Math Core

We write numbers in base ten because we have ten fingers. Computers use base two, and clocks use base sixty for minutes and seconds. MATHCOUNTS regularly asks you to convert between bases, do arithmetic in another base, or find an unknown base. Once you see what the digits really mean, all three are routine.

What the digits mean

In base ten, $2143$ means $2 \cdot 10^3 + 1 \cdot 10^2 + 4 \cdot 10 + 3$ . In base $b$ , the place values are powers of $b$ instead.

Definition

Base-b numeral

In base $b$ , the numeral $\underline{d_3}\,\underline{d_2}\,\underline{d_1}\,\underline{d_0}$ (written with a subscript, like $2143_5$ ) means $d_3 \cdot b^3 + d_2 \cdot b^2 + d_1 \cdot b + d_0.$ Each digit is a whole number from $0$ to $b - 1$ .

Common mistake

Base $b$ has no digit equal to $b$ . In base $5$ the digits are $0$ – $4$ only, so " $2173_5$ " is not a valid numeral. When a problem asks for an unknown base, the base must be larger than every digit that appears.

Converting to base ten

Multiply each digit by its place value and add.

Worked example: Base five to base ten

Convert $2143_5$ to base ten.

$2 \cdot 125 + 1 \cdot 25 + 4 \cdot 5 + 3 = 250 + 25 + 20 + 3 = 298.$

Converting from base ten

Find the largest power of $b$ that fits, see how many times it fits, and repeat with what's left. Alternatively, divide by $b$ repeatedly and read the remainders from bottom to top.

Worked example: Base ten to base six

Write $200$ in base six.

Powers of $6$ : $1, 6, 36, 216$ . Since $216 > 200$ , start with $36$ .

$200 = 5 \cdot 36 + 20$ , so the $36$ s digit is $5$ .
$20 = 3 \cdot 6 + 2$ , so the $6$ s digit is $3$ and the units digit is $2$ .

So $200 = 532_6$ . With repeated division: $200 \div 6 = 33$ R $2$ , $33 \div 6 = 5$ R $3$ , $5 \div 6 = 0$ R $5$ . Reading the remainders upward gives $532$ .

The last digit is a remainder

The units digit of $n$ in base $b$ is the remainder when $n$ is divided by $b$ . For example, $200 = 532_6$ ends in $2$ because $200$ leaves remainder $2$ when divided by $6$ .

Arithmetic in another base

Add column by column just like in base ten, but carry when a column reaches $b$ instead of $10$ .

Worked example: Adding in base seven

Compute $365_7 + 254_7$ . Give your answer in base seven.

Units: $5 + 4 = 9 = 1 \cdot 7 + 2$ . Write $2$ , carry $1$ .
Sevens: $6 + 5 + 1 = 12 = 1 \cdot 7 + 5$ . Write $5$ , carry $1$ .
Forty-nines: $3 + 2 + 1 = 6$ . Write $6$ .

The sum is $652_7$ . Check in base ten: $365_7 = 194$ , $254_7 = 137$ , and $194 + 137 = 331 = 652_7$ .

Finding an unknown base

Write each numeral as a polynomial in $b$ and solve.

Worked example: Solve for the base

In some base $b$ , $(14_b)^2 = 232_b$ . Find $b$ .

$(b + 4)^2 = 2b^2 + 3b + 2$ , so $b^2 + 8b + 16 = 2b^2 + 3b + 2$ , which gives $b^2 - 5b - 14 = 0$ , or $(b - 7)(b + 2) = 0$ . The base must be positive, so $b = 7$ . Check: $14_7 = 11$ , $11^2 = 121$ , and $232_7 = 98 + 21 + 2 = 121$ .

Tip

Always check the answer against the digits. Here the largest digit is $4$ , and $7 > 4$ , so base $7$ is allowed.