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Let-S-1-2-9-0-A-How-many-multiples-of-eight-of-four-digits-can-be-formed-from-S-B-The-same-question-for-different-digits-




Question Number 178338 by Acem last updated on 15/Oct/22
Let S= {1, 2, ..., 9, 0} A: How many multiples   of eight of four digits can be formed from S   B: The same question for different digits
$${Let}\:{S}=\:\left\{\mathrm{1},\:\mathrm{2},\:…,\:\mathrm{9},\:\mathrm{0}\right\}\:\boldsymbol{{A}}:\:{How}\:{many}\:{multiples} \\ $$$$\:{of}\:{eight}\:{of}\:{four}\:{digits}\:{can}\:{be}\:{formed}\:{from}\:{S} \\ $$$$\:\boldsymbol{{B}}:\:{The}\:{same}\:{question}\:{for}\:{different}\:{digits} \\ $$
Commented by Acem last updated on 17/Oct/22
Hint:  Answer of B is 560 numbers with different digits
$$\boldsymbol{{Hint}}: \\ $$$${Answer}\:{of}\:{B}\:{is}\:\mathrm{560}\:{numbers}\:{with}\:{different}\:{digits} \\ $$
Answered by mr W last updated on 16/Oct/22
[A]  ⌊((9999)/8)⌋−⌊((999)/8)⌋=1249−124=1125    solution for A is simple. we just need  to count the number of terms of an  A.P.:  1000, 1008, 1016, ....,9992  that′s ((9992−1000)/8)+1=1125 numbers.
$$\left[\boldsymbol{{A}}\right] \\ $$$$\lfloor\frac{\mathrm{9999}}{\mathrm{8}}\rfloor−\lfloor\frac{\mathrm{999}}{\mathrm{8}}\rfloor=\mathrm{1249}−\mathrm{124}=\mathrm{1125} \\ $$$$ \\ $$$${solution}\:{for}\:{A}\:{is}\:{simple}.\:{we}\:{just}\:{need} \\ $$$${to}\:{count}\:{the}\:{number}\:{of}\:{terms}\:{of}\:{an} \\ $$$${A}.{P}.: \\ $$$$\mathrm{1000},\:\mathrm{1008},\:\mathrm{1016},\:….,\mathrm{9992} \\ $$$${that}'{s}\:\frac{\mathrm{9992}−\mathrm{1000}}{\mathrm{8}}+\mathrm{1}=\mathrm{1125}\:{numbers}. \\ $$
Commented by mr W last updated on 16/Oct/22
maybe there are better ways.
$${maybe}\:{there}\:{are}\:{better}\:{ways}. \\ $$
Commented by mr W last updated on 16/Oct/22
[B]  say such a number is ABCD with  A≠B≠C≠D, A≠0   determinant (((CD=),(08),(16),(24),(32),(40),(48),(56),(64),(72),(80),(96)),(B_n ,3,(3/1),(2/1),(3/1),3,(2/1),4,(2/1),(3/1),3,(3/1)),(A_n ,7,(6/7),(6/7),(6/7),7,(6/7),(6/7),(6/7),(6/7),7,(6/7)),(,(21),(25),(19),(25),(21),(19),(25),(19),(25),(21),(25)))  21×3+25×5+19×3=245   determinant (((CD=),(04),(12),(20),(28),(36),(52),(60),(68),(76),(84),(92)),(B_n ,5,4,5,5,4,4,5,5,4,5,4),(A_n ,7,6,7,6,6,6,7,6,6,6,6),(,(35),(24),(35),(30),(24),(24),(35),(30),(24),(30),(24)))  35×3+24×5+30×3=315    ⇒totally 245+315=560 numbers with   different digits    ================  explanation of the table  say the number is ABCD.  when CD=08, B can be 2, 4, 6, i.e.  3 possibilities (B_n =3). for each B=2,  A can be 1,3,4,5,6,7,9, i.e. 7 possibilities   (A_n =7). ⇒3×7=21 numbers.
