AMC 8 · 2016 · #24

Grade 4 number-theorylogic

Archetype Small-Domain Constrained Search

divisibility-rulesdigit-constraintsdigit-sumsystematic-enumeration caseworksystematic-enumerationdigit-constraints ↑ Prerequisites: divisibility-rulesplace-value

📏 Long solution 💡 4 insights

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Problem

The digits $1$ , $2$ , $3$ , $4$ , and $5$ are each used once to write a five-digit number $PQRST$ . The three-digit number $PQR$ is divisible by $4$ , the three-digit number $QRS$ is divisible by $5$ , and the three-digit number $RST$ is divisible by $3$ . What is $P$ ?

$\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad \textbf{(E) }5$

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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Toolkit + CCSS Solution

Understand

Restated: Place the digits $1, 2, 3, 4, 5$ — each used exactly once — into the five spots $P, Q, R, S, T$ so that the three-digit number $PQR$ is a multiple of $4$, the three-digit number $QRS$ is a multiple of $5$, and the three-digit number $RST$ is a multiple of $3$. What digit ends up in position $P$?

Givens: Five digits $\{1, 2, 3, 4, 5\}$, each used exactly once; $PQR$ is divisible by $4$; $QRS$ is divisible by $5$; $RST$ is divisible by $3$; Answer choices: (A) $1$, (B) $2$, (C) $3$, (D) $4$, (E) $5$

Unknowns: The digit assigned to position $P$

Understand

Plan

Primary tool: #3 Eliminate Possibilities

Secondary: #7 Identify Subproblems

There are only $5! = 120$ ways to arrange the digits, but the three divisibility rules slice that pool down very fast. Tool #3 (Eliminate Possibilities) is the right driver: each rule rules out almost every digit at a specific position, leaving at most one survivor. Tool #7 (Identify Subproblems) tells us the order to apply the rules — pin down the most restrictive position first ($S$ from the $5$-rule), then the next ($R$ from the $4$-rule), then $T$ (from the $3$-rule), and finally $P, Q$ by what is left. This is much cleaner than algebra (#13) on a problem that is really a logic puzzle.

Execute — Answer: A

#3 Eliminate Possibilities 4.OA.B.4 Step 1

Lock down $S$ first.
$QRS$ must be a multiple of $5$, so its last digit $S$ has to be $0$ or $5$.
The digit $0$ is not available, so $S = 5$.
That uses up the digit $5$ and leaves $\{1, 2, 3, 4\}$ for the other four slots.

$$S = 5; \quad \text{remaining digits} = \{1, 2, 3, 4\}$$

💡 The $5$-rule is the strictest constraint in this set: only digits ending in $0$ or $5$ work. With $0$ banned, $S$ is forced — a clean elimination.

#3 Eliminate Possibilities 4.OA.B.4 Step 2

Narrow $R$ next.
$PQR$ must be a multiple of $4$, and a number is divisible by $4$ exactly when its last two digits form a multiple of $4$ — which forces the last digit $R$ to be even.
Among the remaining digits $\{1, 2, 3, 4\}$, the evens are $2$ and $4$.
So $R \in \{2, 4\}$.

$$R \in \{2, 4\}$$

💡 Even-last-digit is a necessary (not sufficient) condition for divisibility by $4$, but it is already enough to cut $R$'s possibilities from four down to two.

#7 Identify Subproblems 4.OA.B.4 Step 3

Use the $3$-rule on $RST$ to choose between $R = 2$ and $R = 4$, and to pin down $T$.
A number is divisible by $3$ exactly when its digit sum is divisible by $3$.
With $S = 5$, we need $R + 5 + T$ to be a multiple of $3$.
Check each candidate for $R$ against the leftover digits available for $T$.

$R + S + T \equiv 0 \pmod 3$, so $R + T \equiv 1 \pmod 3$ (since $5 \equiv 2$).

