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Question-199270




Question Number 199270 by mr W last updated on 31/Oct/23
Commented by mr W last updated on 31/Oct/23
an unsolved old question #199155.
$${an}\:{unsolved}\:{old}\:{question}\:#\mathrm{199155}. \\ $$
Commented by mr W last updated on 31/Oct/23
i got 1111.1111.  can someone comfirm?  nikif99 sir: can you use your fortran  program to check it?
$${i}\:{got}\:\mathrm{1111}.\mathrm{1111}. \\ $$$${can}\:{someone}\:{comfirm}? \\ $$$${nikif}\mathrm{99}\:{sir}:\:{can}\:{you}\:{use}\:{your}\:{fortran} \\ $$$${program}\:{to}\:{check}\:{it}? \\ $$
Commented by nikif99 last updated on 31/Oct/23
I get mean=1234.43210678.  ending digits in blue might be “noise”  of the processor or indeed decimal  places when total divided by (5!×4)=480.  I will check it more.
$${I}\:{get}\:{mean}=\mathrm{1234}.\mathrm{43210678}. \\ $$$${ending}\:{digits}\:{in}\:{blue}\:{might}\:{be}\:“{noise}'' \\ $$$${of}\:{the}\:{processor}\:{or}\:{indeed}\:{decimal} \\ $$$${places}\:{when}\:{total}\:{divided}\:{by}\:\left(\mathrm{5}!×\mathrm{4}\right)=\mathrm{480}. \\ $$$${I}\:{will}\:{check}\:{it}\:{more}. \\ $$
Commented by mr W last updated on 31/Oct/23
thanks sirs!
$${thanks}\:{sirs}! \\ $$
Answered by aleks041103 last updated on 31/Oct/23
let the needed average be A  let the average of all the numbers made  without the decimal point be B.  then:  A=(1/4)(0.1B+0.01B+0.001B+0.0001B)=  =((0.1111)/4)B  for B:  B=(1/(5!))Σ_(σ∈S_5 ) (σ(1)+10σ(2)+10^2 σ(4)+10^3 σ(5)+10^4 σ(8))=  =((4!)/(5!))(1+10+10^2 +10^3 +10^4 )(1+2+4+5+8)=  =((11111)/5)20=44444  ⇒A=((0.1111)/4)44444  ⇒A=1234.4321
$${let}\:{the}\:{needed}\:{average}\:{be}\:{A} \\ $$$${let}\:{the}\:{average}\:{of}\:{all}\:{the}\:{numbers}\:{made} \\ $$$${without}\:{the}\:{decimal}\:{point}\:{be}\:{B}. \\ $$$${then}: \\ $$$${A}=\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{0}.\mathrm{1}{B}+\mathrm{0}.\mathrm{01}{B}+\mathrm{0}.\mathrm{001}{B}+\mathrm{0}.\mathrm{0001}{B}\right)= \\ $$$$=\frac{\mathrm{0}.\mathrm{1111}}{\mathrm{4}}{B} \\ $$$${for}\:{B}: \\ $$$${B}=\frac{\mathrm{1}}{\mathrm{5}!}\underset{\sigma\in{S}_{\mathrm{5}} } {\sum}\left(\sigma\left(\mathrm{1}\right)+\mathrm{10}\sigma\left(\mathrm{2}\right)+\mathrm{10}^{\mathrm{2}} \sigma\left(\mathrm{4}\right)+\mathrm{10}^{\mathrm{3}} \sigma\left(\mathrm{5}\right)+\mathrm{10}^{\mathrm{4}} \sigma\left(\mathrm{8}\right)\right)= \\ $$$$=\frac{\mathrm{4}!}{\mathrm{5}!}\left(\mathrm{1}+\mathrm{10}+\mathrm{10}^{\mathrm{2}} +\mathrm{10}^{\mathrm{3}} +\mathrm{10}^{\mathrm{4}} \right)\left(\mathrm{1}+\mathrm{2}+\mathrm{4}+\mathrm{5}+\mathrm{8}\right)= \\ $$$$=\frac{\mathrm{11111}}{\mathrm{5}}\mathrm{20}=\mathrm{44444} \\ $$$$\Rightarrow{A}=\frac{\mathrm{0}.\mathrm{1111}}{\mathrm{4}}\mathrm{44444} \\ $$$$\Rightarrow{A}=\mathrm{1234}.\mathrm{4321} \\ $$
Commented by aleks041103 last updated on 31/Oct/23
Commented by MathematicalUser2357 last updated on 04/Jan/24
CODING TIME!
