Question Number 114295 by mathdave last updated on 18/Sep/20
Answered by mr W last updated on 19/Sep/20
![say the number is [abcde] let A=a+c+e B=b+d A+B=41 (as given) A−B=0 or ±11, ±22, ... ⇒A=20, B=21 (impossible, since B<2×9=18) ⇒A=26, B=15 ⇒A=37, B=4 (impossible, since A<3×9=27) so we have only one possibility: A=26, B=15 ⇒A=9+9+8 ⇒B=9+6 ⇒smallest number is 86999 ⇒largest number is 99968](https://www.tinkutara.com/question/Q114341.png)
$${say}\:{the}\:{number}\:{is}\:\left[{abcde}\right] \\ $$$${let}\:{A}={a}+{c}+{e} \\ $$$${B}={b}+{d} \\ $$$${A}+{B}=\mathrm{41}\:\left({as}\:{given}\right) \\ $$$${A}−{B}=\mathrm{0}\:{or}\:\pm\mathrm{11},\:\pm\mathrm{22},\:… \\ $$$$\Rightarrow{A}=\mathrm{20},\:{B}=\mathrm{21}\:\left({impossible},\:{since}\:{B}<\mathrm{2}×\mathrm{9}=\mathrm{18}\right) \\ $$$$\Rightarrow{A}=\mathrm{26},\:{B}=\mathrm{15} \\ $$$$\Rightarrow{A}=\mathrm{37},\:{B}=\mathrm{4}\:\:\left({impossible},\:{since}\:{A}<\mathrm{3}×\mathrm{9}=\mathrm{27}\right) \\ $$$${so}\:{we}\:{have}\:{only}\:{one}\:{possibility}: \\ $$$${A}=\mathrm{26},\:{B}=\mathrm{15} \\ $$$$\Rightarrow{A}=\mathrm{9}+\mathrm{9}+\mathrm{8} \\ $$$$\Rightarrow{B}=\mathrm{9}+\mathrm{6} \\ $$$$\Rightarrow{smallest}\:{number}\:{is}\:\mathrm{86999} \\ $$$$\Rightarrow{largest}\:{number}\:{is}\:\mathrm{99968} \\ $$