💡 Two pieces of the puzzle are still free ($R$ and $T$); the $3$-rule ties them together into one tiny subproblem that we can finish by checking a handful of cases.

#3 Eliminate Possibilities 4.OA.B.4 Step 4

Case $R = 2$: leftover digits for $T$ are $\{1, 3, 4\}$, and we need $2 + 5 + T = 7 + T$ divisible by $3$.
Test: $7+1=8$, $7+3=10$, $7+4=11$ — none are multiples of $3$.
So $R = 2$ is impossible.
Case $R = 4$: leftover digits for $T$ are $\{1, 2, 3\}$, and we need $4 + 5 + T = 9 + T$ divisible by $3$.
Since $9$ is already a multiple of $3$, $T$ must be too — only $T = 3$ works.
So $R = 4, T = 3$.

$$R = 4, \; T = 3; \quad \text{remaining digits for } P, Q = \{1, 2\}$$

💡 Two short case-checks beat any algebra here. The $3$-rule with $S = 5$ leaves exactly one survivor: $(R, T) = (4, 3)$.

#3 Eliminate Possibilities 4.OA.B.4 Step 5

Finish with $P$ and $Q$.
The leftover digits are $\{1, 2\}$, and the $4$-rule on $PQR$ — with $R = 4$ now known — demands that $QR = Q4$ be divisible by $4$.
Test: $14 \div 4 = 3.5$ (no); $24 \div 4 = 6$ (yes).
So $Q = 2$, leaving $P = 1$.
The full number is $PQRST = 12453$.

$$Q = 2, \; P = 1; \quad PQRST = 12453 \;\Rightarrow\; P = 1 \;\Rightarrow\; \textbf{(A)}$$

💡 With only two digits left for two slots, one quick divisibility check forces the assignment. $P$ is whatever $Q$ isn't.

[1] #3 4.OA.B.4 Lock down $S$ first. $QRS$ must be a multiple of $5$, so its last digit $S$ has

[2] #3 4.OA.B.4 Narrow $R$ next. $PQR$ must be a multiple of $4$, and a number is divisible by $

[3] #7 4.OA.B.4 Use the $3$-rule on $RST$ to choose between $R = 2$ and $R = 4$, and to pin down

[4] #3 4.OA.B.4 Case $R = 2$: leftover digits for $T$ are $\{1, 3, 4\}$, and we need $2 + 5 + T

[5] #3 4.OA.B.4 Finish with $P$ and $Q$. The leftover digits are $\{1, 2\}$, and the $4$-rule on

Review

Reasonableness: Verify $PQRST = 12453$ against all three rules: $PQR = 124$ and $124 = 4 \times 31$ (divisible by $4$, OK); $QRS = 245$ ends in $5$ (divisible by $5$, OK); $RST = 453$ with digit sum $4 + 5 + 3 = 12 = 3 \times 4$ (divisible by $3$, OK). All five digits $1, 2, 3, 4, 5$ appear exactly once. Everything checks out, so $P = 1$ and the answer is (A).

Alternative: Tool #2 (Make a Systematic List): With $S = 5$ forced and $R \in \{2, 4\}$, list every $(P, Q, R, T)$ permutation of the leftover digits and keep only the ones where $PQR$ is a multiple of $4$ and $RST$ is a multiple of $3$. There are only $2 \times 3! = 12$ permutations to scan, and just one — $(P, Q, R, T) = (1, 2, 4, 3)$ — survives. Same answer (A), reached by brute enumeration instead of staged elimination.

CCSS standards used (min grade 4)

4.OA.B.4 Find factors and multiples; apply divisibility reasoning to whole numbers in the range 1-100 (Applying the divisibility rules for $3$, $4$, and $5$ to test each candidate digit in positions $S$, $R$, $T$, $Q$, and pin down $P$.)

⭐ This AMC 8 problem is a Grade 4 divisibility-rules puzzle — knowing the rules for $3$, $4$, and $5$ is enough to crack it.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: A

Review

CCSS standards used (min grade 4)

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