# Number Bases

## ADDITION AND SUBTRACTION IN NUMBER BASES.

Most students have been finding it so difficult to add or subtract in number bases. The method used on tackling the problems is not different from what we have done here.

### Example 1: (Number Bases)

If you are given 264, 86, 475 all in base nine to be added.

Do not hesitate to convert those numbers first to base ten.

##### For 264nine

= 2 x 9² + 6 x 9¹ + 4 x 9° = 162 + 54 + 4

= 220ten

##### For 86nine

= 8 x 9¹ + 6 x 9° = 72 + 6

= 78ten

##### For 475nine

= 4 x 9² + 7 x 9¹ + 5 x 9° = 324 + 63 + 5

= 392ten

#### Step two: Add up all those numbers in base ten.

220 + 78 + 392 = 690ten.

Final step is to convert it back to base nine.

Learn how to convert from base ten to any required base here.

Therefore, (264 + 86 + 475)nine

= 846nine.

### Example 2. (Number Bases)

Subtract 265 from 317 in base eight.

Solution:

First convert to base ten.

For 265eight

= 2 x 8² + 6 x 8¹ + 5 x 8° = 128 + 48 + 5

= 181ten

For 317eight

= 3 x 8² + 1 x 8¹ + 7 x ° = 192 x 8 x 7

= 207ten

Step two: Now subtract

207 – 181 = 26ten.

Final step is to convert it back to base eight.

Thus, (317 – 265)eight = 32eight.

### Example 3. (Number Bases)

If 123 + 1302 + xxxx in base four is 10122four, find the missing figure.

##### Solution:

First convert the known figures to base ten.

For 123four

= 1 x 4² + 2 x 4¹ + 3 x 4° = 16 + 8 + 3

= 27ten

For 1302four

= 1 x 4³ + 3 x 4² + 0 x 4¹ + 2 x 4°

= 64 + 48 + 0 + 2

= 114ten

For 10122four

= 1 x 4⁴ + 0 x 4³ + 1 x 4² + 2 x 4¹ + 2 x 4°

= 256 + 0 + 16 + 8 + 2

= 282ten

Step 2: Subtract (27 + 114)ten from 282ten

= 282 – 141 = 141ten

Step 3: Convert the 141ten back to the required base .

Therefore, xxxx = 2031.

## MULTIPLICATION AND DIVISION OF NUMBER BASES

This is not different from the one we have done when adding or subtracting number bases. As far as you know what multiplication, division, addition and subtraction are all about. All you have to do first is to convert to base ten, then return the answer gotten to the required base.

### Example 1: (Number Bases)

Evaluate 321six X 25six

Solution:

Step 1: Convert to base ten first.

For 321six

= 3 x 6² + 2 x 6¹ + 1 x 6°

= 108 + 12 + 1

= 121ten

For 25six

= 2 x 6¹ + 5 x 6°

= 12 + 5

= 17ten

Step 2: Multiply 121 by 17

= 121 x 17

= 2057ten

Step 3: Convert back to base six

321six X 25six = 13305six.

### Example 2: (Number Bases)

Find the product of 1011two and 1101two.

Solution:

Step 1: Convert to base ten

For 1011two

= 1 x 2³ + 0 x 2² + 1 x 2¹ + 1 x 2°

= 8 + 0 + 2 + 1

= 11ten

For 1101two

= 1 x 2³ + 1 x 2² + 0 x 2¹ + 1 x 2°

= 8 + 4 + 0 + 1

= 13ten

Step 2: Multiply the two.

11 x 13 = 143ten

Step 3: Convert it back to base two.

1011two X 1101two = 10001111two

### Example 3:

If 231 base four X 10101 base two = “M” base six, Find M.

Solution:

Do you notice that they are of different bases?. Do not panic because you can still solve it😊😊

Step1: Convert the two numbers to base ten

For 231 base four

= 2 x 4² + 3 x 4¹ + 1 x 4°

= 32 + 12 + 1

= 45 base ten

For 10101 base two

= 1 x 2⁴ + 0 x 2³ + 1 x 2² + 0 x 2¹ + 1 x 2°

= 16 + 0 + 4 + 0 + 1

= 21 base ten

Step 2: Multiply the two.

45 x 21 = 945 base ten

Step 3: Convert to the required base.

