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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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