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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?
$${how}\:{many}\:{natural}\:{numbers}\:{with}\:\mathrm{4} \\ $$$${different}\:{digits}\:{are}\:{divisible}\:{by}\:\mathrm{3}? \\ $$
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 ✓
$${A}=\left\{\mathrm{1},\mathrm{4},\mathrm{7}\right\} \\ $$$${B}=\left\{\mathrm{2},\mathrm{5},\mathrm{8}\right\} \\ $$$${C}=\left\{\mathrm{3},\mathrm{6},\mathrm{9}\right\} \\ $$$${to}\:{select}\:\mathrm{3}\:{non}−{zero}\:{digits}\:{such}\:{that} \\ $$$${their}\:{sum}\:{is}\:{divisible}\:{by}\:\mathrm{3},\:{there}\:{are}\: \\ $$$${case}\:\mathrm{1}:\:{all}\:{digits}\:{from}\:{A}\:{or}\:{B}\:{or}\:{C} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\mathrm{3}\:{way} \\ $$$${case}\:\mathrm{2}:\:{one}\:{digit}\:{from}\:{each}\:{group} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\mathrm{3}×\mathrm{3}×\mathrm{3}=\mathrm{27}\:{ways} \\ $$$${totally}\:\mathrm{3}+\mathrm{27}=\mathrm{30}\:{ways} \\ $$$$ \\ $$$${to}\:{select}\:\mathrm{4}\:{non}−{zero}\:{digits}\:{such}\:{that} \\ $$$${their}\:{sum}\:{is}\:{divisible}\:{by}\:\mathrm{3},\:{there}\:{are}\: \\ $$$${case}\:\mathrm{1}:\:\mathrm{2}\:{digits}\:{from}\:{C},\:{one}\:{from}\:{A}\:{and}\:{one}\:{from}\:{B} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\mathrm{3}×\mathrm{3}×\mathrm{3}=\mathrm{27}\:{ways} \\ $$$${case}\:\mathrm{2}:\:\mathrm{2}\:{digits}\:{from}\:{A}\:{and}\:\mathrm{2}\:{from}\:{B} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\mathrm{3}×\mathrm{3}=\mathrm{9}\:{ways} \\ $$$${case}\:\mathrm{3}:\:\mathrm{3}\:{digits}\:{from}\:{A}\:{and}\:{one}\:{from}\:{C} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\mathrm{1}×\mathrm{3}=\mathrm{3}\:{ways} \\ $$$${case}\:\mathrm{4}:\:\mathrm{3}\:{digits}\:{from}\:{B}\:{and}\:{one}\:{from}\:{C} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\mathrm{1}×\mathrm{3}=\mathrm{3}\:{ways} \\ $$$${totally}\:\mathrm{27}+\mathrm{9}+\mathrm{3}+\mathrm{3}=\mathrm{42}\:{ways} \\ $$$$ \\ $$$${numbers}\:{of}\:{form}\:{pqrs}:\:\mathrm{42}×\mathrm{4}!=\mathrm{1008} \\ $$$${numbers}\:{of}\:{form}\:{p}\mathrm{0}{qr}:\:\mathrm{3}×\mathrm{30}×\mathrm{3}!=\mathrm{540} \\ $$$${totally}:\:\mathrm{1008}+\mathrm{540}=\mathrm{1548}\:\checkmark \\ $$

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