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Question Number 197589 by mr W last updated on 23/Sep/23
how many natural numbers with 4 different digits are divisible by 3?
howmanynaturalnumberswith4differentdigitsaredivisibleby3?
Answered by mr W last updated on 24/Sep/23
A={1,4,7} B={2,5,8} C={3,6,9} to select 3 non−zero digits such that their sum is divisible by 3, there are case 1: all digits from A or B or C ⇒3 way case 2: one digit from each group ⇒3×3×3=27 ways totally 3+27=30 ways to select 4 non−zero digits such that their sum is divisible by 3, there are case 1: 2 digits from C, one from A and one from B ⇒3×3×3=27 ways case 2: 2 digits from A and 2 from B ⇒3×3=9 ways case 3: 3 digits from A and one from C ⇒1×3=3 ways case 4: 3 digits from B and one from C ⇒1×3=3 ways totally 27+9+3+3=42 ways numbers of form pqrs: 42×4!=1008 numbers of form p0qr: 3×30×3!=540 totally: 1008+540=1548 ✓
A={1,4,7}B={2,5,8}C={3,6,9}toselect3non−zerodigitssuchthattheirsumisdivisibleby3,therearecase1:alldigitsfromAorBorC⇒3waycase2:onedigitfromeachgroup⇒3×3×3=27waystotally3+27=30waystoselect4non−zerodigitssuchthattheirsumisdivisibleby3,therearecase1:2digitsfromC,onefromAandonefromB⇒3×3×3=27wayscase2:2digitsfromAand2fromB⇒3×3=9wayscase3:3digitsfromAandonefromC⇒1×3=3wayscase4:3digitsfromBandonefromC⇒1×3=3waystotally27+9+3+3=42waysnumbersofformpqrs:42×4!=1008numbersofformp0qr:3×30×3!=540totally:1008+540=1548✓

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