AMC 10 · 2003 · #25

Easy mode Grade 4

Problem

How many four-digit numbers end in $23$ (that is, their last two digits are $2$ and then $3$ ) and are also divisible by $3$ ?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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.

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Toolkit + CCSS Solution

Understand

Restated: A four-digit number that ends in $23$ looks like $\overline{ab23}$: a thousands digit, a hundreds digit, then the fixed digits $2$ and $3$. Count how many such numbers are multiples of $3$.

Givens: The number has four digits and its last two digits are $2$ and $3$.; So the number can be written as $1000a + 100b + 23$, where $a$ is the thousands digit and $b$ is the hundreds digit.; The number must be divisible by $3$.; Answer choices: (A) $27$, (B) $30$, (C) $33$, (D) $81$, (E) $90$.

Unknowns: How many choices of the two free digits $a$ and $b$ make $\overline{ab23}$ a multiple of $3$.

Understand

Plan

Primary tool: #4 Introduce a Variable

Secondary: #5 Look for a Pattern, #7 Identify Subproblems, #2 Make a Systematic List

Only two digits are free — the thousands digit and the hundreds digit — so name them $a$ and $b$ (Tool #4). The divisibility rule for $3$ says a number is a multiple of $3$ exactly when its digits add to a multiple of $3$. That turns the whole problem into one condition on $a+b$. To count the digits $b$ that fit, look at the repeating remainder pattern of $0,1,\dots,9$ when divided by $3$ (Tool #5): the pattern is what makes the count exact instead of guesswork. Because the needed remainder for $b$ depends on $a$, split the nine values of $a$ into three cases by their own remainder (Tool #7) and count each case with a short list (Tool #2). Add the three case-counts for the total.

Execute — Answer: B

#4 Introduce a Variable 4.OA.B.4 Step 1

Turn the number into a digit-sum condition

Write the number as $\overline{ab23}$, so its four digits are $a$, $b$, $2$, and $3$.
A number is divisible by $3$ exactly when the sum of its digits is divisible by $3$.
Here that digit sum is $a + b + 2 + 3 = a + b + 5$.
So the only thing that matters is whether $a + b + 5$ is a multiple of $3$.

$$\text{digit sum} = a + b + 2 + 3 = a + b + 5$$

💡 For divisibility by $3$ only the digit sum matters, so the fixed $2$ and $3$ just add a constant $5$.

#4 Introduce a Variable 4.NBT.B.6 Step 2

Find the remainder the free digits must give

We need $a + b + 5$ to be a multiple of $3$.
Since $5$ leaves remainder $2$ when divided by $3$, adding $a+b$ has to bump that remainder from $2$ back up to a multiple of $3$.
That happens exactly when $a + b$ itself leaves remainder $1$ after dividing by $3$ (because $1 + 2 = 3$).
So the condition becomes: $a + b$ leaves remainder $1$ when divided by $3$.

$$a + b + 5 \equiv 0 \pmod 3 \;\Rightarrow\; a + b \equiv 1 \pmod 3$$

💡 The constant $5$ already contributes a remainder of $2$, so the two free digits must supply the missing remainder of $1$.

#5 Look for a Pattern 4.OA.C.5 Step 3

Count the b-digits for each remainder

For the hundreds digit $b$ from $0$ to $9$, group by remainder when divided by $3$.
Remainder $0$: $\{0,3,6,9\}$ — that is $4$ values.
Remainder $1$: $\{1,4,7\}$ — $3$ values.
Remainder $2$: $\{2,5,8\}$ — $3$ values.
So once we know which remainder $b$ needs, we know how many choices it has: $4$ if it needs remainder $0$, otherwise $3$.

$$b\equiv 0:\{0,3,6,9\}\,(4),\quad b\equiv 1:\{1,4,7\}\,(3),\quad b\equiv 2:\{2,5,8\}\,(3)$$

💡 Ten digits split $4$-$3$-$3$ by remainder because $0$ through $9$ isn't a whole number of full groups of $3$.

