Competition · AMC preparation · step 4 of 4
AMC 8 · 2016 · #24
Grade 4 number-theorylogicPick an answer.
AMC 8 2016 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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.
Pin down S first
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}.
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.
The digit that lands in position S is forced to be 5.
▸ Why?
The number QRS is a multiple of 5, and a whole number is a multiple of 5 only when its last digit is 0 or 5; since 0 is not among the digits 1 to 5, the last digit S has to be 5.
▸ Why?
Whether QRS is a multiple of 5 is settled by its last digit alone: read it as the hundreds-and-tens part plus the ones digit S, and that first part is already an exact number of 5s, so only S can change the remainder.
▸ Why?
Ten ones bundle into one ten and ten tens into one hundred, so the tens and hundreds places are built entirely out of tens, and each ten is two fives that leave nothing over.
Narrow the choices for R
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}.
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.4Eliminate PossibilitiesApply the divisible-by-3 rule
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.
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.4Identify SubproblemsTest both cases for R
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.
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.4Eliminate PossibilitiesFill in P and Q
The leftover {1, 2} must make Q4 a multiple of 4: 14 fails, 24 works, so Q = 2 and P = 1, giving 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.4Eliminate PossibilitiesThis AMC 8 problem is a Grade 4 divisibility-rules puzzle — knowing the rules for 3, 4, and 5 is enough to crack it.
- Pin down S first
- Narrow the choices for R
- Apply the divisible-by-3 rule
- Test both cases for R
- Fill in P and Q
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