AMC 8 · 1999 · #15

Easy mode Grade 5

Problem

In Flatville, a bicycle license plate has three letters in a row.

The first letter comes from this group of 5 letters: {C, H, L, P, R}.
The second letter comes from this group of 3 letters: {A, I, O}.
The third letter comes from this group of 4 letters: {D, M, N, T}.

The town wants more license plates, so it will add 2 brand-new letters. Each new letter has to be placed into one of the three groups. The town can:

put both new letters into the same group, or
put one new letter into one group and the other new letter into a different group.

The town will choose the option that creates the most extra license plates.

How many extra license plates can be made at most, on top of the ones that already exist?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 1999 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.

View mode:

Toolkit + CCSS Solution

Understand

Restated: Flatville license plates are three letters: the 1st from a set of $5$, the 2nd from a set of $3$, the 3rd from a set of $4$. Two new letters get added to these sets (both to one set, or one each to two sets). What is the largest number of ADDITIONAL plates that can be made?

Givens: Original set sizes are $5$, $3$, $4$; Exactly $2$ new letters are added, distributed among the three sets; The total number of plates is (1st-set size) $\times$ (2nd-set size) $\times$ (3rd-set size); Answer choices: (A) $24$, (B) $30$, (C) $36$, (D) $40$, (E) $60$

Unknowns: The largest possible number of ADDITIONAL plates after adding $2$ letters

Understand

Plan

Primary tool: #2 Make a Systematic List

Secondary: #6 Guess and Check

Only a handful of distributions exist for $2$ extra letters across $3$ sets, so Tool #2 (Make a Systematic List) sweeps them all without missing any: $3$ ways to put both letters in one set plus $3$ ways to put one letter in each of two sets — exactly $6$ cases. Tool #6 (Guess and Check) is the natural companion: compute the new product for each case and pick the largest. The size of the case list is small enough that listing beats algebra, which keeps the solution accessible.

Execute — Answer: D

#2 Make a Systematic List 5.NBT.B.5 Step 1

Compute the original number of plates by multiplying the three set sizes.
This is the baseline we will subtract from at the end.

$$5 \times 3 \times 4 = 60 \text{ plates}$$

💡 Three independent choices multiply — the Grade 5 multiplication-as-counting move.

#2 Make a Systematic List 4.OA.A.3 Step 2

List every way to spread $2$ new letters across the three sets.
Either both letters land in one set ($3$ choices for which set), or one letter goes to each of two different sets ($3$ ways to pick which two).
That gives $6$ cases total.

$$\text{Cases: } (7,3,4),\ (5,5,4),\ (5,3,6),\ (6,4,4),\ (6,3,5),\ (5,4,5)$$

💡 When the case count is small, writing the full list is the safest way not to miss the winner.

#6 Guess and Check 5.NBT.B.5 Step 3

Compute the new plate total for each case by multiplying the three updated set sizes.

$$7\times3\times4=84,\ 5\times5\times4=100,\ 5\times3\times6=90,\ 6\times4\times4=96,\ 6\times3\times5=90,\ 5\times4\times5=100$$

💡 One multiplication per row turns the case list into a column of totals you can scan.

#6 Guess and Check 4.OA.A.3 Step 4

The largest new total is $100$ (it shows up twice: $5{\cdot}5{\cdot}4$ and $5{\cdot}4{\cdot}5$).
Subtract the original $60$ to get the number of ADDITIONAL plates.

$$100 - 60 = 40 \;\Rightarrow\; \textbf{(D)}$$

💡 "Additional" means the gain over the starting count — subtract the old total from the best new one.

[1] #2 5.NBT.B.5 Compute the original number of plates by multiplying the three set sizes. This i

[2] #2 4.OA.A.3 List every way to spread $2$ new letters across the three sets. Either both lett

[3] #6 5.NBT.B.5 Compute the new plate total for each case by multiplying the three updated set s

[4] #6 4.OA.A.3 The largest new total is $100$ (it shows up twice: $5{\cdot}5{\cdot}4$ and $5{\c

Review

Reasonableness: Quick sanity check on the winning cases: $5+5+4=14$ and $5+4+5=14$, matching the required total of $5+3+4+2=14$. Both winners have set sizes as close to equal as possible, which fits the well-known rule "for a fixed sum, the product is largest when the numbers are most balanced." The losing cases all have one set much bigger or smaller than the others. Also, $40$ is one of the listed choices, and the bigger choice $60$ would mean the new total was $120$, which no $14$-sum case can reach ($5\cdot5\cdot4=100$ is the cap). So (D) is the only consistent answer.

Alternative: Tool #6 (Guess and Check) with the balance rule: skip the full list and try only the cases that make the three sizes most equal. Starting from $(5,3,4)$, adding both letters to the smallest set gives $(5,5,4)$ with product $100$; adding one each to the two smaller sets gives $(5,4,5)$, also $100$. Any less-balanced case (like $(7,3,4)$) clearly gives a smaller product, so the new total is $100$ and the additional count is $100-60=40$.

CCSS standards used (min grade 5)

5.NBT.B.5 Fluently multiply multi-digit whole numbers using the standard algorithm (Multiplying the three set sizes in each case to get the new plate total (for example, $5 \times 5 \times 4 = 100$).)
4.OA.A.3 Solve multistep word problems posed with whole numbers (Enumerating the $6$ ways to distribute $2$ new letters and then subtracting the original $60$ from the best new total to get the additional plates.)

⭐ Only $6$ ways exist to spread $2$ new letters across $3$ sets — list them, multiply, and the best one is $5 \times 5 \times 4 = 100$, which is $40$ more plates than the original $60$.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 5)

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