Number bases · Number bases

We count in base ten: each place is worth ten times the one to its right (units, tens, hundreds, …). 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 only uses 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 divides by the base repeatedly. Both give the same pair, so each checks 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: $45 \div 3 = 15$ r 0, $15 \div 3 = 5$ r 0, $5 \div 3 = 1$ r 2, $1 \div 3 = 0$ r 1. Reading 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 $314_{\text{five}} = 75 + 5 + 4 = 84$ , then divide 84 by 3 repeatedly to get $10010_{\text{three}}$ .

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, $4 + 3 = 7 = 1 \times 5 + 2$ : 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, and 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
$124_n = 1 \times n^2 + 2 \times n + 4 = n^2 + 2n + 4$ and $232_{\text{five}} = 2 \times 25 + 3 \times 5 + 2 = 67$ .
Think first. What are the place values in base $n$ ?
Solve
$n^2 + 2n + 4 = 67$ , so $n^2 + 2n - 63 = 0$ and $(n + 9)(n - 7) = 0$ .
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 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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