Question Number 177309 by SLVR last updated on 03/Oct/22
$${How}\:{many}\:\mathrm{3}\:{digited}\:{numbers}\: \\ $$$$\left.{which}\:{are}\:{divisible}\:{by}\:\mathrm{1}\right)\mathrm{3} \\ $$$$\left.\mathrm{2}\left.\right)\left.\mathrm{4}\left.\:\left.\:\left.\:\:\mathrm{3}\right)\mathrm{5}\:\:\:\:\mathrm{4}\right)\mathrm{6}\:\:\:\mathrm{5}\right)\mathrm{7}\:\:\:\:\mathrm{6}\right)\mathrm{8}\:\:\mathrm{7}\right)\mathrm{9} \\ $$$${with}\:{repetetion}\:{of}\:{digits}\:{is} \\ $$$${NOT}\:{allowed}…{one}\:{problem}\:{process} \\ $$
Commented by SLVR last updated on 03/Oct/22
$${i}\:{need}\:\:{for}\:{divisible}\:{by}\:\mathrm{7}\: \\ $$
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](https://www.tinkutara.com/question/Q177322.png)
$${N}=\left({abc}\right)=\mathrm{100}{a}+\mathrm{10}{b}+{c}\:\: \\ $$$$\mathrm{0}\leqslant{a},{b},{c}\in\mathbb{N}\leqslant\mathrm{9}\:\:\:\:\:\:{a}\neq\mathrm{0}\: \\ $$$$\Rightarrow{N}\overset{\mathrm{7}} {\equiv}\mathrm{0}\:\:\:\:\:\:{and}\:\:\:\:\:\:{repetetion}\:{of}\:{digits} \\ $$$$\Rightarrow\mathrm{100}{a}+\mathrm{10}{b}+{c}\overset{\mathrm{7}} {\equiv}\mathrm{2}{a}+\mathrm{3}{b}+{c}\overset{\mathrm{7}} {\equiv}\mathrm{0} \\ $$$$\left.{i}\right)\:{a}={b}\:\Rightarrow\:{c}\overset{\mathrm{7}} {\equiv}\mathrm{2}{a}\:\Rightarrow\:\mathrm{13}\:\:{numbers}\: \\ $$$$\left.{ii}\right)\:{a}={c}\:\Rightarrow\:{b}\overset{\mathrm{7}} {\equiv}\mathrm{6}{a}\:\Rightarrow\:\mathrm{12}\: \\ $$$$\left.{iii}\right)\:{b}={c}\:\Rightarrow\:{a}\overset{\mathrm{7}} {\equiv}\mathrm{5}{b}\:\Rightarrow\:\mathrm{12}\:\: \\ $$$$\left.{iv}\right)\:\:{a}={b}={c}\:\Rightarrow\:\mathrm{6}{a}\overset{\mathrm{7}} {\equiv}\mathrm{0}\:\Rightarrow\:\mathrm{1} \\ $$$$\Rightarrow{all}\:{numbers}=\mathrm{13}+\mathrm{12}+\mathrm{12}−\mathrm{2}×\mathrm{1}=\mathrm{35} \\ $$$$\mathrm{3}\:{digit}\:{number}\:{are}\:{divisible}\:{by}\:\mathrm{7}: \\ $$$$\left[\frac{\mathrm{999}}{\mathrm{7}}\right]−\left[\frac{\mathrm{99}}{\mathrm{7}}\right]=\mathrm{128} \\ $$$$\Rightarrow\mathrm{128}−\mathrm{35}=\mathrm{93} \\ $$
Commented by SLVR last updated on 04/Oct/22
$${Thanks}\:{for}\:{your}\:{concern}\:{sir} \\ $$$${but}\:{how}\:\mathrm{4}{th}\:{line}\mathrm{2}{a}+\mathrm{3}{b}+{c}\equiv\mathrm{0}{mod}\mathrm{7} \\ $$$${and}\:{how}\:\mathrm{13}\:{numbers}\:{if}\:{a}={b} \\ $$
Commented by SLVR last updated on 04/Oct/22
$${kindly}\:{explain}\:{what}\:{those}\:\mathrm{34} \\ $$$${numbers}\:{be}\:{deducted}..{please} \\ $$
Commented by mr W last updated on 04/Oct/22
$$\left.{ii}\right)\:{a}={c} \\ $$$$\mathrm{2}{a}+\mathrm{3}{b}+{c}=\mathrm{3}\left({a}+{b}\right)\overset{\mathrm{7}} {\equiv}\mathrm{0} \\ $$$${a}+{b}=\mathrm{7}\:\Rightarrow\mathrm{7} \\ $$$${a}+{b}=\mathrm{14}\:\Rightarrow\mathrm{5} \\ $$$${total}\:\mathrm{12}\:{not}\:\mathrm{11}? \\ $$$$ \\ $$$${i}\:{counted}\:{totally}\:\mathrm{93}\:{valid}\:{numbers}. \\ $$
Commented by mr W last updated on 04/Oct/22
Commented by mahdipoor last updated on 04/Oct/22
$${answer}\:{to}\:{SLVR}: \\ $$$${i}> \\ $$$$\mathrm{100}{a}+\mathrm{10}{b}+{c}\overset{\mathrm{7}} {\equiv}\left(\mathrm{98}{a}+\mathrm{7}{b}\right)+\left(\mathrm{2}{a}+\mathrm{3}{b}+{c}\right)\overset{\mathrm{7}} {\equiv} \\ $$$$\mathrm{2}{a}+\mathrm{3}{b}+{c}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{7}\mid\mathrm{98}{a}+\mathrm{7}{b}=\mathrm{7}\left(\mathrm{14}{a}+{b}\right) \\ $$$${ii}>\:\:{for}\:{example}: \\ $$$${a}={c}\:\Rightarrow\:\mathrm{3}\left({a}+{b}\right)\overset{\mathrm{7}} {\equiv}\left({a}+{b}\right)\overset{\mathrm{7}} {\equiv}\mathrm{0}\:\Rightarrow \\ $$$${b}\overset{\mathrm{7}} {\equiv}−{a}+\mathrm{7}{a}\:\Rightarrow{b}\overset{\mathrm{7}} {\equiv}\mathrm{6}{a}\:\:\:\:\mathrm{0}\leqslant{a},{b}\in{N}\leqslant\mathrm{9}\:\:\:\:\:{a}\neq\mathrm{0}\:\: \\ $$$$\left({a},{b}\right)=\left(\mathrm{1},\mathrm{6}\right),\left(\mathrm{2},\mathrm{5}\right),\left(\mathrm{3},\mathrm{4}\right),\left(\mathrm{4},\mathrm{3}\right),\left(\mathrm{5},\mathrm{2}\right),\left(\mathrm{5},\mathrm{9}\right), \\ $$$$\left(\mathrm{6},\mathrm{1}\right),\left(\mathrm{6},\mathrm{8}\right),\left(\mathrm{7},\mathrm{0}\right),\left(\mathrm{7},\mathrm{7}\right),\left(\mathrm{8},\mathrm{6}\right),\left(\mathrm{9},\mathrm{5}\right) \\ $$$$\Rightarrow\mathrm{12}\:{number}\: \\ $$$${answer}\:{to}\:{Mr}.{W}: \\ $$$${thx},{i}\:{edited}. \\ $$