$$\mathrm{CODING}\:\mathrm{TIME}! \\ $$
Answered by mr W last updated on 31/Oct/23
this is my solution:  generallly,  the sum of some numbers  is the sum of the values which their   digits represent. for example:   123+132+213+231+312+321=1332  but we can also get the sum like this:  the digit “1” can stand on the first  position which represents 100. there  are 2! such possibilities. that means  the sum of digit “1” on the first   position is 1×100×2!. the digit “1”  can also stand on the second and  the third position, so the sum which  the digit “1” represents is  1×(100+10+1)×2!=1×111×2!  and the sum which all didits present  is thus  (1+2+3)×111×2!=6×111×2=1332.    now back to our question.  with the digits “1, 2, 4, 5, 8” and the   decimal point “.” we can form totally  5!×4=480  numbers. each digit “1”   can represent 1000, 100, 10, 1, 0.1,   0.01, 0.001, 0.0001 respectively. there   are 4! possibilities for the digit “1” on  1000s  position, 2×4! possibilities on  100s position, 3×4! on 10s position,  4×4! on 1s position, 4×4! on 0.1s  position, 3×4! on 0.01s position,  2×4! on 0.001position and 1×4!  on 0.0001 position. so the sum  which all digits represent is  (1+2+4+5+8)×(1×1000.0001+2×100.001+3×10.01+4×1.1)×4!  so the average of the numbers is  (((1+2+4+5+8)×(1×1000.0001+2×100.001+3×10.01+4×1.1)×4!)/(5!×4))  =((20×(1234.4321))/(5×4))  =1234.4321 ✓
$${this}\:{is}\:{my}\:{solution}: \\ $$$${generallly},\:\:{the}\:{sum}\:{of}\:{some}\:{numbers} \\ $$$${is}\:{the}\:{sum}\:{of}\:{the}\:{values}\:{which}\:{their}\: \\ $$$${digits}\:{represent}.\:{for}\:{example}:\: \\ $$$$\mathrm{123}+\mathrm{132}+\mathrm{213}+\mathrm{231}+\mathrm{312}+\mathrm{321}=\mathrm{1332} \\ $$$${but}\:{we}\:{can}\:{also}\:{get}\:{the}\:{sum}\:{like}\:{this}: \\ $$$${the}\:{digit}\:“\mathrm{1}''\:{can}\:{stand}\:{on}\:{the}\:{first} \\ $$$${position}\:{which}\:{represents}\:\mathrm{100}.\:{there} \\ $$$${are}\:\mathrm{2}!\:{such}\:{possibilities}.\:{that}\:{means} \\ $$$${the}\:{sum}\:{of}\:{digit}\:“\mathrm{1}''\:{on}\:{the}\:{first}\: \\ $$$${position}\:{is}\:\mathrm{1}×\mathrm{100}×\mathrm{2}!.\:{the}\:{digit}\:“\mathrm{1}'' \\ $$$${can}\:{also}\:{stand}\:{on}\:{the}\:{second}\:{and} \\ $$$${the}\:{third}\:{position},\:{so}\:{the}\:{sum}\:{which} \\ $$$${the}\:{digit}\:“\mathrm{1}''\:{represents}\:{is} \\ $$$$\mathrm{1}×\left(\mathrm{100}+\mathrm{10}+\mathrm{1}\right)×\mathrm{2}!