Question Number 191775 by Mingma last updated on 30/Apr/23
Answered by deleteduser1 last updated on 30/Apr/23
![ABCDEF^(____________) =10000ABCD^(_______) +EF^(___) 17[10000AB^(___) +CDEF^(_______) ]=12[100CDEF^(_______) +AB^(___) ] 170000AB^(___) −12AB^(___) =1200CDEF^(_______) −17CDEF^(________) ⇒((AB^(___) )/(CDEF^(________) ))=((1183)/(169988))=((13)/(1868)) Note that AB^(___) and CDEF^(_______) can still be increased by a factor of 5 while ABCDEF^(___________) remains a 6-digit number ⇒The largest 6-digit number=659340 The smallest 6-digit number=131868](Q191781.png)
$$\overset{\_\_\_\_\_\_\_\_\_\_\_\_} {{ABCDEF}}=\mathrm{10000}\overset{\_\_\_\_\_\_\_} {{ABCD}}+\overset{\_\_\_} {{EF}} \\ $$$$\mathrm{17}\left[\mathrm{10000}\overset{\_\_\_} {{AB}}+\overset{\_\_\_\_\_\_\_} {{CDEF}}\right]=\mathrm{12}\left[\mathrm{100}\overset{\_\_\_\_\_\_\_} {{CDEF}}+\overset{\_\_\_} {{AB}}\right] \\ $$$$\mathrm{170000}\overset{\_\_\_} {{AB}}−\mathrm{12}\overset{\_\_\_} {{AB}}=\mathrm{1200}\overset{\_\_\_\_\_\_\_} {{CDEF}}−\mathrm{17}\overset{\_\_\_\_\_\_\_\_} {{CDEF}} \\ $$$$\Rightarrow\frac{\overset{\_\_\_} {{AB}}}{\overset{\_\_\_\_\_\_\_\_} {{CDEF}}}=\frac{\mathrm{1183}}{\mathrm{169988}}=\frac{\mathrm{13}}{\mathrm{1868}} \\ $$$${Note}\:{that}\:\overset{\_\_\_} {{AB}}\:{and}\:\overset{\_\_\_\_\_\_\_} {{CDEF}}\:{can}\:{still}\:{be}\:{increased} \\ $$$${by}\:{a}\:{factor}\:{of}\:\mathrm{5}\:{while}\:\:\overset{\_\_\_\_\_\_\_\_\_\_\_} {{ABCDEF}}\:{remains}\:{a} \\ $$$$\mathrm{6}-{digit}\:{number} \\ $$$$\Rightarrow{The}\:{largest}\:\mathrm{6}-{digit}\:{number}=\mathrm{659340} \\ $$$$\:\:\:\:\:{The}\:{smallest}\:\mathrm{6}-{digit}\:{number}=\mathrm{131868} \\ $$
Commented by Mingma last updated on 30/Apr/23
Excellent!
Answered by Frix last updated on 30/Apr/23
$$\mathrm{This}\:\mathrm{leads}\:\mathrm{to} \\ $$$$\frac{{AB}}{\mathrm{13}}=\frac{{CDEF}}{\mathrm{1868}} \\ $$$$\mathrm{With}\:\mathrm{all}\:\mathrm{digits}\:\mathrm{different}\:\mathrm{we}\:\mathrm{have}\:\mathrm{only}\:\mathrm{2} \\ $$$$\mathrm{possibilities}: \\ $$$$\frac{\mathrm{39}}{\mathrm{13}}=\frac{\mathrm{5604}}{\mathrm{1868}}=\mathrm{3} \\ $$$$\frac{\mathrm{65}}{\mathrm{13}}=\frac{\mathrm{9340}}{\mathrm{1868}}=\mathrm{5} \\ $$$$\Rightarrow \\ $$$$\mathrm{Answer}\:\mathrm{is}\:\mathrm{659340} \\ $$
Commented by Mingma last updated on 30/Apr/23
Excellent!