Asked by Jeremy
How many positive three-digit integers with each digit greater than $4$ are divisible by $6$?
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GPT-5 mini
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Digits must be from {5,6,7,8,9}. Divisibility by 2 forces the unit digit to be 6 or 8.
Residues mod 3: 5≡2, 6≡0, 7≡1, 8≡2, 9≡0. There are n0=2 digits ≡0, n1=1 digit ≡1, n2=2 digits ≡2.
- If unit = 6 (residue 0), the first two digits must sum to 0 mod 3. Ordered pairs count = n0^2 + 2·n1·n2 = 2^2 + 2·1·2 = 8.
- If unit = 8 (residue 2), the first two digits must sum to 1 mod 3. Ordered pairs count = 2·n0·n1 + n2^2 = 2·2·1 + 2^2 = 8.
Total = 8 + 8 = 16.
Residues mod 3: 5≡2, 6≡0, 7≡1, 8≡2, 9≡0. There are n0=2 digits ≡0, n1=1 digit ≡1, n2=2 digits ≡2.
- If unit = 6 (residue 0), the first two digits must sum to 0 mod 3. Ordered pairs count = n0^2 + 2·n1·n2 = 2^2 + 2·1·2 = 8.
- If unit = 8 (residue 2), the first two digits must sum to 1 mod 3. Ordered pairs count = 2·n0·n1 + n2^2 = 2·2·1 + 2^2 = 8.
Total = 8 + 8 = 16.
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