Question Number 191831 by Shlock last updated on 01/May/23
Answered by mehdee42 last updated on 02/May/23
$${suppose}\:,\:\:{n}=<{abcdefghij}>\:,{is}\:{an}\:<{i}.{n}> \\ $$$${a}+{b}+{c}+…+{i}+{j}\overset{\mathrm{9}} {\equiv}\mathrm{0}\Rightarrow\mathrm{9}\mid{n} \\ $$$$\mathrm{9}\mid{n}\:,\:\mathrm{11111}\mid{n}\:\:,\:\left(\mathrm{9},\mathrm{11111}\right)=\mathrm{1}\Rightarrow\mathrm{99999}\mid{n} \\ $$$${let}\:\:{x}=<{abcde}>\:\:\&\:\:{y}={f}<{fghij}> \\ $$$$\Rightarrow{n}=\mathrm{10}^{\mathrm{5}} {x}+{y}\overset{\mathrm{99999}} {\equiv}\:{x}+{y}\:\Rightarrow{x}+{y}\overset{\mathrm{99999}} {\equiv}\:\mathrm{0} \\ $$$$\mathrm{0}<{x}+{y}<\mathrm{2}×\mathrm{99999}\Rightarrow{x}+{y}=\mathrm{99999} \\ $$$$\left({a}+{f}={b}+{g}=…={e}+{j}=\mathrm{9}\right) \\ $$$${there}\:{are}\:\mathrm{5}!=\mathrm{120}\:{ways}\:\:{to}\:{move}\:<{abcde}>. \\ $$$${also}\:{there}\:{are}\:\mathrm{2}^{\mathrm{5}} =\mathrm{32}\:{ways}\:\:{move}\:{for}\:{pair}\:<\left({a},{f}\right),…,\left({e},{j}\right)> \\ $$$${on}\:{thr}\:{other}\:{hand}\:,\:{a}\neq\mathrm{0}. \\ $$$${so}\:{the}\:{total}\:{number}\:{intersting}\:\::\:\frac{\mathrm{9}}{\mathrm{10}}×\mathrm{32}×\mathrm{120}=\mathrm{3456}\: \\ $$