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Question-41754

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Question Number 41754 by Tawa1 last updated on 12/Aug/18
Answered by candre last updated on 12/Aug/18
we have from 1 to 1000 number that can have 3 3×× ×3× ××3 separating cases ××3: 10×10=100(9×9=81) ×3×: 10×10=100(9×9=81) 3××: 10×10=100(9×9=81) analysying the intersection ×33:10(9) 3×3:10(9) 33×:10(9) 333:1 so, by inclusion exclusion principe 100+100+100−10−10−10+1=271 or, counting alredy excluding overlaping (() one) 81×3+9×3+1=271
$$\mathrm{we}\:\mathrm{have}\:\mathrm{from}\:\mathrm{1}\:\mathrm{to}\:\mathrm{1000}\:\mathrm{number}\:\mathrm{that}\:\mathrm{can}\:\mathrm{have}\:\mathrm{3} \\ $$$$\mathrm{3}×× \\ $$$$×\mathrm{3}× \\ $$$$××\mathrm{3} \\ $$$$\mathrm{separating}\:\mathrm{cases} \\ $$$$××\mathrm{3}: \\ $$$$\mathrm{10}×\mathrm{10}=\mathrm{100}\left(\mathrm{9}×\mathrm{9}=\mathrm{81}\right) \\ $$$$×\mathrm{3}×: \\ $$$$\mathrm{10}×\mathrm{10}=\mathrm{100}\left(\mathrm{9}×\mathrm{9}=\mathrm{81}\right) \\ $$$$\mathrm{3}××: \\ $$$$\mathrm{10}×\mathrm{10}=\mathrm{100}\left(\mathrm{9}×\mathrm{9}=\mathrm{81}\right) \\ $$$$\mathrm{analysying}\:\mathrm{the}\:\mathrm{intersection} \\ $$$$×\mathrm{33}:\mathrm{10}\left(\mathrm{9}\right) \\ $$$$\mathrm{3}×\mathrm{3}:\mathrm{10}\left(\mathrm{9}\right) \\ $$$$\mathrm{33}×:\mathrm{10}\left(\mathrm{9}\right) \\ $$$$\mathrm{333}:\mathrm{1} \\ $$$$\mathrm{so},\:{by}\:{inclusion}\:{exclusion}\:{principe} \\ $$$$\mathrm{100}+\mathrm{100}+\mathrm{100}−\mathrm{10}−\mathrm{10}−\mathrm{10}+\mathrm{1}=\mathrm{271} \\ $$$$\mathrm{or},\:\mathrm{counting}\:\mathrm{alredy}\:\mathrm{excluding}\:\mathrm{overlaping}\:\left(\left(\right)\:\mathrm{one}\right) \\ $$$$\mathrm{81}×\mathrm{3}+\mathrm{9}×\mathrm{3}+\mathrm{1}=\mathrm{271} \\ $$
Commented by Tawa1 last updated on 12/Aug/18
God bless you sir
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 12/Aug/18
1)single digit number using 3 is =1 2)two digit number unit place is 3 then tength place can be filled by using (1,2,3,4,5,6,7,8,9) 9 ways so =9 3)two digit number tength place is 3 then unit place can be (0 ,1,2,3,4,5,6,7,8,9)=10 4)three digit number hundred place is 3 tenth place (0, 1, 2 ,3, 4, 5, 6, 7, 8, 9) 10was similarly unit place 10ways=1×10×10=100 5)three digit number tength place is 3 hundred place(1 ,2 ,3 ,4, 5 ,6 ,7 ,8 ,9)=9ways unit place (0 ,1 ,2 ,3 ,4 ,5 ,6 ,7 ,8 ,9)=10 ways 1×9×10=90 6)three digit number unit place is 3 tength place (0 ,1 ,2 ,3 ,4 ,5 ,6 ,7 ,8 ,9) hundred place(1, 2^� 3, ^� 4 ,5 ,6 ,7 ,8 ,9) 1×10×9=90 so total =1+9+10+100+90+90=300
$$ \\ $$$$\left.\mathrm{1}\right){single}\:{digit}\:{number}\:{using}\:\mathrm{3}\:{is}\:=\mathrm{1} \\ $$$$\left.\mathrm{2}\right){two}\:{digit}\:{number}\:{unit}\:{place}\:{is}\:\mathrm{3}\:{then}\:{tength}\: \\ $$$${place}\:{can}\:{be}\:{filled}\:{by}\:{using}\:\left(\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5},\mathrm{6},\mathrm{7},\mathrm{8},\mathrm{9}\right) \\ $$$$\mathrm{9}\:{ways}\:{so}\:=\mathrm{9} \\ $$$$\left.