AMC 8 · 2016 · #24

Grade 4 number-theorylogic

Archetype Small-Domain Constrained Search └ Sub-archetype Factor / Divisibility Constraint

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

📏 Long solution 💡 4 insights

📘 View easy version →

Problem

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?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2016 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Eliminate Possibilities

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.

1STEP 1

QRS is a multiple of 5, so its last digit S is 0 or 5; zero is unavailable, so S = 5, leaving {1, 2, 3, 4}.

S = 5; remaining digits = {1, 2, 3, 4}

2STEP 2

PQR is a multiple of 4 means QR is too, so the last digit R must be even; among {1, 2, 3, 4} that gives R ∈ {2, 4}.

R ∈ {2, 4}

3STEP 3

The 3-rule on RST needs the digit sum divisible by 3; with S = 5, that means R + 5 + T is a multiple of 3 — so test R = 2 versus R = 4.

R + S + T ≡ 0 (mod 3), so R + T ≡ 1 (mod 3) (since 5 ≡ 2).

4STEP 4

R = 2 gives 7 + T with T ∈ {1, 3, 4} — never a multiple of 3; R = 4 gives 9 + T, forcing T = 3, so R = 4, T = 3.

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

5STEP 5

The leftover {1, 2} must make Q4 a multiple of 4: 14 fails, 24 works, so Q = 2 and P = 1, giving PQRST = 12453.

Q = 2, P = 1; PQRST = 12453 → P = 1 → (A)

Answer

Verify PQRST = 12453 against all three rules: PQR = 124 and 124 = 4 × 31 (divisible by 4, OK); QRS = 245 ends in 5 (divisible by 5, OK); RST = 453 with digit sum 4 + 5 + 3 = 12 = 3 × 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).

💡Key takeaway

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

More like this