AMC 8 · 2016 · #24
Easy mode Grade 4Problem
Take the digits 1, 2, 3, 4, 5 and write each one exactly once in a row. That makes a five-digit number PQRST, where P is the first digit, Q the second, and so on.
Now look at three smaller numbers inside it:
- PQR (the first three digits) must be divisible by 4.
- QRS (the middle three digits) must be divisible by 5.
- RST (the last three digits) must be divisible by 3.
What is the value of P?
Pick an answer.
AMC 8 2016 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: 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
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) $5$
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
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.
💡 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.
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 $R \in \{2, 4\}$.
💡 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.
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 $T$.
💡 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.
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 $R = 4, T = 3$.
💡 Two short case-checks beat any algebra here. The $3$-rule with $S = 5$ leaves exactly one survivor: $(R, T) = (4, 3)$.
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 $PQRST = 12453$.
💡 With only two digits left for two slots, one quick divisibility check forces the assignment. $P$ is whatever $Q$ isn't.
4.OA.B.4 Lock down $S$ first. $QRS$ must be a multiple of $5$, so its last digit $S$ has 4.OA.B.4 Narrow $R$ next. $PQR$ must be a multiple of $4$, and a number is divisible by $ 4.OA.B.4 Use the $3$-rule on $RST$ to choose between $R = 2$ and $R = 4$, and to pin down 4.OA.B.4 Case $R = 2$: leftover digits for $T$ are $\{1, 3, 4\}$, and we need $2 + 5 + T 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.4Find 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.
⭐ This AMC 8 problem is a Grade 4 divisibility-rules puzzle — knowing the rules for $3$, $4$, and $5$ is enough to crack it.
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