\mathrm{3}\right){two}\:{digit}\:{number}\:{tength}\:{place}\:{is}\:\mathrm{3}\:{then}\:{unit} \\ $$$${place}\:{can}\:{be}\:\left(\mathrm{0}\:,\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5},\mathrm{6},\mathrm{7},\mathrm{8},\mathrm{9}\right)=\mathrm{10} \\ $$$$ \\ $$$$\left.\mathrm{4}\right){three}\:{digit}\:\:{number}\:{hundred}\:{place}\:{is}\:\mathrm{3} \\ $$$${tenth}\:{place}\:\left(\mathrm{0},\:\mathrm{1},\:\mathrm{2}\:,\mathrm{3},\:\mathrm{4},\:\mathrm{5},\:\mathrm{6},\:\mathrm{7},\:\mathrm{8},\:\mathrm{9}\right)\:\mathrm{10}{was} \\ $$$${similarly}\:{unit}\:{place}\:\mathrm{10}{ways}=\mathrm{1}×\mathrm{10}×\mathrm{10}=\mathrm{100} \\ $$$$\left.\mathrm{5}\right){three}\:{digit}\:{number}\:{tength}\:{place}\:{is}\:\mathrm{3} \\ $$$${hundred}\:{place}\left(\mathrm{1}\:,\mathrm{2}\:,\mathrm{3}\:,\mathrm{4},\:\mathrm{5}\:,\mathrm{6}\:,\mathrm{7}\:,\mathrm{8}\:,\mathrm{9}\right)=\mathrm{9}{ways} \\ $$$${unit}\:{place}\:\left(\mathrm{0}\:,\mathrm{1}\:,\mathrm{2}\:,\mathrm{3}\:,\mathrm{4}\:,\mathrm{5}\:,\mathrm{6}\:,\mathrm{7}\:,\mathrm{8}\:,\mathrm{9}\right)=\mathrm{10}\:{ways} \\ $$$$\mathrm{1}×\mathrm{9}×\mathrm{10}=\mathrm{90} \\ $$$$\left.\mathrm{6}\right){three}\:{digit}\:{number}\:{unit}\:{place}\:{is}\:\mathrm{3} \\ $$$${tength}\:{place}\:\left(\mathrm{0}\:,\mathrm{1}\:,\mathrm{2}\:,\mathrm{3}\:,\mathrm{4}\:,\mathrm{5}\:,\mathrm{6}\:,\mathrm{7}\:,\mathrm{8}\:,\mathrm{9}\right) \\ $$$${hundred}\:{place}\left(\mathrm{1},\:\bar {\mathrm{2}}\:\mathrm{3},\bar {\:}\mathrm{4}\:,\mathrm{5}\:,\mathrm{6}\:,\mathrm{7}\:,\mathrm{8}\:,\mathrm{9}\right) \\ $$$$\mathrm{1}×\mathrm{10}×\mathrm{9}=\mathrm{90} \\ $$$${so}\:{total}\:=\mathrm{1}+\mathrm{9}+\mathrm{10}+\mathrm{100}+\mathrm{90}+\mathrm{90}=\mathrm{300}\: \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by Tawa1 last updated on 12/Aug/18
God bless you sir
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Answered by MrW3 last updated on 12/Aug/18
3: ⇒1 3X: ⇒9 X3: ⇒8 33: ⇒1×2=2 Σ=19 3XY: ⇒9×9=81 X3Y: ⇒8×9=72 XY3: ⇒8×9=72 33X: ⇒2×9=18 3X3: ⇒2×9=18 X33: ⇒2×8=16 333: ⇒1×3=3 Σ=280 ⇒1+19+280=300
$$\mathrm{3}:\:\Rightarrow\mathrm{1} \\ $$$$ \\ $$$$\mathrm{3}{X}:\:\Rightarrow\mathrm{9} \\ $$$${X}\mathrm{3}:\:\Rightarrow\mathrm{8} \\ $$$$\mathrm{33}:\:\Rightarrow\mathrm{1}×\mathrm{2}=\mathrm{2} \\ $$$$\Sigma=\mathrm{19} \\ $$$$ \\ $$$$\mathrm{3}{XY}:\:\Rightarrow\mathrm{9}×\mathrm{9}=\mathrm{81} \\ $$$${X}\mathrm{3}{Y}:\:\Rightarrow\mathrm{8}×\mathrm{9}=\mathrm{72} \\ $$$${XY}\mathrm{3}:\:\Rightarrow\mathrm{8}×\mathrm{9}=\mathrm{72} \\ $$$$\mathrm{33}{X}:\:\Rightarrow\mathrm{2}×\mathrm{9}=\mathrm{18} \\ $$$$\mathrm{3}{X}\mathrm{3}:\:\Rightarrow\mathrm{2}×\mathrm{9}=\mathrm{18} \\ $$$${X}\mathrm{33}:\:\Rightarrow\mathrm{2}×\mathrm{8}=\mathrm{16} \\ $$$$\mathrm{333}:\:\Rightarrow\mathrm{1}×\mathrm{3}=\mathrm{3} \\ $$$$\Sigma=\mathrm{280} \\ $$$$ \\ $$$$\Rightarrow\mathrm{1}+\mathrm{19}+\mathrm{280}=\mathrm{300} \\ $$
Commented by Tawa1 last updated on 12/Aug/18
God bless you sir
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}\: \\ $$

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