Bases & Digits

What you need to know

• What the digits mean. In base $b$, the places stand for $1, b, b^2, b^3, \dots$ from right to left. So the digits $d_k \dots d_1 d_0$ mean $d_k b^k+\dots+d_1 b+d_0$. Each digit is a whole number from $0$ to $b-1$, and the first digit is not $0$. Example: $1203_4 = 1\cdot64+2\cdot16+0\cdot4+3 = 99$.
• How many digits. A positive whole number $N$ has exactly $k$ digits in base $b$ exactly when $b^{k-1} \le N < b^k$. Example: the $5$-digit numbers in base $2$ are $16$ through $31$, since $2^4 = 16$ and $2^5 = 32$.
• Converting to base $b$. Divide by $b$ again and again; the remainders, read from last to first, are the digits. Or take away the largest power of $b$ that fits, as often as it fits, and repeat with smaller powers. Example: $50 = 16\cdot3+2$, $16 = 5\cdot3+1$, $5 = 1\cdot3+2$, $1 = 0\cdot3+1$, so $50 = 1212_3$.
• Base-$10$ numbers with letters. A two-digit number with digits $a$ and $b$ is $10a+b$ (here $b$ is a digit, not a base); a three-digit one is $100a+10b+c$. Reversing $10a+b$ gives $10b+a$, and the difference is $9(a-b)$, always a multiple of $9$. Example: $72-27 = 45 = 9\cdot5$.
• The digit-sum trick. A number and its digit sum leave the same remainder when divided by $9$. We write $N \equiv (\text{digit sum}) \pmod 9$, where "$\pmod 9$" means we compare remainders after dividing by $9$. In base $b$, use $b-1$ instead of $9$. Example: $1234$ and $1+2+3+4 = 10$ both leave remainder $1$. In base $5$, $213_5 = 58$ and $2+1+3 = 6$ both leave remainder $2$ when divided by $4$.
• Palindromes. A palindrome reads the same both ways. A four-digit palindrome with digits $a, b, b, a$ equals $1001a+110b$. Since $1001 = 11\cdot91$ and $110 = 11\cdot10$, it is always divisible by $11$. Example: $3443 = 11\cdot313$.

How AMC 10 tests it

• "$N$ is written one way in base $b$ and another way in base $b+1$ (or base $10$); find $b$ or $N$." Write each form as a polynomial in $b$ and solve (problems 10–18).
• Digit puzzles: a number equals the product of its digits plus their sum, or reversing its digits multiplies it by a fixed number. Write it as $10a+b$ and solve.
• Counting palindromes, or numbers with a digit property, in a range, sometimes in base $2$ or $3$ (problems 12–20).
• Digits of powers or factorials in another base, such as how many zeros $n!$ ends with in base $12$. Since $12 = 2^2\cdot3$, each zero uses up two $2$s and one $3$: count the $2$s and $3$s in $n!$, halve the count of $2$s (rounding down), and take the smaller number. Example: $9!$ has seven $2$s and four $3$s, so it ends in the smaller of $3$ and $4$, which is $3$ zeros in base $12$.

Standard approaches

1. Turn every digit string into algebra at once. The base becomes an unknown whole number with limits on it.
2. Use $b^{k-1} \le N < b^k$ to narrow down $b$ before solving anything.
3. Look at remainders after dividing by $9$, $b-1$, or $11$ to rule out cases quickly.
4. To count numbers, build them digit by digit and multiply the choices. Treat the first digit separately, since it can't be $0$.
5. When the bases are small, convert and look. A table of small powers of $2$ and $3$ beats theory.

Worked example

When $2024$ is written in base $b$, it has exactly three digits and its last digit is $0$. How many possible values of $b$ are there?

(A) $1$ (B) $2$ (C) $3$ (D) $4$ (E) $5$

Step 1: three digits. We need $b^2 \le 2024 < b^3$.

Step 2: limits on $b$. Since $12^3 = 1728$ is less than $2024$ and $13^3 = 2197$ is more, $b \ge 13$. Since $44^2 = 1936 \le 2024$ but $45^2 = 2025$, $b \le 44$.

Step 3: last digit $0$. The last digit is the remainder when $2024$ is divided by $b$. It is $0$ exactly when $b$ divides $2024$.

Step 4: count. $2024 = 2^3\cdot11\cdot23$ has divisors $1, 2, 4, 8, 11, 22, 23, 44, 46, \dots$. The ones from $13$ to $44$ are $22$, $23$, and $44$. For example, $2024 = 4\cdot22^2+4\cdot22+0 = 440_{22}$.

The answer is $\boxed{\textbf{(C)}\ 3}$.

Pitfalls

• Allowing a digit that is $b$ or more, or a first digit of $0$; either makes a "solution" invalid. Example: if you get $b = 4$ for a number written $352_b$, throw it out: base $4$ has no digit $5$.
• Reading the remainders in the wrong order: the first remainder is the last (units) digit. Example: $50$ in base $3$ gives remainders $2, 1, 2, 1$; written in that order, $2121_3 = 70$, not $50$.
• Mixing up "exactly three digits" with "at most three digits." "At most" drops $b^2 \le N$, which is what puts an upper limit on $b$. Example: base $46$ writes $2024$ with only two digits, so it must not count above.
• Assuming base-$10$ digit facts work in every base. The digit-sum test for $3$ works only because $10$ leaves remainder $1$ when divided by $3$. Example: $12_8 = 10$ has digit sum $3$, but $10$ is not divisible by $3$.

Traps that recur

• Testing n = 1 alone (which yields many digit triples) or using C_n with n digits instead of 2n digits. (2018 AMC 10A #25)
• Treating the base-5 and base-6 strings as base-10 numbers without separating the units-digit condition from the tens-digit condition, or forgetting carries are impossible here. (2013 AMC 10B #25)
• Forgetting to drop b = 3 (below the required b ≥ 5) from the residue class 3 mod 8, which gives K = 759 and digit sum 21. (2024 AMC 10A #18)
• Counting the 17 alternating terms as 17 powers of two, or getting 8 * 17 = 136 and forgetting the final +1. (2020 AMC 10A #21)