Number bases

Place value in any base

Our usual numbers are base 10: the columns are worth $1, 10, 10^2, 10^3, \dots$ and the digits run from 0 to 9. The same idea works in any base $b$: columns are worth powers of $b$, and the digits run from 0 to $b - 1$. A subscript shows the base, so $234_5$ means base 5.

That digit rule matters: $158_5$ is meaningless, because 8 is not a base-5 digit.

Converting to decimal

Multiply each digit by its column value and add:

$$234_5 = 2 \times 5^2 + 3 \times 5 + 4 = 50 + 15 + 4 = 69$$

The same works for any base — for binary, the columns are $1, 2, 4, 8, 16, \dots$, so $1011_2 = 8 + 0 + 2 + 1 = 11$.

Converting from decimal

Divide by the base repeatedly, keeping track of remainders, until the quotient reaches 0. The remainders, read in reverse (last to first), are the digits. Converting 45 to base 3:

• $45 \div 3 = 15$ remainder $0$
• $15 \div 3 = 5$ remainder $0$
• $5 \div 3 = 1$ remainder $2$
• $1 \div 3 = 0$ remainder $1$

Reading upwards gives $45 = 1200_3$. Check it: $1 \times 27 + 2 \times 9 = 45$. Always convert back as a check — it takes seconds.

Arithmetic and bigger bases

Column arithmetic works in any base — you just carry at the base instead of at ten. In base 7, $5 + 4$ is nine, which is one seven and two: $12_7$. Subtraction borrows a seven, not a ten.

Hexadecimal (base 16) needs six extra digit symbols: A to F stand for 10 to 15.

When a puzzle hides the base, call it $N$ and use place value: if $44_N = 36$, then $4N + 4 = 36$, so $N = 8$. Remember the constraints — the base must be a whole number larger than every digit that appears.