Question and Answers Forum

All Questions      Topic List

Permutation and Combination Questions

Previous in All Question      Next in All Question      

Previous in Permutation and Combination      Next in Permutation and Combination      

Question Number 177309 by SLVR last updated on 03/Oct/22

How many 3 digited numbers which are divisible by 1)3 2)4 3)5 4)6 5)7 6)8 7)9 with repetetion of digits is NOT allowed...one problem process

$${How}\:{many}\:\mathrm{3}\:{digited}\:{numbers}\: \\ $$$$\left.{which}\:{are}\:{divisible}\:{by}\:\mathrm{1}\right)\mathrm{3} \\ $$$$\left.\mathrm{2}\left.\right)\left.\mathrm{4}\left.\:\left.\:\left.\:\:\mathrm{3}\right)\mathrm{5}\:\:\:\:\mathrm{4}\right)\mathrm{6}\:\:\:\mathrm{5}\right)\mathrm{7}\:\:\:\:\mathrm{6}\right)\mathrm{8}\:\:\mathrm{7}\right)\mathrm{9} \\ $$$${with}\:{repetetion}\:{of}\:{digits}\:{is} \\ $$$${NOT}\:{allowed}...{one}\:{problem}\:{process} \\ $$

Commented by SLVR last updated on 03/Oct/22

i need for divisible by 7

$${i}\:{need}\:\:{for}\:{divisible}\:{by}\:\mathrm{7}\: \\ $$

Answered by mahdipoor last updated on 04/Oct/22

N=(abc)=100a+10b+c 0≤a,b,c∈N≤9 a≠0 ⇒N≡^7 0 and repetetion of digits ⇒100a+10b+c≡^7 2a+3b+c≡^7 0 i) a=b ⇒ c≡^7 2a ⇒ 13 numbers ii) a=c ⇒ b≡^7 6a ⇒ 12 iii) b=c ⇒ a≡^7 5b ⇒ 12 iv) a=b=c ⇒ 6a≡^7 0 ⇒ 1 ⇒all numbers=13+12+12−2×1=35 3 digit number are divisible by 7: [((999)/7)]−[((99)/7)]=128 ⇒128−35=93

$${N}=\left({abc}\right)=\mathrm{100}{a}+\mathrm{10}{b}+{c}\:\: \\ $$$$\mathrm{0}\leqslant{a},{b},{c}\in\mathbb{N}\leqslant\mathrm{9}\:\:\:\:\:\:{a}\neq\mathrm{0}\: \\ $$$$\Rightarrow{N}\overset{\mathrm{7}} {\equiv}\mathrm{0}\:\:\:\:\:\:{and}\:\:\:\:\:\:{repetetion}\:{of}\:{digits} \\ $$$$\Rightarrow\mathrm{100}{a}+\mathrm{10}{b}+{c}\overset{\mathrm{7}} {\equiv}\mathrm{2}{a}+\mathrm{3}{b}+{c}\overset{\mathrm{7}} {\equiv}\mathrm{0} \\ $$$$\left.{i}\right)\:{a}={b}\:\Rightarrow\:{c}\overset{\mathrm{7}} {\equiv}\mathrm{2}{a}\:\Rightarrow\:\mathrm{13}\:\:{numbers}\: \\ $$$$\left.{ii}\right)\:{a}={c}\:\Rightarrow\:{b}\overset{\mathrm{7}} {\equiv}\mathrm{6}{a}\:\Rightarrow\:\mathrm{12}\: \\ $$$$\left.{iii}\right)\:{b}={c}\:\Rightarrow\:{a}\overset{\mathrm{7}} {\equiv}\mathrm{5}{b}\:\Rightarrow\:\mathrm{12}\:\: \\ $$$$\left.{iv}\right)\:\:{a}={b}={c}\:\Rightarrow\:\mathrm{6}{a}\overset{\mathrm{7}} {\equiv}\mathrm{0}\:\Rightarrow\:\mathrm{1} \\ $$$$\Rightarrow{all}\:{numbers}=\mathrm{13}+\mathrm{12}+\mathrm{12}−\mathrm{2}×\mathrm{1}=\mathrm{35} \\ $$$$\mathrm{3}\:{digit}\:{number}\:{are}\:{divisible}\:{by}\:\mathrm{7}: \\ $$$$\left[\frac{\mathrm{999}}{\mathrm{7}}\right]−\left[\frac{\mathrm{99}}{\mathrm{7}}\right]=\mathrm{128} \\ $$$$\Rightarrow\mathrm{128}−\mathrm{35}=\mathrm{93} \\ $$

