# Binary Arithmetic Operations – Addition | Subtraction in Other Bases

**a.** 3042_{five} + 2103_{five}

**Explanation:**

**1st Column (from right to left)** = 2 + 3 = 5. How many fives are there in 5? \( \frac{5}{5} \)= 1 remainder 0. Write 0 and carry 1 to column 2.

**2nd Column** = 1 + 4 + 0 = 5. How many fives are there in 5? \( \frac{5}{5} \)= 1 remainder 0. Write 0 and carry 1 to column 3.

**3rd Column** = 1 + 0 + 1 = 2. 2 is less than 5 so we write 2 down.

**4th column** = 3 + 2 = 5. How many fives are there in 5? \( \frac{5}{5} \)= 1 remainder 0. Write 0 and carry 1 to column 5.

**5th column** = We add 1 to the 5th column.

**Answer** = 10200_{5}

**b.** 2001_{three} + 1022_{three}

**Explanation**

**1st Column (from right to left)** = 1 + 2 = 3. How many threes are there in 3? \( \frac{3}{3} \)= 1 remainder 0. Write 0 and carry 1 to column 2.

**2nd Column** = 1 + 0 + 2 = 3. How many threes are there in 3? \( \frac{3}{3} \)= 1 remainder 0. Write 0 and carry 1 to column 3.

**3rd Column** = 1 + 0 + 0 = 1. 1 is less than 3 so we write 1 down.

**4th column** = 2 + 1 = 3. How many threes are there in 3? \( \frac{3}{3} \)= 1 remainder 0. Write 0 and carry 1 to column 5.

**5th column** = We add 1 to the 5th column.

**Answer** = 10100_{5}

**c.** 151.24_{six} + 12.34_{six} + 20.11_{six}

**Explanation**

**1st Column (from right to left)** = 4 + 4 + 1 = 9. How many six’s are there in 9? \( \frac{9}{6} \) = 1 remainder 3. Write 3 and carry 1 to column 2.

**2nd Column** = 1 + 2 + 3 + 1 = 7. How many six’s are there in 7? \( \frac{7}{6} \) = 1 remainder 1. Write 1 and carry 1 to column 3.

**3rd Column** = 1 + 1 + 2 + 0 = 4. 4 is less than 6 so we write 4 down.

**4th column** = 5 + 1 + 2 = 8. How many six’s are there in 8? \( \frac{8}{6} \) = 1 remainder 2. Write 2 and carry 1 to column 5.

**5th column** = 1 + 1 = 2. 2 is less than 6 so we write 6 down.

**Answer** = 224.13_{6}

**d.** 5346_{8} – 637_{8}

**1st Column (from right to left) **= 6 is less than 7 so we will have to borrow from the 2nd column. Since it is in base 8 any number borrowed is 8. So borrowing 8 from the second column will give (8 + 6) â€“ 7 = 14 – 7 = 7

**2nd Column** = This reduces 4 to 3 in the second column. 3 – 3 = 0. We write 0 down.

**3rd Column** = 3 is less than 6 so we will have to borrow from the 4th column. Since it is in base 8 any number borrowed is 8. So borrowing 8 from the fourth column will give (8 + 3) â€“ 6 = 11 – 6 = 5

**4th Column **= This reduces 5 to 4 in the fourth column. 4 – 0 = 4. We write 4 down.

**Answer** = 4507_{8}

**e.** 3002_{four} â€“ 2213_{four}

**Explanation**

**1st Column (from right to left) **= 2 is less than 3 so we will have to borrow from the 2nd column. Since it is in base 4 any number borrowed is 4. So borrowing 4 from the second column will give (4 + 2) â€“ 3 = 6 – 3 = 3

**2nd Column** = This reduces 0 to -1 in the second column. We will have to borrow (4) from the third column. 4 – 1 – 1 = 2. We write 2 down.

**3rd Column** = This reduces 0 to -1 in the third column. We will have to borrow (4) from the 4th column. 4 – 1 – 2 = 1. We write down 1.

**4th Column **= This reduces 3 to 2 in the fourth column. 2 – 2 = 0.

**Answer** = 123_{4}

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