Question Number 43587 by peter frank last updated on 12/Sep/18
$${a}\:{and}\:{b}\:\:{are}\:{the}\:{digit}\:{in}\:{a}\:{four}\:{digit} \\ $$$${number}\:\mathrm{12}{ab}.{if}\:\mathrm{12}{ab}\:{is}\:{divisble}\:{by}\: \\ $$$$\mathrm{5}\:{and}\:\mathrm{9}\:.{find}\:{the}\:{sum}\:{of}\:{all}\:{possible} \\ $$$${value}\:{of}\:\:{a}. \\ $$
Answered by MrW3 last updated on 12/Sep/18
$${b}=\mathrm{0}\:{or}\:\mathrm{5} \\ $$$${with}\:{b}=\mathrm{0}: \\ $$$$\mathrm{1200}+\mathrm{10}{a}=\mathrm{9}{n} \\ $$$$\left(\mathrm{133}+{a}\right)×\mathrm{9}+{a}+\mathrm{3}=\mathrm{9}{n} \\ $$$$\Rightarrow{a}=\mathrm{6} \\ $$$${with}\:{b}=\mathrm{5}: \\ $$$$\mathrm{1200}+\mathrm{10}{a}+\mathrm{5}=\mathrm{9}{n} \\ $$$$\left(\mathrm{133}+{a}\right)×\mathrm{9}+{a}+\mathrm{8}=\mathrm{9}{n} \\ $$$$\Rightarrow{a}=\mathrm{1} \\ $$$$\Rightarrow\Sigma{a}=\mathrm{6}+\mathrm{1}=\mathrm{7} \\ $$$${the}\:{numer}\:{is}\:\mathrm{1260}\:{or}\:\mathrm{1215} \\ $$
Commented by peter frank last updated on 12/Sep/18
$$\mathrm{1200}\:?\:{where}\:{do}\:{u}\:{get} \\ $$
Commented by peter frank last updated on 12/Sep/18
$${okay}\:{sir}\:{how}\:{about}\:\mathrm{9}{n}\:{and}\:{b}=\mathrm{0}\:{came}\:{from}. \\ $$
Commented by MrW3 last updated on 12/Sep/18
$${in}\:{the}\:{four}\:{digit}\:{number}\:\mathrm{12}{ab} \\ $$$$\mathrm{1}\:{means}\:\mathrm{1000} \\ $$$$\mathrm{2}\:{means}\:\mathrm{200} \\ $$$${a}\:{means}\:\mathrm{10}{a} \\ $$$${b}\:{means}\:{b} \\ $$$$\mathrm{12}{ab}\equiv\mathrm{1000}+\mathrm{200}+\mathrm{10}{a}+{b}\equiv\mathrm{1200}+\mathrm{10}{a}+{b} \\ $$
Commented by MrW3 last updated on 12/Sep/18
$${if}\:{a}\:{number}\:{is}\:{divisible}\:{by}\:\mathrm{5},\:{its}\:{last} \\ $$$${digit}\:{must}\:{be}\:\mathrm{0}\:{or}\:\mathrm{5},\:{i}.{e}.\:{b}=\mathrm{0}\:{or}\:\mathrm{5}. \\ $$$${case}\:\mathrm{1}:\:{b}=\mathrm{0} \\ $$$$\left(\mathrm{12}{a}\mathrm{0}\right)=\mathrm{1200}+\mathrm{10}{a} \\ $$$${this}\:{number}\:{is}\:{divisible}\:{by}\:\mathrm{9},\:{i}.{e}. \\ $$$${it}\:{is}\:{a}\:{multiple}\:{of}\:\mathrm{9},\:{i}.{e}. \\ $$$$\mathrm{1200}+\mathrm{10}{a}=\mathrm{9}{n}\:{with}\:{n}={some}\:{integer} \\ $$$$\mathrm{1200}+\mathrm{10}{a}=\mathrm{133}×\mathrm{9}+\mathrm{3}+\mathrm{9}{a}+{a} \\ $$$$=\mathrm{9}×\left(\mathrm{133}+{a}\right)+\left({a}+\mathrm{3}\right) \\ $$$${such}\:{that}\:{it}\:{is}\:{a}\:{multiple}\:{of}\:\mathrm{9}, \\ $$$$\left({a}+\mathrm{3}\right)\:{must}\:{be}\:\mathrm{9},\:{i}.{e}.\:{a}=\mathrm{6} \\ $$$$ \\ $$$${case}\:\mathrm{2}:\:{b}=\mathrm{5} \\ $$$${now}\:{you}\:{can}\:{figure}\:{it}\:{out}\:{by}\:{yourself}. \\ $$