AMC 10 · 2003 · #10
Grade 6 countingNebraska, the home of the AMC, changed its license plate scheme. Each old license plate consisted of a letter followed by four digits. Each new license plate consists of three letters followed by three digits. By how many times has the number of possible license plates increased?
Pick an answer.
AMC 10 2003 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: An old license plate is $1$ letter then $4$ digits; a new one is $3$ letters then $3$ digits. Find how many times as many plates the new scheme allows compared to the old scheme — that is, the new count divided by the old count.
Givens: Old plate: one letter followed by four digits; New plate: three letters followed by three digits; There are $26$ possible letters and $10$ possible digits ($0$–$9$); Answer choices: (A) $\frac{26}{10}$, (B) $\frac{26^2}{10^2}$, (C) $\frac{26^2}{10}$, (D) $\frac{26^3}{10^3}$, (E) $\frac{26^3}{10^2}$
Unknowns: How many times as large the number of new plates is compared to the number of old plates (new $\div$ old)
Understand
Restated: An old license plate is $1$ letter then $4$ digits; a new one is $3$ letters then $3$ digits. Find how many times as many plates the new scheme allows compared to the old scheme — that is, the new count divided by the old count.
Givens: Old plate: one letter followed by four digits; New plate: three letters followed by three digits; There are $26$ possible letters and $10$ possible digits ($0$–$9$); Answer choices: (A) $\frac{26}{10}$, (B) $\frac{26^2}{10^2}$, (C) $\frac{26^2}{10}$, (D) $\frac{26^3}{10^3}$, (E) $\frac{26^3}{10^2}$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #16 Change Focus / Count the Complement, #3 Eliminate Possibilities
A plate is built slot by slot, and each slot is chosen independently, so tool #7 (Identify Subproblems) says: count the choices for each slot and multiply them to get the total number of plates. Do this once for the old scheme and once for the new. Then tool #16 (Change Focus) reframes "increased by how many times" as a single division — new total over old total — rather than a subtraction. Because the answer choices are all ratios of powers of $26$ and $10$, tool #3 (Eliminate Possibilities) lets the simplified exponents point straight at the matching choice without any large multiplication.
Execute — Answer: C
3.OA.A.1 Step 1 Count the old plates
- The old plate has five slots: one letter, then four digits.
- The letter slot has $26$ choices, and each digit slot has $10$ choices.
- Since the slots are filled independently, multiply the choices together.
- That gives $26$ times four copies of $10$, which is $26 \cdot 10^4$ possible old plates.
💡 Filling independent slots multiplies the choices, so a letter slot and four digit slots give $26 \cdot 10^4$.
6.EE.A.1 Step 2 Count the new plates
- The new plate has three letter slots and three digit slots.
- Each letter slot has $26$ choices and each digit slot has $10$ choices, all filled independently.
- Multiplying gives three copies of $26$ and three copies of $10$, which is $26^3 \cdot 10^3$ possible new plates.
💡 Three letter slots and three digit slots multiply into $26^3 \cdot 10^3$.
4.OA.A.2 Step 3 Turn "how many times" into a division
- "By how many times has the number increased" means: the new count is how many times the old count.
- That is answered by dividing the new total by the old total, not by subtracting.
- So set up the single fraction of new over old.
💡 "How many times as many" is always a division of the bigger amount by the smaller.
6.EE.A.3 Step 4 Simplify the ratio
- Cancel common factors of $26$ and of $10$ between the top and bottom.
- On top there are three $26$s and on the bottom one, leaving $26^{3-1} = 26^2$.
- On top there are three $10$s and on the bottom four, leaving one extra $10$ in the bottom, so a factor of $\frac{1}{10}$.
- The ratio is $\dfrac{26^2}{10}$, which is choice (C).
💡 Dividing like bases subtracts their counts: the $26$s drop from $3$ to $2$ and the $10$s from $3$ to $-1$.
3.OA.A.1 The old plate has five slots: one letter, then four digits. The letter slot has 6.EE.A.1 The new plate has three letter slots and three digit slots. Each letter slot has 4.OA.A.2 "By how many times has the number increased" means: the new count is how many ti 6.EE.A.3 Cancel common factors of $26$ and of $10$ between the top and bottom. On top the Review
Reasonableness: Going from one letter to three letters multiplies the count by $26^2$, and going from four digits to three digits divides it by $10$. So the plate count should grow by exactly $26^2$ and shrink by $10$, giving $\frac{26^2}{10}$ — matching (C). The result is bigger than $1$ (about $67.6$), which makes sense: gaining two extra letter slots ($\times 676$) far outweighs losing one digit slot ($\div 10$), so the new scheme really does allow more plates.
Alternative: Track each slot's change separately instead of writing full totals. Letters go from $1$ slot to $3$ slots, a factor of $26^3 / 26^1 = 26^2$. Digits go from $4$ slots to $3$ slots, a factor of $10^3 / 10^4 = 1/10$. Multiply the two factors: $26^2 \cdot \frac{1}{10} = \frac{26^2}{10}$, again (C).
CCSS standards used (min grade 6)
3.OA.A.1Interpret products of whole numbers as total number of objects in groups (Multiplying the independent slot choices ($26$ and four $10$s) to count the old plates.)6.EE.A.1Write and evaluate numerical expressions involving whole-number exponents (Writing the repeated slot products as $26^3 \cdot 10^3$ for the new plates.)4.OA.A.2Multiply or divide to solve word problems involving multiplicative comparison (Reading "by how many times has it increased" as the division new $\div$ old.)6.EE.A.3Apply the properties of operations to generate equivalent expressions (Cancelling common factors of $26$ and $10$ to reduce the ratio to $\frac{26^2}{10}$.)
⭐ Count each plate by multiplying the choices in every slot, then divide the new count by the old to see how many times bigger it got.
⭐ Count each plate by multiplying the choices in every slot, then divide the new count by the old to see how many times bigger it got.
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