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Question-51287

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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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