=\mathrm{1}×\mathrm{111}×\mathrm{2}! \\ $$$${and}\:{the}\:{sum}\:{which}\:{all}\:{didits}\:{present} \\ $$$${is}\:{thus} \\ $$$$\left(\mathrm{1}+\mathrm{2}+\mathrm{3}\right)×\mathrm{111}×\mathrm{2}!=\mathrm{6}×\mathrm{111}×\mathrm{2}=\mathrm{1332}. \\ $$$$ \\ $$$${now}\:{back}\:{to}\:{our}\:{question}. \\ $$$${with}\:{the}\:{digits}\:“\mathrm{1},\:\mathrm{2},\:\mathrm{4},\:\mathrm{5},\:\mathrm{8}''\:{and}\:{the}\: \\ $$$${decimal}\:{point}\:“.''\:{we}\:{can}\:{form}\:{totally} \\ $$$$\mathrm{5}!×\mathrm{4}=\mathrm{480}\:\:{numbers}.\:{each}\:{digit}\:“\mathrm{1}''\: \\ $$$${can}\:{represent}\:\mathrm{1000},\:\mathrm{100},\:\mathrm{10},\:\mathrm{1},\:\mathrm{0}.\mathrm{1},\: \\ $$$$\mathrm{0}.\mathrm{01},\:\mathrm{0}.\mathrm{001},\:\mathrm{0}.\mathrm{0001}\:{respectively}.\:{there}\: \\ $$$${are}\:\mathrm{4}!\:{possibilities}\:{for}\:{the}\:{digit}\:“\mathrm{1}''\:{on} \\ $$$$\mathrm{1000}{s}\:\:{position},\:\mathrm{2}×\mathrm{4}!\:{possibilities}\:{on} \\ $$$$\mathrm{100}{s}\:{position},\:\mathrm{3}×\mathrm{4}!\:{on}\:\mathrm{10}{s}\:{position}, \\ $$$$\mathrm{4}×\mathrm{4}!\:{on}\:\mathrm{1}{s}\:{position},\:\mathrm{4}×\mathrm{4}!\:{on}\:\mathrm{0}.\mathrm{1}{s} \\ $$$${position},\:\mathrm{3}×\mathrm{4}!\:{on}\:\mathrm{0}.\mathrm{01}{s}\:{position}, \\ $$$$\mathrm{2}×\mathrm{4}!\:{on}\:\mathrm{0}.\mathrm{001}{position}\:{and}\:\mathrm{1}×\mathrm{4}! \\ $$$${on}\:\mathrm{0}.\mathrm{0001}\:{position}.\:{so}\:{the}\:{sum} \\ $$$${which}\:{all}\:{digits}\:{represent}\:{is} \\ $$$$\left(\mathrm{1}+\mathrm{2}+\mathrm{4}+\mathrm{5}+\mathrm{8}\right)×\left(\mathrm{1}×\mathrm{1000}.\mathrm{0001}+\mathrm{2}×\mathrm{100}.\mathrm{001}+\mathrm{3}×\mathrm{10}.\mathrm{01}+\mathrm{4}×\mathrm{1}.\mathrm{1}\right)×\mathrm{4}! \\ $$$${so}\:{the}\:{average}\:{of}\:{the}\:{numbers}\:{is} \\ $$$$\frac{\left(\mathrm{1}+\mathrm{2}+\mathrm{4}+\mathrm{5}+\mathrm{8}\right)×\left(\mathrm{1}×\mathrm{1000}.\mathrm{0001}+\mathrm{2}×\mathrm{100}.\mathrm{001}+\mathrm{3}×\mathrm{10}.\mathrm{01}+\mathrm{4}×\mathrm{1}.\mathrm{1}\right)×\mathrm{4}!}{\mathrm{5}!×\mathrm{4}} \\ $$$$=\frac{\mathrm{20}×\left(\mathrm{1234}.\mathrm{4321}\right)}{\mathrm{5}×\mathrm{4}} \\ $$$$=\mathrm{1234}.\mathrm{4321}\:\checkmark \\ $$
Commented by aleks041103 last updated on 31/Oct/23
The poblem in your solution comes from the fact  that yoy assume the digit can always represent  0.1, which is incorrect.  For example, if 1 is the first digit, it cannot  have a factor 0.1, because then the  number would be 0.1..., which is not  allowed in the statement.