$$\left[\boldsymbol{{B}}\right] \\ $$$${say}\:{such}\:{a}\:{number}\:{is}\:{ABCD}\:{with} \\ $$$${A}\neq{B}\neq{C}\neq{D},\:{A}\neq\mathrm{0} \\ $$$$\begin{array}{|c|c|c|c|}{{CD}=}&\hline{\mathrm{08}}&\hline{\mathrm{16}}&\hline{\mathrm{24}}&\hline{\mathrm{32}}&\hline{\mathrm{40}}&\hline{\mathrm{48}}&\hline{\mathrm{56}}&\hline{\mathrm{64}}&\hline{\mathrm{72}}&\hline{\mathrm{80}}&\hline{\mathrm{96}}\\{{B}_{{n}} }&\hline{\mathrm{3}}&\hline{\mathrm{3}/\mathrm{1}}&\hline{\mathrm{2}/\mathrm{1}}&\hline{\mathrm{3}/\mathrm{1}}&\hline{\mathrm{3}}&\hline{\mathrm{2}/\mathrm{1}}&\hline{\mathrm{4}}&\hline{\mathrm{2}/\mathrm{1}}&\hline{\mathrm{3}/\mathrm{1}}&\hline{\mathrm{3}}&\hline{\mathrm{3}/\mathrm{1}}\\{{A}_{{n}} }&\hline{\mathrm{7}}&\hline{\mathrm{6}/\mathrm{7}}&\hline{\mathrm{6}/\mathrm{7}}&\hline{\mathrm{6}/\mathrm{7}}&\hline{\mathrm{7}}&\hline{\mathrm{6}/\mathrm{7}}&\hline{\mathrm{6}/\mathrm{7}}&\hline{\mathrm{6}/\mathrm{7}}&\hline{\mathrm{6}/\mathrm{7}}&\hline{\mathrm{7}}&\hline{\mathrm{6}/\mathrm{7}}\\{}&\hline{\mathrm{21}}&\hline{\mathrm{25}}&\hline{\mathrm{19}}&\hline{\mathrm{25}}&\hline{\mathrm{21}}&\hline{\mathrm{19}}&\hline{\mathrm{25}}&\hline{\mathrm{19}}&\hline{\mathrm{25}}&\hline{\mathrm{21}}&\hline{\mathrm{25}}\\\hline\end{array} \\ $$$$\mathrm{21}×\mathrm{3}+\mathrm{25}×\mathrm{5}+\mathrm{19}×\mathrm{3}=\mathrm{245} \\ $$$$\begin{array}{|c|c|c|c|}{{CD}=}&\hline{\mathrm{04}}&\hline{\mathrm{12}}&\hline{\mathrm{20}}&\hline{\mathrm{28}}&\hline{\mathrm{36}}&\hline{\mathrm{52}}&\hline{\mathrm{60}}&\hline{\mathrm{68}}&\hline{\mathrm{76}}&\hline{\mathrm{84}}&\hline{\mathrm{92}}\\{{B}_{{n}} }&\hline{\mathrm{5}}&\hline{\mathrm{4}}&\hline{\mathrm{5}}&\hline{\mathrm{5}}&\hline{\mathrm{4}}&\hline{\mathrm{4}}&\hline{\mathrm{5}}&\hline{\mathrm{5}}&\hline{\mathrm{4}}&\hline{\mathrm{5}}&\hline{\mathrm{4}}\\{{A}_{{n}} }&\hline{\mathrm{7}}&\hline{\mathrm{6}}&\hline{\mathrm{7}}&\hline{\mathrm{6}}&\hline{\mathrm{6}}&\hline{\mathrm{6}}&\hline{\mathrm{7}}&\hline{\mathrm{6}}&\hline{\mathrm{6}}&\hline{\mathrm{6}}&\hline{\mathrm{6}}\\{}&\hline{\mathrm{35}}&\hline{\mathrm{24}}&\hline{\mathrm{35}}&\hline{\mathrm{30}}&\hline{\mathrm{24}}&\hline{\mathrm{24}}&\hline{\mathrm{35}}&\hline{\mathrm{30}}&\hline{\mathrm{24}}&\hline{\mathrm{30}}&\hline{\mathrm{24}}\\\hline\end{array} \\ $$$$\mathrm{35}×\mathrm{3}+\mathrm{24}×\mathrm{5}+\mathrm{30}×\mathrm{3}=\mathrm{315} \\ $$$$ \\ $$$$\Rightarrow{totally}\:\mathrm{245}+\mathrm{315}=\mathrm{560}\:{numbers}\:{with}\: \\ $$$${different}\:{digits} \\ $$$$ \\ $$$$================ \\ $$$${explanation}\:{of}\:{the}\:{table} \\ $$$${say}\:{the}\:{number}\:{is}\:{ABCD}. \\ $$$${when}\:{CD}=\mathrm{08},\:{B}\:{can}\:{be}\:\mathrm{2},\:\mathrm{4},\:\mathrm{6},\:{i}.