Commented by SLVR last updated on 04/Oct/22

Thanks for your concern sir but how 4th line2a+3b+c≡0mod7 and how 13 numbers if a=b

$${Thanks}\:{for}\:{your}\:{concern}\:{sir} \\ $$$${but}\:{how}\:\mathrm{4}{th}\:{line}\mathrm{2}{a}+\mathrm{3}{b}+{c}\equiv\mathrm{0}{mod}\mathrm{7} \\ $$$${and}\:{how}\:\mathrm{13}\:{numbers}\:{if}\:{a}={b} \\ $$

Commented by SLVR last updated on 04/Oct/22

kindly explain what those 34 numbers be deducted..please

$${kindly}\:{explain}\:{what}\:{those}\:\mathrm{34} \\ $$$${numbers}\:{be}\:{deducted}..{please} \\ $$

Commented by mr W last updated on 04/Oct/22

ii) a=c 2a+3b+c=3(a+b)≡^7 0 a+b=7 ⇒7 a+b=14 ⇒5 total 12 not 11? i counted totally 93 valid numbers.

$$\left.{ii}\right)\:{a}={c} \\ $$$$\mathrm{2}{a}+\mathrm{3}{b}+{c}=\mathrm{3}\left({a}+{b}\right)\overset{\mathrm{7}} {\equiv}\mathrm{0} \\ $$$${a}+{b}=\mathrm{7}\:\Rightarrow\mathrm{7} \\ $$$${a}+{b}=\mathrm{14}\:\Rightarrow\mathrm{5} \\ $$$${total}\:\mathrm{12}\:{not}\:\mathrm{11}? \\ $$$$ \\ $$$${i}\:{counted}\:{totally}\:\mathrm{93}\:{valid}\:{numbers}. \\ $$

Commented by mr W last updated on 04/Oct/22

Commented by mahdipoor last updated on 04/Oct/22

answer to SLVR: i> 100a+10b+c≡^7 (98a+7b)+(2a+3b+c)≡^7 2a+3b+c 7∣98a+7b=7(14a+b) ii> for example: a=c ⇒ 3(a+b)≡^7 (a+b)≡^7 0 ⇒ b≡^7 −a+7a ⇒b≡^7 6a 0≤a,b∈N≤9 a≠0 (a,b)=(1,6),(2,5),(3,4),(4,3),(5,2),(5,9), (6,1),(6,8),(7,0),(7,7),(8,6),(9,5) ⇒12 number answer to Mr.W: thx,i edited.

$${answer}\:{to}\:{SLVR}: \\ $$$${i}> \\ $$$$\mathrm{100}{a}+\mathrm{10}{b}+{c}\overset{\mathrm{7}} {\equiv}\left(\mathrm{98}{a}+\mathrm{7}{b}\right)+\left(\mathrm{2}{a}+\mathrm{3}{b}+{c}\right)\overset{\mathrm{7}} {\equiv} \\ $$$$\mathrm{2}{a}+\mathrm{3}{b}+{c}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{7}\mid\mathrm{98}{a}+\mathrm{7}{b}=\mathrm{7}\left(\mathrm{14}{a}+{b}\right) \\ $$$${ii}>\:\:{for}\:{example}: \\ $$$${a}={c}\:\Rightarrow\:\mathrm{3}\left({a}+{b}\right)\overset{\mathrm{7}} {\equiv}\left({a}+{b}\right)\overset{\mathrm{7}} {\equiv}\mathrm{0}\:\Rightarrow \\ $$$${b}\overset{\mathrm{7}} {\equiv}−{a}+\mathrm{7}{a}\:\Rightarrow{b}\overset{\mathrm{7}} {\equiv}\mathrm{6}{a}\:\:\:\:\mathrm{0}\leqslant{a},{b}\in{N}\leqslant\mathrm{9}\:\:\:\:\:{a}\neq\mathrm{0}\:\: \\ $$$$\left({a},{b}\right)=\left(\mathrm{1},\mathrm{6}\right),\left(\mathrm{2},\mathrm{5}\right),\left(\mathrm{3},\mathrm{4}\right),\left(\mathrm{4},\mathrm{3}\right),\left(\mathrm{5},\mathrm{2}\right),\left(\mathrm{5},\mathrm{9}\right), \\ $$$$\left(\mathrm{6},\mathrm{1}\right),\left(\mathrm{6},\mathrm{8}\right),\left(\mathrm{7},\mathrm{0}\right),\left(\mathrm{7},\mathrm{7}\right),\left(\mathrm{8},\mathrm{6}\right),\left(\mathrm{9},\mathrm{5}\right) \\ $$$$\Rightarrow\mathrm{12}\:{number}\: \\ $$$${answer}\:{to}\:{Mr}.{W}: \\ $$$${thx},{i}\:{edited}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com