231 base four X 10101 base two  =  4213 base six

Therefore, M = 4213

### Example 4:

Evaluate 2115 base seven ÷ 12 base seven

Solution:

Step 1: Convert to base ten

For 2115 base seven

= 2 x 7³ + 1 x 7² + 1 x 7¹ + 5 x 7°

= 686 + 49 + 7 + 5

= 747 base ten

For 12 base seven

= 1 x 7¹ + 2 x 7°

= 7 + 2 = 9 base ten

Step 2: Now divide.

747/9 = 83 base ten

Step 3: Convert back to base seven.

83 base ten = 146 base seven.

So, (2115 ÷ 12) base ten = 146 base seven.

### Example 5:

Divide 1423 by 24 in base five.

Solution:

Step 1: Convert to base ten

For 1423 base five

= 1 x 5³ + 4 x 5² + 2 x 5¹ + 3 x 5°

= 125 + 100 + 10 + 3

= 238 base ten

For 24 base five

= 2 x 5¹ + 4 x 5°

= 10 + 4 = 14 base ten

Step 2: Divide

238/14 = 17 base ten

Step 3: Convert back to base five

Therefore:

1423 base five ÷ 24 base five = 32 base five

## OTHER CASES OF NUMBER BASES.

Number base sometimes have unknown that must be made known.

### Example 1:

Calculate “y” if 134 base “y” = 54 base eight

Step 1: Convert both sides to base ten

For 134 base “y”

= 1 x y² + 3 x y¹ + 4 x y°

= (y² + 3y + 4) base ten

For 54 base eight

= 5 x 8¹ + 4 x 8°

= 40 + 4 = 44 base ten

##### Step 2: Equate the two terms.

(y² + 3y + 4) = 44

y² + 3y + 4 = 44

y² + 3y + 4 -44 = 0

y² + 3y – 40 = 0

y² + 8y – 5y – 40 = 0

y(y + 8) – 5(y + 8) = 0

(y – 5)(y + 8) = 0

Either (y – 5) = 0

or (y + 8) = 0

When y – 5 = 0

y = 5

And when y + 8 = 0

y = – 8

Number base cannot take a negative value,

so y = 5.

### Example 2:

If 251 base “b” = 100 base two, find the value of “b”.

Solution:

Similar thing we did in example 1 above will be done here too.

Step 1: Convert to base ten.

For 251 base “b”

= 2 x b² + 5 x b¹ + 1 x b°

= (2b² + 5b + 1) base ten

For 100 base two

= 1 x 2² + 0 x 2¹ + 0 x 2°

= 4 + 0 + 0 = 4 base ten

##### Equating both of them

(2b² + 5b + 1) = 4

2b² + 5b + 1 = 4

2b² + 5b + 1 – 4 = 0

2b² + 5b – 3 = 0

2b² + 6b – b – 3 = 0

2b(b + 3) – 1(b + 3) = 0

(2b – 1)(b + 3) = 0

So, either 2b – 1 = 0 or

b + 3 = 0

If 2b – 1 = 0

b = 1/2

And if b + 3 = 0

b = -3

Base cannot have negative sign, hence b = 1/2

## QUESTIONS AND SOLUTIONS (Number Bases):

### Q1. Evaluate 3112 base five X 34 base five.

Solution:

Step 1: Convert to base ten

For 3112 base five

= 3 x 5³ + 1 x 5² + 1 x 5¹ + 2 x 5°

= 375 + 25 + 5 + 2 = 407 base ten

For 34 base five

= 3 x 5¹ + 4 x 5°

= 15 + 4 = 19 base ten

Step 2: Multiply both

407 x 19 = 7733 base ten

Step 3: Convert back to base five

(3112 x 34) base five = 221413 base five

### Q2. Divide 2173 by 16 in base nine

Solution:

Convert to base ten.

For 2173 base nine

= 2 x 9³ + 1 x 9² + 7 x 9¹ + 3 x 9°

= 1458 + 81 + 63 + 3

= 1605 base ten

For 16 base nine

= 1 x 9¹ + 6 x 9°

= 9 + 6 = 15 base ten

Divide the numbers

1605/15 = 107 base ten

2173 base nine ÷ 16 base nine = 128 base nine.

