Problem 24 · 2016 AMC 8
Hard
Number Theory
divisibilitycasework
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?
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Answer: A — P = 1.
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Hint 1 of 3
Attack the TIGHTEST rule first. Divisible-by-5 is the strictest: the last digit must be 0 or 5, and the only one of {1,2,3,4,5} available is 5 — so QRS divisible by 5 instantly forces S = 5. One clue, one digit pinned.
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Hint 2 of 3
Next-tightest: PQR divisible by 4 means its last two digits QR form a multiple of 4. With 5 used up, QR is a two-digit number from {1, 2, 3, 4} that's a multiple of 4 — only 12, 24, 32 qualify. Just three cases left.
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Hint 3 of 3
Finish with the gentlest rule (divisible by 3 = digit sum divisible by 3) to test the survivors. Notice R is shared by all three numbers, so it's heavily constrained — track it.
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Approach: apply the divisibility rules from most-restrictive to least, then test the few survivors
- Divisible-by-5 is strictest: QRS ends in S, which must be 0 or 5, so S = 5.
- Divisible-by-4: QR (the last two digits of PQR) must be a multiple of 4 built from {1, 2, 3, 4} — only 12, 24, 32 work. Three cases.
- QR = 12: leftovers {3, 4} for P, T; RST = 25T has digit sum 7 + T, and T ∈ {3, 4} gives 10 or 11 — not divisible by 3. Out.
- QR = 24: leftovers {1, 3}; RST = 45T has digit sum 9 + T, and T = 3 gives 12, divisible by 3. ✓ That forces P = 1, so PQRST = 12435 and P = 1.
- QR = 32: leftovers {1, 4}; RST = 25T sum 7 + T with T ∈ {1, 4} gives 8 or 11 — not divisible by 3. Out. Only one solution survives.
- Why this transfers: for digit puzzles with several divisibility rules, always resolve the most restrictive constraint first (it pins digits outright), which shrinks the casework before the looser rules ever come up.
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