#7 Identify Subproblems 4.OA.B.4 Step 4

Split the a-digits into three cases

The thousands digit $a$ runs from $1$ to $9$.
Group these by remainder too.
Remainder $1$: $a\in\{1,4,7\}$; then $b$ must have remainder $1-1=0$, so $b$ has $4$ choices → $3\times 4 = 12$.
Remainder $2$: $a\in\{2,5,8\}$; then $b$ must have remainder $1-2$, which is remainder $2$, so $3$ choices → $3\times 3 = 9$.
Remainder $0$: $a\in\{3,6,9\}$; then $b$ must have remainder $1$, so $3$ choices → $3\times 3 = 9$.

$$a\equiv1:3\times4=12,\quad a\equiv2:3\times3=9,\quad a\equiv0:3\times3=9$$

💡 Each remainder-group of $a$ pins down exactly which remainder $b$ needs, so each case is a clean multiply.

#7 Identify Subproblems 4.OA.A.3 Step 5

Add the three cases

The three cases are separate — each thousands digit $a$ lands in exactly one remainder group — so no number is counted twice.
Add the counts: $12 + 9 + 9 = 30$.
So there are $30$ four-digit numbers ending in $23$ that are divisible by $3$.
Among the choices, $30$ is option (B).

$$12 + 9 + 9 = 30 \;\Rightarrow\; \textbf{(B)}$$

💡 The three remainder cases don't overlap, so their counts simply add.

[1] #4 4.OA.B.4 Write the number as $\overline{ab23}$, so its four digits are $a$, $b$, $2$, and

[2] #4 4.NBT.B.6 We need $a + b + 5$ to be a multiple of $3$. Since $5$ leaves remainder $2$ when

[3] #5 4.OA.C.5 For the hundreds digit $b$ from $0$ to $9$, group by remainder when divided by $

[4] #7 4.OA.B.4 The thousands digit $a$ runs from $1$ to $9$. Group these by remainder too. Rema

[5] #7 4.OA.A.3 The three cases are separate — each thousands digit $a$ lands in exactly one rem

Review

Reasonableness: Sanity check the size: the free digits $a$ (nine choices) and $b$ (ten choices) give $9\times 10 = 90$ candidate numbers, and divisibility by $3$ should keep roughly one in three, so about $30$ — matching the answer and ruling out the far-larger $81$ and $90$. Spot-check two members: $1023$ has digit sum $1+0+2+3=6$, a multiple of $3$ ✓ (here $a=1,b=0$, remainder-$1$ case); $9923$ has digit sum $9+9+2+3=23$, not a multiple of $3$, and indeed $a=9,b=9$ gives $a+b=18\equiv 0$, not $1$, so it is correctly excluded ✓.

Alternative: Look only at the two free digits as a two-digit prefix. The prefixes run over all of $10,11,\dots,99$ — exactly $90$ consecutive whole numbers. A prefix and its digit sum $a+b$ leave the same remainder mod $3$, so we want the prefixes that leave remainder $1$. Among any $90$ consecutive numbers, each remainder $0,1,2$ occurs exactly $90\div 3 = 30$ times. So exactly $30$ prefixes work — the same answer with almost no casework.

CCSS standards used (min grade 4)

4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite (Applying the divisibility rule for $3$ (a number is a multiple of $3$ when its digit sum is) and splitting the thousands digit into remainder cases.)
4.NBT.B.6 Find whole-number quotients and remainders with up to four-digit dividends (Reducing 'divisible by $3$' to the remainder condition $a+b\equiv 1$ using remainders on division by $3$.)
4.OA.C.5 Generate a number or shape pattern following a given rule (Using the repeating $4$-$3$-$3$ remainder pattern of the digits $0$ through $9$ to count the valid hundreds digits.)
4.OA.A.3 Solve multi-step word problems using four operations with whole numbers (Combining the three non-overlapping case counts, $12+9+9=30$.)

⭐ Divisibility by $3$ only cares about the digit sum, so freeze the fixed digits and count how the two free digits can land in the right remainder group.

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Turn the number into a digit-sum condition

Find the remainder the free digits must give

Count the b-digits for each remainder

Split the a-digits into three cases

Add the three cases

Review

CCSS standards used (min grade 4)

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