{e}. \\ $$$$\mathrm{3}\:{possibilities}\:\left({B}_{{n}} =\mathrm{3}\right).\:{for}\:{each}\:{B}=\mathrm{2}, \\ $$$${A}\:{can}\:{be}\:\mathrm{1},\mathrm{3},\mathrm{4},\mathrm{5},\mathrm{6},\mathrm{7},\mathrm{9},\:{i}.{e}.\:\mathrm{7}\:{possibilities} \\ $$$$\:\left({A}_{{n}} =\mathrm{7}\right).\:\Rightarrow\mathrm{3}×\mathrm{7}=\mathrm{21}\:{numbers}. \\ $$
Commented by Acem last updated on 16/Oct/22
  I have another way for B, i hope it′s simple  but   our results are different
$$ \\ $$$${I}\:{have}\:{another}\:{way}\:{for}\:{B},\:{i}\:{hope}\:{it}'{s}\:{simple}\:\:{but} \\ $$$$\:{our}\:{results}\:{are}\:{different} \\ $$
Commented by Acem last updated on 16/Oct/22
 My answer for B is 541,    There are 19 diff. num. between us,    The solution is below
$$\:{My}\:{answer}\:{for}\:{B}\:{is}\:\mathrm{541},\: \\ $$$$\:{There}\:{are}\:\mathrm{19}\:{diff}.\:{num}.\:{between}\:{us},\: \\ $$$$\:{The}\:{solution}\:{is}\:{below} \\ $$
Commented by mr W last updated on 16/Oct/22
please recheck your solution sir.  the correct answer is 560.
$${please}\:{recheck}\:{your}\:{solution}\:{sir}. \\ $$$${the}\:{correct}\:{answer}\:{is}\:\mathrm{560}. \\ $$
Commented by Acem last updated on 17/Oct/22
Yes, because one of the excluded numbers   i subtract it twice, ooops. Anyway my method   is less difficult of yours. I tried yours and it   need for each CD must write or figure   2 subgroups for AB, while mine need only   simply compensation values odd & even.   Good luck!
$${Yes},\:{because}\:{one}\:{of}\:{the}\:{excluded}\:{numbers} \\ $$$$\:{i}\:{subtract}\:{it}\:{twice},\:{ooops}.\:{Anyway}\:{my}\:{method} \\ $$$$\:{is}\:{less}\:{difficult}\:{of}\:{yours}.\:{I}\:{tried}\:{yours}\:{and}\:{it} \\ $$$$\:{need}\:{for}\:{each}\:{CD}\:{must}\:{write}\:{or}\:{figure} \\ $$$$\:\mathrm{2}\:{subgroups}\:{for}\:{AB},\:{while}\:{mine}\:{need}\:{only} \\ $$$$\:{simply}\:{compensation}\:{values}\:{odd}\:\&\:{even}. \\ $$$$\:{Good}\:{luck}! \\ $$
Commented by mr W last updated on 18/Oct/22
i never praise myself.
