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Question-51287




Question Number 51287 by aseerimad last updated on 25/Dec/18
Answered by mr W last updated on 26/Dec/18
let′s look at an example:  the number 1234 consists of 4 digits:  1 represents here the value 1×1000  2 represents here the value 2×100  3 represents here the value 3×10  4 represents here the value 4×1  ⇒1234=1×1000+2×100+3×10+4×1  so when we want to get the sum of some  numbers, we only need to calculate  the sum of the values which their  digits represent. for example:  1234+4321  =(1+4)×1000+(2+3)×100+(3+2)×10+(4+1)×1  =5×(1000+100+10+1)  =5×1111  =5555  we can use this method to solve our  problem. its advantage is that we  don′t need to know the numbers   themself. we only need to know the  digits from which the numbers are  formed. and we know the digits, they  are 1,2,3 and 4.    let′s look at the digit 1:   it can be at the thousands place as  1XYZ. we know there are 3! numbers  in which the digit 1 is at the thousands  place. the sum of all values which the  digit 1 represent in the thousands place  is then 1×1000×3!.  similarly the digit 1 can be in the  hundreds place. there are also 3! such  numbers. the sum of all values which the  digit 1 represent in the hundreds place  is 1×100×3!.  similarly the digit 1 can be in the  tens place. there are also 3! such  numbers. the sum of all values which the  digit 1 represent in the tens place  is 1×10×3!.  finally the digit 1 can be in the  units place. there are also 3! such  numbers. the sum of all values which the  digit 1 represent in the units place  is 1×1×3!.  in this way we get the sum of all  values which the digit 1 represent in  all possible places is:  1×(1000+100+10+1)×3!  this applies also for the digits 2,3,4.    the sum of all values which all of  these 4 digits in all possible places  represent is then  (1+2+3+4)×(1000+100+10+1)×3!  this is also the sum of all numbers  formed by these 4 digits.    so the answer for the question is  (1+2+3+4)×(1000+100+10+1)×3!  =10×(1000+100+10+1)×3!  =10×1111×3!  =10×6666  =66660
$${let}'{s}\:{look}\:{at}\:{an}\:{example}: \\ $$$${the}\:{number}\:\mathrm{1234}\:{consists}\:{of}\:\mathrm{4}\:{digits}: \\ $$$$\mathrm{1}\:{represents}\:{here}\:{the}\:{value}\:\mathrm{1}×\mathrm{1000} \\ $$$$\mathrm{2}\:{represents}\:{here}\:{the}\:{value}\:\mathrm{2}×\mathrm{100} \\ $$$$\mathrm{3}\:{represents}\:{here}\:{the}\:{value}\:\mathrm{3}×\mathrm{10} \\ $$$$\mathrm{4}\:{represents}\:{here}\:{the}\:{value}\:\mathrm{4}×\mathrm{1} \\ $$$$\Rightarrow\mathrm{1234}=\mathrm{1}×\mathrm{1000}+\mathrm{2}×\mathrm{100}+\mathrm{3}×\mathrm{10}+\mathrm{4}×\mathrm{1} \\ $$$${so}\:{when}\:{we}\:{want}\:{to}\:{get}\:{the}\:{sum}\:{of}\:{some} \\ $$$${numbers},\:{we}\:{only}\:{need}\:{to}\:{calculate} \\ $$$${the}\:{sum}\:{of}\:{the}\:{values}\:{which}\:{their} \\ $$$${digits}\:{represent}.\:{for}\:{example}: \\ $$$$\mathrm{1234}+\mathrm{4321} \\ $$$$=\left(\mathrm{1}+\mathrm{4}\right)×\mathrm{1000}+\left(\mathrm{2}+\mathrm{3}\right)×\mathrm{100}+\left(\mathrm{3}+\mathrm{2}\right)×\mathrm{10}+\left(\mathrm{4}+\mathrm{1}\right)×\mathrm{1} \\ $$$$=\mathrm{5}×\left(\mathrm{1000}+\mathrm{100}+\mathrm{10}+\mathrm{1}\right) \\ $$$$=\mathrm{5}×\mathrm{1111} \\ $$$$=\mathrm{5555} \\ $$$${we}\:{can}\:{use}\:{this}\:{method}\:{to}\:{solve}\:{our} \\ $$$${problem}.\:{its}\:{advantage}\:{is}\:{that}\:{we} \\ $$$${don}'{t}\:{need}\:{to}\:{know}\:{the}\:{numbers}\: \\ $$$${themself}.\:{we}\:{only}\:{need}\:{to}\:{know}\:{the} \\ $$$${digits}\:{from}\:{which}\:{the}\:{numbers}\:{are} \\ $$$${formed}.