Number bases · Number bases

We count in base ten: each place is worth ten times the one to its right (units, tens, hundreds and so on). In base five, each place is worth five times the one to its right: units, fives, twenty-fives, one hundred and twenty-fives. A base-five number uses only the digits 0 to 4.

Try it

Number basesChange the number and the base

place value	5³ = 125	5² = 25	5¹ = 5	5⁰ = 1
digit	1	1	0	1
value	125	25	0	1

1101_fivein base five151in base ten

number (base ten) = 151base = 5

In base five the place values are powers of 5: 125, 25, 5, 1. Multiply each digit by its place value and add: 1 × 125 + 1 × 25 + 0 × 5 + 1 × 1 = 151. Every digit must be less than 5.

Pick a number and a base. To base ten multiplies each digit by its place value. From base ten keeps dividing by the base. The two methods undo each other, so each is a check on the other.

To base ten: multiply by place values

\begin{aligned} &3021_{\text{five}} \\ &= 3 \times 5^3 + 0 \times 5^2 + 2 \times 5 + 1 \\ &= 375 + 0 + 10 + 1 \\ &= 386 \end{aligned}

Place values in base fiveEach place is 5 times the one to its right

More: changing to base ten

From base ten: divide and read the remainders upwards

To write 45 in base three, keep dividing by 3 and note each remainder:

${45 \div 3 = 15}$ remainder 0;
${15 \div 3 = 5}$ remainder 0;
${5 \div 3 = 1}$ remainder 2;
${1 \div 3 = 0}$ remainder 1.

Read the remainders from the bottom up: $1200_{\text{three}}$ .

Divide, then read upwardsThe first remainder is the units digit

More: changing from base ten

Between two other bases: go through base ten

To change $314_{\text{five}}$ to base three:

first to base ten: $314_{\text{five}} = 3 \times 25 + 1 \times 5 + 4 = 84$ ;
then keep dividing 84 by 3, and read the remainders upwards: $10010_{\text{three}}$ .

More: between two other bases

Arithmetic in a base

Add column by column as usual, but carry whenever a column reaches the base. In base five, $4 + 3 = 7$ , which is one 5 and 2 over: write 2 and carry 1.

Carrying in base fiveCarry 1 each time a column reaches 5

To subtract, borrow one from the next column. In base five, a borrowed 1 is worth 5 in the column to its right, not 10.

To multiply or divide, the safest way is to change both numbers to base ten, work there, then change the answer back.

More: arithmetic in a base

An unknown base

Write both numbers in base ten, with the unknown base as a letter, and solve the equation.

Worked example · WAEC 2014

WAEC 2014 · Paper 2 · Q2 (b)

If $124_n = 232_{\text{five}}$ , find $n$ .

Base ten on each side
- ${124_n = 1 \times n^2 + 2 \times n + 4}$ , which is ${n^2 + 2n + 4}$ .
- ${232_{\text{five}} = 2 \times 25 + 3 \times 5 + 2 = 67}$ .
Think first. What are the place values in base $n$ ?
Solve
- ${n^2 + 2n + 4 = 67}$ .
- Take 67 from both sides: ${n^2 + 2n - 63 = 0}$ .
- Factorise: ${(n + 9)(n - 7) = 0}$ , so ${n = -9}$ or ${n = 7}$ .
Choose the base
A base must be a positive whole number bigger than every digit used (here 4), so $n = 7$ .
Think first. Which root can be a base?

An unknown digit works the same way. Call it $p$ , write the number in base ten with $p$ in it, and solve. Remember that a digit must be less than the base.

More: an unknown base or digit

Your turn

WAEC 2019 · Paper 2 · Q3 (b)

(b)
If $1342_{\text{five}} - 241_{\text{five}} = x_{\text{ten}}$ , find the value of $x$ .

Worked solution (try it first)

(b)

Change both to base ten: $1342_{\text{five}} = 1 \times 125 + 3 \times 25 + 4 \times 5 + 2$
$= 222$ and $241_{\text{five}} = 2 \times 25 + 4 \times 5 + 1$
$= 71$ .
So $x = 222 - 71 = 151$ .

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