$${i}\:{never}\:{praise}\:{myself}. \\ $$
Commented by Acem last updated on 18/Oct/22
You do indirectly. But me never.     Anyway i said that my method is easier not to   praise myself “didn′t talk my skin color”, but   for all of us to usefull from each other,   and take from all the better idea, especially the   easier, whatever whose!    By the way_1  our methods and ideas about    this qusetion are very close    By the way_2  when i composed this question   i was sure that you will like it and   solve it correctly.    Sincerely
$${You}\:{do}\:{indirectly}.\:{But}\:{me}\:{never}.\: \\ $$$$ \\ $$$${Anyway}\:{i}\:{said}\:{that}\:{my}\:{method}\:{is}\:{easier}\:{not}\:{to} \\ $$$$\:{praise}\:{myself}\:“{didn}'{t}\:{talk}\:{my}\:{skin}\:{color}'',\:{but} \\ $$$$\:{for}\:{all}\:{of}\:{us}\:{to}\:{usefull}\:{from}\:{each}\:{other}, \\ $$$$\:{and}\:{take}\:{from}\:{all}\:{the}\:{better}\:{idea},\:{especially}\:{the} \\ $$$$\:{easier},\:{whatever}\:{whose}! \\ $$$$ \\ $$$${By}\:{the}\:{way}_{\mathrm{1}} \:{our}\:{methods}\:{and}\:{ideas}\:{about}\: \\ $$$$\:{this}\:{qusetion}\:{are}\:{very}\:{close} \\ $$$$ \\ $$$${By}\:{the}\:{way}_{\mathrm{2}} \:{when}\:{i}\:{composed}\:{this}\:{question} \\ $$$$\:{i}\:{was}\:{sure}\:{that}\:{you}\:{will}\:{like}\:{it}\:{and} \\ $$$$\:\boldsymbol{{solve}}\:\boldsymbol{{it}}\:\boldsymbol{{correctly}}. \\ $$$$ \\ $$$${Sincerely} \\ $$$$\: \\ $$$$ \\ $$
Answered by Acem last updated on 19/Oct/22
   A:                               Last        two         digits   determinant ((T_(th) ,( Th),,,,,,,n,),(1,(   0^(ev) ),(00),(08),(16),(...),(88),(96),(13),(even_(Th) )),(2,(   1^(odd) ),(04),(12),(20),(...),(84),(92),(12),(odd_(Th) )),(3,(   2^(ev) ),,,,(...),,,(13),(even_(Th) )),(⋮,(  ⋮),,,,,,,,),(9,(   9^(odd) ),(04),(12),(20),(...),(84),(92),(12),(odd_(Th) )))   determinant (((The 2dig_(Th) . has 5_(even) & 5_(odd) )),((9_T_(th)  ×5_(even) ×13 + 9_T_(th)  ×5_(odd) ×12)))     NumNumb._(divisible by  8)  = 9×5(13+12)= 1 125   Note: This method is for who didn′t get the   two methods above. And it′s usefull for next (B)     B: is next soon
$$ \\ $$$$\:\boldsymbol{{A}}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{Last}\:\:\:\:\:\:\:\:{two}\:\:\:\:\:\:\:\:\:{digits} \\ $$$$\begin{array}{|c|c|c|c|c|c|}{{T}_{{th}} }&\hline{\:\boldsymbol{{Th}}}&\hline{}&\hline{}&\hline{}&\hline{}&\hline{}&\hline{}&\hline{{n}}&\hline{}\\{\mathrm{1}}&\hline{\:\:\:\mathrm{0}^{{ev}} }&\hline{\mathrm{00}}&\hline{\mathrm{08}}&\hline{\mathrm{16}}&\hline{…}&\hline{\mathrm{88}}&\hline{\mathrm{96}}&\hline{\mathrm{13}}&\hline{{even}_{{Th}} }\\{\mathrm{2}}&\hline{\:\:\:\mathrm{1}^{{odd}} }&\hline{\mathrm{04}}&\hline{\mathrm{12}}&\hline{\mathrm{20}}&\hline{…}&\hline{\mathrm{84}}&\hline{\mathrm{92}}&\hline{\mathrm{12}}&\hline{{odd}_{{Th}} }\\{\mathrm{3}}&\hline{\:\:\:\mathrm{2}^{{ev}} }&\hline{}&\hline{}&\hline{}&\hline{…}&\hline{}&\hline{}&\hline{\mathrm{13}}&\hline{{even}_{{Th}} }\\{\vdots}&\hline{\:\:\vdots}&\hline{}&\hline{}&\hline{}&\hline{}&\hline{}&\hline{}&\hline{}&\hline{}\\{\mathrm{9}}&\hline{\:\:\:\mathrm{9}^{{odd}} }&\hline{\mathrm{04}}&\hline{\mathrm{12}}&\hline{\mathrm{20}}&\hline{…}&\hline{\mathrm{84}}&\hline{\mathrm{92}}&\hline{\mathrm{12}}&\hline{{odd}_{{Th}} }\\\hline\end{array} \\ $$$$\begin{array}{|c|c|}{{The}\:\mathrm{2}{dig}_{{Th}} .\:{has}\:\mathrm{5}_{{even}} \&\:\mathrm{5}_{{odd}} }\\{\mathrm{9}_{{T}_{{th}} } ×\mathrm{5}_{{even}} ×\mathrm{13}\:+\:\mathrm{9}_{{T}_{{th}} } ×\mathrm{5}_{{odd}} ×\mathrm{12}}\\\hline\end{array} \\ $$$$ \\ $$$$\:{NumNumb}._{{divisible}\:{by}\:\:\mathrm{8}} \:=\:\mathrm{9}×\mathrm{5}\left(\mathrm{13}+\mathrm{12}\right)=\:\mathrm{1}\:\mathrm{125} \\ $$$$\:\boldsymbol{{Note}}:\:{This}\:{method}\:{is}\:{for}\:{who}\:{didn}'{t}\:{get}\:{the} \\ $$$$\:{two}\:{methods}\:{above}.\:{And}\:{it}'{s}\:{usefull}\:{for}\:{next}\:\left({B}\right) \\ $$$$ \\ $$$$\:{B}:\:{is}\:{next}\:{soon} \\ $$$$ \\ $$
Commented by Acem last updated on 16/Oct/22
Good Sir!  as you said it′s easy, but it might not   come to mind
$${Good}\:{Sir}!\:\:{as}\:{you}\:{said}\:{it}'{s}\:{easy},\:{but}\:{it}\:{might}\:{not} \\ $$$$\:{come}\:{to}\:{mind} \\ $$
Answered by Acem last updated on 19/Oct/22
   B:   NumNumb._(divis. by  8, diff.dig) =  N_(even_(Th) ) + N_(odd_(Th) ) ...(1)   N_(even_(Th) )  ; last two dig are: 00, 08, 16, 24, ..., 88, 96   determinant ((,(even)),(T_(Th) ,(Th)),(1^(⌣5) ,0^(⌣9) ),(2^(⌣4) ,2^(⌣8) ),(3,4),(4,6),(5,8^(⌣8) ),(6,),(7,),(8,),(9,))N_(two 1st dig) = 5_(odd) ×5^(Th) +4_(even) ×4^(Th) = 41   Tips:   0: must subtract 9   even: must subtract 11_(To figure how i said 11) ^(Try making a num. xy_08)    odd: must subtract 5   xx: is not valid, must subtract 41   08: 9+11⇒20, 32⇒ 16, 48⇒22...etc     determinant ((,(00),(08),(16),(24),(32),(40),(48),(56),(64),(72),(80),(88),(96),(n=13)),((Sum),(41),(20),(16),(22),(16),(20),(22),(16),(22),(16),(20),(41),(16),))   Sum= 41×2+ 20×3+ 16×5+ 22×3= 288  N_(even_(Th) ) = 41×13−288= 245 ...