$${The}\:{poblem}\:{in}\:{your}\:{solution}\:{comes}\:{from}\:{the}\:{fact} \\ $$$${that}\:{yoy}\:{assume}\:{the}\:{digit}\:{can}\:{always}\:{represent} \\ $$$$\mathrm{0}.\mathrm{1},\:{which}\:{is}\:{incorrect}. \\ $$$${For}\:{example},\:{if}\:\mathrm{1}\:{is}\:{the}\:{first}\:{digit},\:{it}\:{cannot} \\ $$$${have}\:{a}\:{factor}\:\mathrm{0}.\mathrm{1},\:{because}\:{then}\:{the} \\ $$$${number}\:{would}\:{be}\:\mathrm{0}.\mathrm{1}…,\:{which}\:{is}\:{not} \\ $$$${allowed}\:{in}\:{the}\:{statement}. \\ $$
Commented by mr W last updated on 31/Oct/23
i didn′t assume that. it just meant  that “1” can stand  for 0.1, for  example in the numbers like  2458.1, 4528.1, etc.
$${i}\:{didn}'{t}\:{assume}\:{that}.\:{it}\:{just}\:{meant} \\ $$$${that}\:“\mathrm{1}''\:{can}\:{stand}\:\:{for}\:\mathrm{0}.\mathrm{1},\:{for} \\ $$$${example}\:{in}\:{the}\:{numbers}\:{like} \\ $$$$\mathrm{2458}.\mathrm{1},\:\mathrm{4528}.\mathrm{1},\:{etc}. \\ $$
Commented by aleks041103 last updated on 31/Oct/23
yes. you are right.  actually there is another problem.  you assume there are 4! ways to have ′1′  at the 10s position, for example.  this is true in the case of integers,   but the addition of the decimal dot   changes things.  example  1 in 10s position  1) ∗∗∗1∗.  2) ∗∗1∗.∗  3) ∗1∗.∗∗  4) 1∗.∗∗∗  So actually for the 10s, there are not 4!  ways, but actually 4×4! ways.
$${yes}.\:{you}\:{are}\:{right}. \\ $$$${actually}\:{there}\:{is}\:{another}\:{problem}. \\ $$$${you}\:{assume}\:{there}\:{are}\:\mathrm{4}!\:{ways}\:{to}\:{have}\:'\mathrm{1}' \\ $$$${at}\:{the}\:\mathrm{10}{s}\:{position},\:{for}\:{example}. \\ $$$${this}\:{is}\:{true}\:{in}\:{the}\:{case}\:{of}\:{integers},\: \\ $$$${but}\:{the}\:{addition}\:{of}\:{the}\:{decimal}\:{dot}\: \\ $$$${changes}\:{things}. \\ $$$${example} \\ $$$$\mathrm{1}\:{in}\:\mathrm{10}{s}\:{position} \\ $$$$\left.\mathrm{1}\right)\:\ast\ast\ast\mathrm{1}\ast. \\ $$$$\left.\mathrm{2}\right)\:\ast\ast\mathrm{1}\ast.\ast \\ $$$$\left.\mathrm{3}\right)\:\ast\mathrm{1}\ast.\ast\ast \\ $$$$\left.\mathrm{4}\right)\:\mathrm{1}\ast.\ast\ast\ast \\ $$$${So}\:{actually}\:{for}\:{the}\:\mathrm{10}{s},\:{there}\:{are}\:{not}\:\mathrm{4}! \\ $$$${ways},\:{but}\:{actually}\:\mathrm{4}×\mathrm{4}!\:{ways}. \\ $$
Commented by mr W last updated on 31/Oct/23
 i got it! thank you sir! i have fixed  my solution and got also 1234.4321  now.
$$\:{i}\:{got}\:{it}!\:{thank}\:{you}\:{sir}!\:{i}\:{have}\:{fixed} \\ $$$${my}\:{solution}\:{and}\:{got}\:{also}\:\mathrm{1234}.\mathrm{4321} \\ $$$${now}. \\ $$

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