### Q3. 111 x 101 in base two

Solution:

Convert to base ten

For 111 base two

= 1 x 2² + 1 x 2¹ + 1 x 2°

= 4 + 2 + 1 = 7 base ten

For 101 base two

= 1 x 2² + 0 x 2¹ + 1 x 2°

= 4 + 0 + 1 = 5 base ten

Multiply the both

7 x 5 = 35 base ten

Return the number base ten to base two

111 x 101 in base two = 100011 base two

### Q4. 3132 base eight + 6 base eight

Solution:

Convert to base ten

For 3132 base eight

= 3 x 8³ + 1 x 8² + 3 x 8¹ + 2 x 8°

= 1536 + 64 + 24 + 2

= 1626 base ten

For 6 base eight

= 6 x 8° = 6 base ten

1626 + 6 = 1632 base ten

Return it to base  eight

3140 base eight.

### Q5. If 25 x 14 = 374, find the number base used.

Solution:

Let the number base used be “y”

(25 base y) X  (14 base y) = 374 base y

Convert all to base ten

For 25 base y

= 2 x y¹ + 5 x y°

= (2y + 5) base ten

For 14 base y

= 1 x y¹ + 4 x y°

= (y + 4) base ten

For 374 base y

= 3 x y² + 7 x y¹ + 4 x y°

= (3y² + 7y + 4) base ten

If (25 base y) X  (14 base y) = 374 base y

##### Then,

(2y + 5)(y + 4) = 3y² + 7y + 4

Expand the equation

2y² + 8y + 5y + 20 = 3y² + 7y + 4

Collection of like terms

2y² – 3y² + 13y – 7y + 20 – 4 = 0

-y² + 6y + 16 = 0

y² – 6y – 16 = 0

y² – 8y + 2y – 16 = 0

y(y – 8) + 2(y – 8) = 0

(y + 2)(y – 8) = 0

y = – 2 or 8

Therefore y = 8

### Q6. Given that 132 base five = P base six, find P.

Solution:

Convert to base ten

For 132 base five

= 1 x 5² + 3 x 5¹ + 2 x 5°

= 25 + 15 + 2

= 42 base ten

For P base six

= P x 6° = P base ten

Equating both sides

P = 42 base ten

Convert back to base six

P = 110 base six

### Q7. If 244 base n = 1022 base four, find n.

Solution:

Convert to base ten

For 244 base n

= 2 x n² + 4 x n¹ + 4 x n °

= (2n² + 4n + 4) base ten

For 1022 base four

= 1 x 4³ + 0 x 4² + 2 x 4² + 2 x 4°

= 64 + 0 + 8 + 2

= 74 base ten

##### Equating both of them

(2n² + 4n + 4) = 74

2n² + 4n – 70 = 0

n² + 2n – 35 = 0

(n – 5)(n + 7) = 0

n = 5 or – 7

Therefore n = 5

because we do not have negative base

### Q8. A number is written as 52 base x. Four times the number is written as 301 base x. What is x?Solution:

A number = 52 base x

Four times the number = 4 X 52 base x = 301 base x

Convert to base ten

For 4

= 4 X x° = 4 base ten

#### For 52 base x

= 5 X x¹ + 2 X x°

= (5x + 2) base ten

For 301 base x

= 3 X x² + 0 X x¹ + 1 X x °

= 3x² + 0 + 1

= (3x² + 1) base ten

4(5x + 2) = 3x² + 1

20x + 8 = 3x² + 1

3x² – 20x – 7 = 0

Let us use quadratic formula to find the value of “x”

x = 7

### Q9. Arrange the following in ascending order.

1101 base two, 26 base eight, 113 base four and 1024 base six.

Solution:

Convert all to base ten.

For 1101 base two

= 1 x 2³ + 1 x 2² + 0 x 2¹ + 1 x 2°

= 8 + 4 + 0 + 1

= 13 base ten

For 26 base eight

= 2 x 8¹ + 6 x 8°

= 16 + 6 = 22 base ten

##### For 113 base four

= 1 x 4² + 1 x 4¹ + 3 x 4°

= 16 + 4 + 3 = 23 base ten

For 1024 base six

= 1 x 6³ + 0 x  6² + 2 x 6¹ + 4 x 6°

= 216 + 0 + 12 + 4

= 232 base ten

Therefore, the ascending order of magnitude

= 1101, 26, 113 and 1024 to their respective bases.

### Q10. If 321 base n = 232 base seven. Find the value of n.

Solution:

Convert all to base ten first.