\:{and}\:{we}\:{know}\:{the}\:{digits},\:{they} \\ $$$${are}\:\mathrm{1},\mathrm{2},\mathrm{3}\:{and}\:\mathrm{4}. \\ $$$$ \\ $$$${let}'{s}\:{look}\:{at}\:{the}\:{digit}\:\mathrm{1}:\: \\ $$$${it}\:{can}\:{be}\:{at}\:{the}\:{thousands}\:{place}\:{as} \\ $$$$\mathrm{1}{XYZ}.\:{we}\:{know}\:{there}\:{are}\:\mathrm{3}!\:{numbers} \\ $$$${in}\:{which}\:{the}\:{digit}\:\mathrm{1}\:{is}\:{at}\:{the}\:{thousands} \\ $$$${place}.\:{the}\:{sum}\:{of}\:{all}\:{values}\:{which}\:{the} \\ $$$${digit}\:\mathrm{1}\:{represent}\:{in}\:{the}\:{thousands}\:{place} \\ $$$${is}\:{then}\:\mathrm{1}×\mathrm{1000}×\mathrm{3}!. \\ $$$${similarly}\:{the}\:{digit}\:\mathrm{1}\:{can}\:{be}\:{in}\:{the} \\ $$$${hundreds}\:{place}.\:{there}\:{are}\:{also}\:\mathrm{3}!\:{such} \\ $$$${numbers}.\:{the}\:{sum}\:{of}\:{all}\:{values}\:{which}\:{the} \\ $$$${digit}\:\mathrm{1}\:{represent}\:{in}\:{the}\:{hundreds}\:{place} \\ $$$${is}\:\mathrm{1}×\mathrm{100}×\mathrm{3}!. \\ $$$${similarly}\:{the}\:{digit}\:\mathrm{1}\:{can}\:{be}\:{in}\:{the} \\ $$$${tens}\:{place}.\:{there}\:{are}\:{also}\:\mathrm{3}!\:{such} \\ $$$${numbers}.\:{the}\:{sum}\:{of}\:{all}\:{values}\:{which}\:{the} \\ $$$${digit}\:\mathrm{1}\:{represent}\:{in}\:{the}\:{tens}\:{place} \\ $$$${is}\:\mathrm{1}×\mathrm{10}×\mathrm{3}!. \\ $$$${finally}\:{the}\:{digit}\:\mathrm{1}\:{can}\:{be}\:{in}\:{the} \\ $$$${units}\:{place}.\:{there}\:{are}\:{also}\:\mathrm{3}!\:{such} \\ $$$${numbers}.\:{the}\:{sum}\:{of}\:{all}\:{values}\:{which}\:{the} \\ $$$${digit}\:\mathrm{1}\:{represent}\:{in}\:{the}\:{units}\:{place} \\ $$$${is}\:\mathrm{1}×\mathrm{1}×\mathrm{3}!. \\ $$$${in}\:{this}\:{way}\:{we}\:{get}\:{the}\:{sum}\:{of}\:{all} \\ $$$${values}\:{which}\:{the}\:{digit}\:\mathrm{1}\:{represent}\:{in} \\ $$$${all}\:{possible}\:{places}\:{is}: \\ $$$$\mathrm{1}×\left(\mathrm{1000}+\mathrm{100}+\mathrm{10}+\mathrm{1}\right)×\mathrm{3}! \\ $$$${this}\:{applies}\:{also}\:{for}\:{the}\:{digits}\:\mathrm{2},\mathrm{3},\mathrm{4}. \\ $$$$ \\ $$$${the}\:{sum}\:{of}\:{all}\:{values}\:{which}\:{all}\:{of} \\ $$$${these}\:\mathrm{4}\:{digits}\:{in}\:{all}\:{possible}\:{places} \\ $$$${represent}\:{is}\:{then} \\ $$$$\left(\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}\right)×\left(\mathrm{1000}+\mathrm{100}+\mathrm{10}+\mathrm{1}\right)×\mathrm{3}! \\ $$$${this}\:{is}\:{also}\:{the}\:{sum}\:{of}\:{all}\:{numbers} \\ $$$${formed}\:{by}\:{these}\:\mathrm{4}\:{digits}. \\ $$$$ \\ $$$${so}\:{the}\:{answer}\:{for}\:{the}\:{question}\:{is} \\ $$$$\left(\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}\right)×\left(\mathrm{1000}+\mathrm{100}+\mathrm{10}+\mathrm{1}\right)×\mathrm{3}! \\ $$$$=\mathrm{10}×\left(\mathrm{1000}+\mathrm{100}+\mathrm{10}+\mathrm{1}\right)×\mathrm{3}! \\ $$$$=\mathrm{10}×\mathrm{1111}×\mathrm{3}! \\ $$$$=\mathrm{10}×\mathrm{6666} \\ $$$$=\mathrm{66660} \\ $$
Commented by aseerimad last updated on 26/Dec/18
can you pls explain the method?
$${can}\:{you}\:{pls}\:{explain}\:{the}\:{method}? \\ $$
Commented by mr W last updated on 26/Dec/18
i have added some explanation into  the workings. hope it′s clear now.
$${i}\:{have}\:{added}\:{some}\:{explanation}\:{into} \\ $$$${the}\:{workings}.\:{hope}\:{it}'{s}\:{clear}\:{now}. \\ $$
Commented by aseerimad last updated on 26/Dec/18
Thank you Thank you very much.Thank you for explaining in such a lucid way.  God Bless you.
$${Thank}\:{you}\:{Thank}\:{you}\:{very}\:{much}.{Thank}\:{you}\:{for}\:{explaining}\:{in}\:{such}\:{a}\:{lucid}\:{way}. \\ $$$${God}\:{Bless}\:{you}. \\ $$

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