(2)     N_(odd_(Th) )  ; last two dig are: 04, 12, 20, 28, ..., 84, 92   determinant ((,(odd)),(T_(Th) ,(Th)),(1^(⌣4) ,1^(⌣8) ),(2^(⌣5) ,3^(⌣8) ),(3,5),(4,7),(5,9^(⌣8) ),(6,),(7,),(8,),(9,))N_(two 1st dig) = 5_(odd) ×4^(Th) +4_(even) ×5^(Th) = 40   Tips:   even: must subtract 5   odd: must subtract 11   xx: is not valid, must subtract 40   determinant ((,(04),(12),(20),(28),(36),(44),(52),(60),(68),(76),(84),(92),(n=12)),((Sum),5,(16),5,(10),(16),(40),(16),5,(10),(16),(10),(16),))  Sum= 5×3+ 16×5+ 10×3+ 40×1= 165  N_(odd_(Th) ) = 40×12−165= 315 ...(3)     By compensate (2) & (3) in (1)    NumNumb._(divis. by  8, diff.dig) = 245+ 315= 560
$$ \\ $$$$\:\boldsymbol{{B}}: \\ $$$$\:{NumNumb}._{{divis}.\:{by}\:\:\mathrm{8},\:{diff}.{dig}} =\:\:{N}_{{even}_{{Th}} } +\:{N}_{{odd}_{{Th}} } …\left(\mathrm{1}\right) \\ $$$$\:{N}_{{even}_{{Th}} } \:;\:{last}\:{two}\:{dig}\:{are}:\:\mathrm{00},\:\mathrm{08},\:\mathrm{16},\:\mathrm{24},\:…,\:\mathrm{88},\:\mathrm{96} \\ $$$$\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|}{}&\hline{\boldsymbol{{even}}}\\{{T}_{{Th}} }&\hline{{Th}}\\{\mathrm{1}^{\smile\mathrm{5}} }&\hline{\mathrm{0}^{\smile\mathrm{9}} }\\{\mathrm{2}^{\smile\mathrm{4}} }&\hline{\mathrm{2}^{\smile\mathrm{8}} }\\{\mathrm{3}}&\hline{\mathrm{4}}\\{\mathrm{4}}&\hline{\mathrm{6}}\\{\mathrm{5}}&\hline{\mathrm{8}^{\smile\mathrm{8}} }\\{\mathrm{6}}&\hline{}\\{\mathrm{7}}&\hline{}\\{\mathrm{8}}&\hline{}\\{\mathrm{9}}&\hline{}\\\hline\end{array}{N}_{{two}\:\mathrm{1}{st}\:{dig}} =\:\mathrm{5}_{{odd}} ×\mathrm{5}^{{Th}} +\mathrm{4}_{{even}} ×\mathrm{4}^{{Th}} =\:\mathrm{41} \\ $$$$\:\boldsymbol{{Tips}}: \\ $$$$\:\mathrm{0}:\:{must}\:{subtract}\:\mathrm{9} \\ $$$$\:{even}:\:{must}\:{subtract}\:\mathrm{11}_{{To}\:{figure}\:{how}\:{i}\:{said}\:\mathrm{11}} ^{{Try}\:{making}\:{a}\:{num}.\:{xy\_}\mathrm{08}} \\ $$$$\:{odd}:\:{must}\:{subtract}\:\mathrm{5} \\ $$$$\:{xx}:\:{is}\:{not}\:{valid},\:{must}\:{subtract}\:\mathrm{41} \\ $$$$\:\mathrm{08}:\:\mathrm{9}+\mathrm{11}\Rightarrow\mathrm{20},\:\mathrm{32}\Rightarrow\:\mathrm{16},\:\mathrm{48}\Rightarrow\mathrm{22}…{etc} \\ $$$$ \\ $$$$\begin{array}{|c|c|}{}&\hline{\mathrm{00}}&\hline{\mathrm{08}}&\hline{\mathrm{16}}&\hline{\mathrm{24}}&\hline{\mathrm{32}}&\hline{\mathrm{40}}&\hline{\mathrm{48}}&\hline{\mathrm{56}}&\hline{\mathrm{64}}&\hline{\mathrm{72}}&\hline{\mathrm{80}}&\hline{\mathrm{88}}&\hline{\mathrm{96}}&\hline{{n}=\mathrm{13}}\\{{Sum}}&\hline{\mathrm{41}}&\hline{\mathrm{20}}&\hline{\mathrm{16}}&\hline{\mathrm{22}}&\hline{\mathrm{16}}&\hline{\mathrm{20}}&\hline{\mathrm{22}}&\hline{\mathrm{16}}&\hline{\mathrm{22}}&\hline{\mathrm{16}}&\hline{\mathrm{20}}&\hline{\mathrm{41}}&\hline{\mathrm{16}}&\hline{}\\\hline\end{array} \\ $$$$\:{Sum}=\:\mathrm{41}×\mathrm{2}+\:\mathrm{20}×\mathrm{3}+\:\mathrm{16}×\mathrm{5}+\:\mathrm{22}×\mathrm{3}=\:\mathrm{288} \\ $$$${N}_{{even}_{{Th}} } =\:\mathrm{41}×\mathrm{13}−\mathrm{288}=\:\mathrm{245}\:…\left(\mathrm{2}\right) \\ $$$$ \\ $$$$\:{N}_{{odd}_{{Th}} } \:;\:{last}\:{two}\:{dig}\:{are}:\:\mathrm{04},\:\mathrm{12},\:\mathrm{20},\:\mathrm{28},\:…,\:\mathrm{84},\:\mathrm{92} \\ $$$$\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|}{}&\hline{\boldsymbol{{odd}}}\\{{T}_{{Th}} }&\hline{{Th}}\\{\mathrm{1}^{\smile\mathrm{4}} }&\hline{\mathrm{1}^{\smile\mathrm{8}} }\\{\mathrm{2}^{\smile\mathrm{5}} }&\hline{\mathrm{3}^{\smile\mathrm{8}} }\\{\mathrm{3}}&\hline{\mathrm{5}}\\{\mathrm{4}}&\hline{\mathrm{7}}\\{\mathrm{5}}&\hline{\mathrm{9}^{\smile\mathrm{8}} }\\{\mathrm{6}}&\hline{}\\{\mathrm{7}}&\hline{}\\{\mathrm{8}}&\hline{}\\{\mathrm{9}}&\hline{}\\\hline\end{array}{N}_{{two}\:\mathrm{1}{st}\:{dig}} =\:\mathrm{5}_{{odd}} ×\mathrm{4}^{{Th}} +\mathrm{4}_{{even}} ×\mathrm{5}^{{Th}} =\:\mathrm{40} \\ $$$$\:\boldsymbol{{Tips}}: \\ $$$$\:{even}:\:{must}\:{subtract}\:\mathrm{5} \\ $$$$\:{odd}:\:{must}\:{subtract}\:\mathrm{11} \\ $$$$\:{xx}:\:{is}\:{not}\:{valid},\:{must}\:{subtract}\:\mathrm{40} \\ $$$$\begin{array}{|c|c|}{}&\hline{\mathrm{04}}&\hline{\mathrm{12}}&\hline{\mathrm{20}}&\hline{\mathrm{28}}&\hline{\mathrm{36}}&\hline{\mathrm{44}}&\hline{\mathrm{52}}&\hline{\mathrm{60}}&\hline{\mathrm{68}}&\hline{\mathrm{76}}&\hline{\mathrm{84}}&\hline{\mathrm{92}}&\hline{{n}=\mathrm{12}}\\{{Sum}}&\hline{\mathrm{5}}&\hline{\mathrm{16}}&\hline{\mathrm{5}}&\hline{\mathrm{10}}&\hline{\mathrm{16}}&\hline{\mathrm{40}}&\hline{\mathrm{16}}&\hline{\mathrm{5}}&\hline{\mathrm{10}}&\hline{\mathrm{16}}&\hline{\mathrm{10}}&\hline{\mathrm{16}}&\hline{}\\\hline\end{array} \\ $$$${Sum}=\:\mathrm{5}×\mathrm{3}+\:\mathrm{16}×\mathrm{5}+\:\mathrm{10}×\mathrm{3}+\:\mathrm{40}×\mathrm{1}=\:\mathrm{165} \\ $$$${N}_{{odd}_{{Th}} } =\:\mathrm{40}×\mathrm{12}−\mathrm{165}=\:\mathrm{315}\:…\left(\mathrm{3}\right) \\ $$$$ \\ $$$$\:{By}\:{compensate}\:\left(\mathrm{2}\right)\:\&\:\left(\mathrm{3}\right)\:{in}\:\left(\mathrm{1}\right) \\ $$$$ \\ $$$${NumNumb}._{{divis}.\:{by}\:\:\mathrm{8},\:{diff}.{dig}} =\:\mathrm{245}+\:\mathrm{315}=\:\mathrm{560} \\ $$$$ \\ $$

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