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.

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

= 220_ten

= 8 x 9¹ + 6 x 9° = 72 + 6

= 78_ten

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

= 392_ten

220 + 78 + 392 = 690_ten.

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

= 846_nine.

Subtract 265 from 317 in base eight.

Solution:

First convert to base ten.

For 265_eight 

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

= 181_ten

For 317_eight

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

= 207_ten

Step two: Now subtract

207 – 181 = 26_ten.

Final step is to convert it back to base eight.

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

Solution:

First convert the known figures to base ten.

For 123_four

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

= 27_ten

For 1302_four

= 1 x 4³ + 3 x 4² + 0 x 4¹ + 2 x 4° 

= 64 + 48 + 0 + 2

= 114_ten

For 10122_four

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

= 256 + 0 + 16 + 8 + 2

= 282_ten

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

= 282 – 141 = 141_ten

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

Evaluate 321_six X 25_six

Solution:

Step 1: Convert to base ten first.

For 321_six

= 3 x 6² + 2 x 6¹ + 1 x 6°

= 108 + 12 + 1

= 121_ten

For 25_six

= 2 x 6¹ + 5 x 6°

= 12 + 5 

= 17_ten

Step 2: Multiply 121 by 17

 = 121 x 17

= 2057_ten

Step 3: Convert back to base six

Find the product of 1011_two and 1101_two.

Solution:

Step 1: Convert to base ten

For 1011_two

= 1 x 2³ + 0 x 2² + 1 x 2¹ + 1 x 2°

= 8 + 0 + 2 + 1

= 11_ten

For 1101_two

= 1 x 2³ + 1 x 2² + 0 x 2¹ + 1 x 2°

= 8 + 4 + 0 + 1

= 13_ten

Step 2: Multiply the two.

11 x 13 = 143_ten

Step 3: Convert it back to base two.

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

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.

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

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

Solving the quadratic equation

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.

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

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 

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

Return to base nine

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

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

Add the two numbers 

1626 + 6 = 1632 base ten

Return it to base  eight

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

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

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

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”

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.

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”

(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.

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

Add all

25 + 15 + 22 = 62 base ten

Convert it back to binary

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 

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

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.

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.

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.