read the remainders from last to firstQ5
8 + 4 + 1 = 13 so 13 is 1101Section E: Spot the Error
Q1
B — 6 is even, so it divides by 2 exactly and the remainder is 0, not 1. The line should be 6 / 2 = 3 remainder 0.
Q2
B — 5 divided by 2 is 2 with a remainder of 1, not 3. The line should be 5 / 2 = 2 remainder 1.
Q3
A — When you divide by 2 the remainder can only be 0 or 1. Since 6 divides exactly, the line should be 6 / 2 = 3 remainder 0.
Q4
D — 1 divided by 2 is 0 with a remainder of 1, because 2 goes into 1 zero times. The line should be 1 / 2 = 0 remainder 1.
Q5
B — You must keep dividing until the quotient is 0. Stopping when n is 1 would miss the last remainder. It should be WHILE n > 0.