For 321 base n

= 3 x n² + 2 x n¹ + 1 x n°

= (3n² + 2n + 1) base ten

For 232 base seven

= 2 x 7² + 3 x 7¹ + 2 x 7°

= 98 + 21 + 2

= 121 base ten

##### Equate the two.

3n² + 2n + 1 = 121

3n² + 2n + 1 – 121 = 0

3n² + 2n – 120 = 0

Using quadratic method to find the value of “n”

n = 6

### Q11. Simplify the following binary numbers.

(a). 101 + 111

Solution:

Convert to base ten

For 101

= 1 x 2² + 0 x 2¹ + 1 x 2°

= 4 + 0 + 1 = 5 base ten

For 111

= 1 x 2² + 1 x 2¹ + 1 x 2°

= 4 + 2 + 1 = 7 base ten

101 + 111 = 5 base ten + 7 base ten

= 12 base ten

Convert 12 base ten back to binary.

Therefore 101 + 111 = 1100

(b). 11001 + 1111 + 10110

Solution:

For 11001

= 1 x 2⁴ + 1 x 2³ + 0 x 2² + 0 x 2¹ + 1 x 2°

= 16 + 8 + 0 + 0 + 1

= 25 base ten

For 1111

= 1 x 2³ + 1 x 2² + 1 x 2¹ + 1 x 2°

= 8 + 4 + 2 + 1

= 15 base ten

For 10110

= 1 x 2⁴ + 0 x 2³ + 1 x 2² + 1 x 2¹ + 0 x 2°

= 16 + 0 + 4 + 2 + 0

= 22 base ten

25 + 15 + 22 = 62 base ten

Convert it back to binary

Therefore 11001 + 1111 + 10110 = 111110

(c). 11111 – 1010

Solution:

For 11111

= 1 x 2⁴ + 1 x 2³ + 1 x 2² + 1 x 2¹ + 1 x 2°

= 16 + 8 + 4 + 2 + 1

= 31 base ten

For 1010

= 1 x 2³ + 0 x 2² + 1 x 2¹ + 0 x 2°

= 8 + 0 + 2 + 0

= 10 base ten

Subtract 10 base ten from 31 base ten

= 21 base ten.

Convert it back to binary

So, 11111 – 1010 = 10101

(d). 1111 x 110

Solution:

Convert both to base ten

For 1111

= 1 x 2³ + 1 x 2² + 1 x 2¹ + 1 x 2°

= 8 + 4 + 2 +1

= 15 base ten

For 110

= 1 x 2² + 1 x 2¹ + 0 x 2°

= 4 + 2 + 0

= 6 base ten

15 x 6 = 90 base ten

Return 90 base ten to binary

Thus, 1111 x 110 = 1011010

### Q12. Simplify 342 base five + 134 base five + 223 base five and leave your answer in base

Solution:

Always convert to base ten first before any other thing.

For 342 base five

= 3 x 5² + 4 x 5¹ + 2 x 5°

= 75 + 20 + 2

= 97 base ten

For 134 base five

= 1 x 5² + 3 x 5¹ + 4 x 5°

= 25 + 15 + 4

= 44 base ten

For 223 base five

= 2 x 5² + 2 x 5¹ + 3 x 5°

= 50 + 10 + 3

= 63 base ten

Then, 97 + 44 – 63 = 78 base ten.

Convert 78 base ten to the required base, which is base five.

So, 342 base five + 134 base five + 223 base five = 303 base five.

### Q13. Multiply (12012) base three by (201) base three.

Solution:

For 12012 base three

= 1 x 3⁴ + 2 x 3³ + 0 x 3² + 1 x 3¹ + 2 x 3°

= 81 + 54 + 0 + 3 + 2

= 140 base ten

For 201 base three

= 2 x 3² + 0 x 3¹ + 1 x 3°

= 18 + 0 + 1

= 19 base ten

140 by 19 = 140 x 19 = 2660 base ten

Convert the 2660 base ten back to base three.

Therefore, (12012) base three by (201) base three = 10122112 base three.

Q14. Divide (100001) base two by (11) base two

Solution:

Convert to base ten.

For 100001 base two

1 x 2^5 + 0 x 2⁴ + 0 x 2³ + 0 x 2² + 0 x 2¹ + 1 x 2°

= 32 + 0 + 0 + 0 + 0 + 1

= 33 base ten

For 11 base two

= 1 x 2¹ + 1 x 2°

= 2 + 1 = 3 base ten

Now divide

33 ÷ 3 = 11 base ten

Convert it back to base two.

Therefore,  100001 base two divided by 11 base two = 1011 base two.