AMC 10 · 2013 · #19
Grade 6 number-theoryIn base 10, the number 2013 ends in the digit 3. In base 9, on the other hand, 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?
Pick an answer.
AMC 10 2013 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: In base $b$, the number $2013$ is written with some string of digits. Count the positive integers $b$ for which that base-$b$ representation ends in the digit $3$.
Givens: The number is $2013$ (its value in base $10$).; In base $10$ it ends in $3$; in base $9$ it is $(2676)_9$ and ends in $6$.; We look at the final (units) digit of $2013$ written in base $b$.; Answer choices: (A) $6$, (B) $9$, (C) $13$, (D) $16$, (E) $18$
Unknowns: How many positive integers $b$ make the last base-$b$ digit of $2013$ equal to $3$.
Understand
Restated: In base $b$, the number $2013$ is written with some string of digits. Count the positive integers $b$ for which that base-$b$ representation ends in the digit $3$.
Givens: The number is $2013$ (its value in base $10$).; In base $10$ it ends in $3$; in base $9$ it is $(2676)_9$ and ends in $6$.; We look at the final (units) digit of $2013$ written in base $b$.; Answer choices: (A) $6$, (B) $9$, (C) $13$, (D) $16$, (E) $18$
Plan
Primary tool: #13 Convert to Algebra
Secondary: #2 Make a Systematic List, #3 Eliminate Possibilities
The words "ends in the digit $3$" hide a clean arithmetic statement: the last digit of a number in base $b$ is just its remainder on division by $b$. Tool #13 (Convert to Algebra) turns "$2013$ ends in $3$ in base $b$" into the congruence $2013 \equiv 3 \pmod{b}$, which rearranges to "$b$ divides $2010$." That converts a base-representation puzzle into a divisor-counting problem. Tool #2 (Make a Systematic List) then counts the divisors of $2010$ from its prime factorization, and Tool #3 (Eliminate Possibilities) removes the bases that are too small for the digit $3$ to be legal.
Execute — Answer: C
4.NBT.B.6 Step 1 Last digit is a remainder
- When you write a number in base $b$, its final (units) digit is the remainder you get when you divide the number by $b$.
- For example $2013 = 9 \cdot 223 + 6$, and indeed $(2013)_{10} = (2676)_9$ ends in $6$.
- So "$2013$ ends in the digit $3$ in base $b$" means exactly that dividing $2013$ by $b$ leaves remainder $3$.
💡 The units digit is whatever is left over after pulling out full groups of $b$ — it is the remainder.
4.OA.B.4 Step 2 Turn it into divisibility
- Remainder $3$ means $2013$ is exactly $3$ more than a multiple of $b$.
- Subtract that $3$: $2013 - 3 = 2010$ must be a whole multiple of $b$.
- So the condition becomes "$b$ is a factor of $2010$." Every base that ends $2013$ in $3$ is a divisor of $2010$, and every divisor of $2010$ makes the remainder $3$.
💡 Peel off the leftover $3$ and what remains, $2010$, has to split evenly by $b$.
4.OA.B.4 Step 3 Count the divisors of 2010
- Factor $2010$ into primes: $2010 = 2 \cdot 3 \cdot 5 \cdot 67$.
- Each prime appears once, so a divisor is built by independently choosing to include or exclude each of the four primes — $2$ choices per prime.
- That gives $2 \cdot 2 \cdot 2 \cdot 2 = 16$ divisors in total: $1, 2, 3, 5, 6, 10, 15, 30, 67, 134, 201, 335, 402, 670, 1005, 2010$.
💡 With no repeated prime, each prime is a yes/no switch, so the divisor count doubles four times.
6.EE.B.5 Step 4 Drop bases too small for digit 3
- A digit must be smaller than its base, so the digit $3$ can only exist when $b > 3$.
- Among the $16$ divisors, that rules out $b = 1$, $b = 2$, and $b = 3$: in those bases a "digit $3$" is impossible (base $1$ is not even a real positional base, and bases $2$ and $3$ only use digits up to $2$).
- All other divisors, $5$ through $2010$, are valid.
💡 You cannot write the digit $3$ in a base that only has digits $0$ up to $b-1 < 3$.
4.OA.B.4 Step 5 Final count
- Start with $16$ divisors of $2010$ and remove the $3$ forbidden small bases ($1, 2, 3$).
- That leaves $16 - 3 = 13$ valid bases: $5, 6, 10, 15, 30, 67, 134, 201, 335, 402, 670, 1005, 2010$.
- So $2013$ ends in the digit $3$ in exactly $13$ bases, which is answer $\textbf{(C)}$.
💡 Total divisors minus the three illegal tiny bases.
4.NBT.B.6 When you write a number in base $b$, its final (units) digit is the remainder yo 4.OA.B.4 Remainder $3$ means $2013$ is exactly $3$ more than a multiple of $b$. Subtract 4.OA.B.4 Factor $2010$ into primes: $2010 = 2 \cdot 3 \cdot 5 \cdot 67$. Each prime appea 6.EE.B.5 A digit must be smaller than its base, so the digit $3$ can only exist when $b > 4.OA.B.4 Start with $16$ divisors of $2010$ and remove the $3$ forbidden small bases ($1, Review
Reasonableness: Spot-check the endpoints of the valid list. Base $5$: $2013 = 5 \cdot 402 + 3$, remainder $3$ — ends in $3$. Base $2010$: $2013 = 2010 \cdot 1 + 3$, so $(2013)_{10} = (1\,3)_{2010}$ — ends in $3$. A non-divisor like $b = 4$ gives $2013 = 4 \cdot 503 + 1$, remainder $1$, correctly excluded. The three thrown-out divisors $1, 2, 3$ are exactly the ones where digit $3$ is illegal, so removing $3$ from $16$ is right. The count $13$ matches (C), and it sensibly sits between the raw divisor count $16$ and the smaller distractors.
Alternative: Tool #2 (Make a Systematic List) alone: after establishing $b \mid 2010$, list the divisor pairs of $2010$ directly — $(1,2010), (2,1005), (3,670), (5,402), (6,335), (10,201), (15,134), (30,67)$ — which shows $16$ divisors, then cross off $1, 2, 3$ by hand to reach $13$. Same answer without invoking the exponent-plus-one divisor formula.
CCSS standards used (min grade 6)
4.NBT.B.6Find whole-number quotients and remainders with up to four-digit dividends (Recognizing that the final base-$b$ digit of $2013$ is the remainder when $2013$ is divided by $b$, so "ends in $3$" means remainder $3$.)4.OA.B.4Find all factor pairs and recognize multiples; determine prime or composite (Rewriting the condition as $b \mid 2010$, prime-factoring $2010 = 2 \cdot 3 \cdot 5 \cdot 67$, counting its $16$ divisors, and doing the final divisor tally.)6.EE.B.5Understand solving an equation or inequality as a process of finding values (Applying the inequality $b > 3$ (a digit must be less than its base) to decide which divisors are legal bases and which must be discarded.)
⭐ The last digit of a number in base $b$ is just the remainder after dividing by $b$; "ends in $3$" means $b$ divides $2013 - 3 = 2010$, and $2010$ has $16$ divisors, so tossing out the bases $1, 2, 3$ (too small to hold a digit $3$) leaves $\textbf{(C)}\ 13$.
⭐ The last digit of a number in base $b$ is just the remainder after dividing by $b$; "ends in $3$" means $b$ divides $2013 - 3 = 2010$, and $2010$ has $16$ divisors, so tossing out the bases $1, 2, 3$ (too small to hold a digit $3$) leaves $\textbf{(C)}\ 13$.
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