Overview

We normally write numbers in base $10$, using ten digits $0$ through $9$. But there is nothing special about ten. Any positive integer $b \ge 2$ can serve as a base, and understanding how bases work opens up a family of techniques that appear regularly on the AMC, AIME, etc.

The main idea is that a number's representation in base $b$ directly encodes its remainders and quotients when divided by powers of $b$. This connects base representations to modular arithmetic.

What Is a Base?

In base $b$, we write a number using digits $0, 1, 2, \ldots, b-1$ and place values that are powers of $b$.

The base $b$ representation

stands for the value

Each digit $a_i$ satisfies $0 \le a_i \le b-1$.

For example, in base $10$:

In base $2$ (binary):

Common Bases

A few bases come up often in competition math:

• Base $2$ (binary): digits are $0$ and $1$. Useful in combinatorics and problems about subsets.
• Base $3$ (ternary): digits are $0, 1, 2$. Appears in balanced ternary and weighing problems.
• Base $8$ (octal) and base $16$ (hexadecimal): occasionally appear in digit manipulation problems.

However, contest problems can use any base, including large or variable ones.

Converting Base $b$ to Base $10$

Expand using the definition:

Another example:

where $A$ represents the digit $10$ in hexadecimal.

Converting Base $10$ to Base $b$

Repeatedly divide by $b$ and record the remainders. The remainders, read from bottom to top, give the base $b$ digits.

Convert $83$ to base $3$:

Reading the remainders from bottom to top: $83 = 10002_3$.

We can verify: $1(3^4) + 0 + 0 + 0 + 2 = 81 + 2 = 83$.

The Last Digit Rule

This is one of the most useful facts about bases in competition math:

The last digit of $N$ in base $b$ is the remainder when $N$ is divided by $b$.

In symbols,

where $a_0$ is the units digit of the base $b$ representation of $N$.

This follows directly from the expansion, every term except $a_0$ is a multiple of $b$.

Similarly, the last two digits of $N$ in base $b$ encode $N \pmod{b^2}$, and so on.

The Digit Sum and Casting Out

In base $10$, we know a number is divisible by $9$ when its digit sum is divisible by $9$. This generalises:

In base $b$, a number is divisible by $b-1$ if and only if the sum of its base-$b$ digits is divisible by $b-1$.

This is because $b \equiv 1 \pmod{b-1}$, so every power $b^k \equiv 1 \pmod{b-1}$, and therefore

Counting Digits

A positive integer $N$ has exactly $d$ digits in base $b$ if and only if

Equivalently,

For example, $100$ in base $2$ has $\lfloor \log_2 100 \rfloor + 1 = 6 + 1 = 7$ digits, since $2^6 = 64 \le 100 < 128 = 2^7$.

Worked Examples

Example 1

Convert $345$ to base $7$.

Reading remainders bottom to top: $345 = 1002_7$.

Check: $1(7^3) + 0 + 0 + 2 = 343 + 2 = 345$.

Example 2 (2013 AMC 10A Problem 19)

In base $10$, the number $2013$ ends in the digit $3$. In base $9$, the same number is written as $(2676)_9$ and ends in the digit $6$. For how many positive integers $b$ does the base-$b$ representation of $2013$ end in the digit $3$?

By the last digit rule, the base-$b$ representation of $2013$ ends in $3$ if and only if

This means $b \mid 2010$. We also need $b \ge 4$ since the digit $3$ must be a valid digit in base $b$ (digits go from $0$ to $b-1$, so $b > 3$).

Now $2010 = 2 \times 3 \times 5 \times 67$. The number of positive divisors of $2010$ is $(1+1)^4 = 16$. Three of these divisors ($1, 2, 3$) are less than $4$, so the answer is $16 - 3 = 13$.

Example 3

What is the sum of the digits of $999$ in base $8$?

First convert, $999 = 124(8) + 7$, then $124 = 15(8) + 4$, then $15 = 1(8) + 7$, then $1 = 0(8) + 1$.

So $999 = 1747_8$.

The digit sum is $1 + 7 + 4 + 7 = 19$.

Key Contest Insight

The connection between bases and modular arithmetic is the single most important idea:

• The last digit of $N$ in base $b$ equals $N \bmod b$.
• The last $k$ digits of $N$ in base $b$ encode $N \bmod b^k$.
• The digit sum in base $b$ relates to $N \bmod (b-1)$.

Whenever a problem mentions a specific base, ask yourself what modular information the base representation is encoding.

Strategy Tips

• When a problem asks about a number's representation in an unknown base $b$, translate digit conditions into divisibility or modular conditions on $b$.
• To count integers whose base $b$ digits satisfy some restriction, think of each digit position as an independent choice (possibly with constraints on the leading digit).
• Base $2$ representations connect naturally to subsets, each subset of $\{1, 2, \ldots, n\}$ corresponds to an $n$ digit binary string.
• If a problem involves both base $b$ and base $c$, rewrite everything in base $10$ to find common ground, then work modularly.

Common Pitfalls

• Forgetting that digits in base $b$ must satisfy $0 \le a_i \le b-1$. The digit $3$ does not exist in base $3$.
• Reading remainders in the wrong order when converting from base $10$ to base $b$. Remainders are read bottom to top.
• Confusing 'the number $123$ in base $5$' (which equals $38$ in base $10$) with 'the number $123$ converted to base $5$.'
• Assuming a number has the same number of digits in every base.

Practice Problems