Question Number 81647 by john santu last updated on 14/Feb/20
$$\mathrm{how}\:\mathrm{to}\:\mathrm{prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{number}\: \\ $$$$\mathrm{is}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{3},\:\mathrm{then}\:\mathrm{the}\:\mathrm{number} \\ $$$$\mathrm{of}\:\mathrm{numbers}\:\mathrm{is}\:\mathrm{a}\:\mathrm{multiple}\:\mathrm{of}\:\mathrm{3} \\ $$
Commented by mr W last updated on 14/Feb/20
$${did}\:{you}\:{want}\:{to}\:{say}: \\ $$$${how}\:{to}\:{prove}\:{that}\:{a}\:{number}\:{is}\:{divisible} \\ $$$${by}\:\mathrm{3}\:{when}\:{the}\:{sum}\:{of}\:{its}\:{digits}\:{is}\:{a} \\ $$$${multiple}\:{of}\:\mathrm{3}. \\ $$$${or} \\ $$$${how}\:{to}\:{prove}\:{that},\:{if}\:{a}\:{number}\:{is}\:{divisible} \\ $$$${by}\:\mathrm{3},\:{then}\:{the}\:{sum}\:{of}\:{its}\:{digits}\:{is}\:{a} \\ $$$${multiple}\:{of}\:\mathrm{3}. \\ $$
Commented by Kunal12588 last updated on 14/Feb/20
$$\mathrm{100}{a}+\mathrm{10}{b}+{c}=\mathrm{3}{q}\:\:\:{for}\:{a}\:\mathrm{3}\:{digit}\:{no}. \\ $$$$\Rightarrow\mathrm{3}{q}=\mathrm{99}{a}+\mathrm{9}{b}+\left({a}+{b}+{c}\right) \\ $$$$\Rightarrow{q}=\mathrm{33}{a}+\mathrm{3}{b}+\frac{{a}+{b}+{c}}{\mathrm{3}} \\ $$$${q}\:\in\:{N}\:\Rightarrow\:{a}+{b}+{c}\:\mid\:\mathrm{3} \\ $$$${for}\:{a}\:'{n}'\:{digit}\:{no}. \\ $$$$\mathrm{3}{m}=\underset{{i}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\mathrm{10}^{{i}} {a}_{{i}} \:;\:{a}_{{i}} \:{is}\:\left({i}+\mathrm{1}\right)^{{th}} \:{digit} \\ $$$$\Rightarrow{m}=\mathrm{3}{k}\:+\:\frac{\mathrm{1}}{\mathrm{3}}\underset{{i}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}{a}_{{i}} \: \\ $$$$\Rightarrow\underset{{i}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}{a}_{{i}} \:\:{is}\:{divisible}\:{by}\:\mathrm{3}. \\ $$$${please}\:{help}\:{me}\:{to}\:{find}\:{k} \\ $$
Commented by Prithwish Sen 1 last updated on 14/Feb/20
$${any}\:{number}\:{can}\:{be}\:{represented}\:{by}\: \\ $$$$\boldsymbol{{N}}=\mathrm{10}^{{n}} {x}_{{n}} +\mathrm{10}^{{n}−\mathrm{1}} {x}_{{n}−\mathrm{1}} +……..+\mathrm{10}^{\mathrm{2}} {x}_{\mathrm{2}} \:+\mathrm{10}{x}_{\mathrm{1}} +{x}_{\mathrm{0}} \\ $$$${now} \\ $$$${x}_{{n}} +{x}_{{n}−\mathrm{1}} +………..+{x}_{\mathrm{1}} +{x}_{\mathrm{0}} =\:\mathrm{3}\boldsymbol{{k}} \\ $$$$\therefore\:\boldsymbol{{N}}=\:\left(\mathrm{999}…{ntimes}+\mathrm{1}\right){x}_{{n}} +……+\left(\mathrm{99}+\mathrm{1}\right){x}_{\mathrm{2}} +\left(\mathrm{9}+\mathrm{1}\right){x}_{\mathrm{1}} +{x}_{\mathrm{0}} \\ $$$$=\:\left(\mathrm{999}…{n}\:{times}\:.{x}_{{n}} +\:….+\mathrm{99}{x}_{\mathrm{2}} +\mathrm{9}{x}_{\mathrm{1}} \right)+\left({x}_{{n}} +{x}_{{n}−\mathrm{1}} +…..+{x}_{\mathrm{2}} +{x}_{\mathrm{1}} +{x}_{\mathrm{0}} \right) \\ $$$$=\mathrm{3}\boldsymbol{{m}}+\mathrm{3}\boldsymbol{{k}} \\ $$$$\boldsymbol{{i}}.\boldsymbol{{e}}\:\boldsymbol{{N}}\:\boldsymbol{{is}}\:\boldsymbol{{also}}\:\boldsymbol{{divisible}}\:\boldsymbol{{by}}\:\mathrm{3}\:\left(\boldsymbol{{proved}}\right) \\ $$$$ \\ $$
Commented by mr W last updated on 14/Feb/20
$$\mathrm{3}{m}=\underset{{i}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\mathrm{10}^{{i}} {a}_{{i}} \\ $$$$\mathrm{3}{m}=\underset{{i}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\left(\mathrm{1}+\mathrm{9}\right)^{{i}} {a}_{{i}} \\ $$$$\mathrm{3}{m}=\underset{{i}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\left[\underset{{k}=\mathrm{0}} {\overset{{i}} {\sum}}{C}_{{k}} ^{{i}} \mathrm{9}^{{k}} \right]{a}_{{i}} \\ $$$$\mathrm{3}{m}=\underset{{i}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\left[\mathrm{1}+\underset{{k}=\mathrm{1}} {\overset{{i}} {\sum}}{C}_{{k}} ^{{i}} \mathrm{9}^{{k}} \right]{a}_{{i}} \\ $$$$\mathrm{3}{m}=\underset{{i}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}{a}_{{i}} +\underset{{i}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\sum}}\left(\underset{{k}=\mathrm{1}} {\overset{{i}} {\sum}}{C}_{{k}} ^{{i}} \mathrm{3}^{\mathrm{2}{k}} \right){a}_{{i}} \\ $$$$\mathrm{3}{m}=\underset{{i}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}{a}_{{i}} +\mathrm{3}{h} \\ $$$$\Rightarrow\underset{{i}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}{a}_{{i}} =\mathrm{3}\left({m}−{h}\right)={multiple}\:{of}\:\mathrm{3} \\ $$
Commented by Prithwish Sen 1 last updated on 14/Feb/20
$${How}\:{are}\:{you}\:{sir}\:? \\ $$
Commented by mr W last updated on 14/Feb/20
$${every}\:{thing}\:{is}\:{ok}\:{with}\:{me},\:{thanks}!\: \\ $$$${you}\:{too}\:{sir}? \\ $$
Commented by john santu last updated on 14/Feb/20
$$\mathrm{yes}.\: \\ $$
Commented by Prithwish Sen 1 last updated on 15/Feb/20
$${fine}\:{sir}. \\ $$