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Question Number 51465 by mr W last updated on 27/Dec/18

Commented by Rasheed.Sindhi last updated on 27/Dec/18

Sir 4-digit numbers I think?

$${Sir}\:\mathrm{4}-{digit}\:{numbers}\:{I}\:{think}? \\ $$

Commented by mr W last updated on 27/Dec/18

Yes, 4−digits numbers only.

$${Yes},\:\mathrm{4}−{digits}\:{numbers}\:{only}. \\ $$

Commented by Rasheed.Sindhi last updated on 27/Dec/18

711040?

$$\mathrm{711040}? \\ $$

Commented by mr W last updated on 27/Dec/18

correct! thank you sir!

$${correct}!\:{thank}\:{you}\:{sir}! \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 27/Dec/18

total four digit numbers formed=4×4×4×4=256  during addition  we have to add each digit 64 times.  1)so unit place of the sum of all four digit is  [64×(1+2+3+4)=640]→0  2)when we add all digit of second place +carry  forward 64 [640+64=704] so tenth digit of sum→4  3)hundred place [640+70=710]→0  4)thousand place [640+71=711]→1  so answer is 711040

$${total}\:{four}\:{digit}\:{numbers}\:{formed}=\mathrm{4}×\mathrm{4}×\mathrm{4}×\mathrm{4}=\mathrm{256} \\ $$$${during}\:{addition} \\ $$$${we}\:{have}\:{to}\:{add}\:{each}\:{digit}\:\mathrm{64}\:{times}. \\ $$$$\left.\mathrm{1}\right){so}\:{unit}\:{place}\:{of}\:{the}\:{sum}\:{of}\:{all}\:{four}\:{digit}\:{is} \\ $$$$\left[\mathrm{64}×\left(\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}\right)=\mathrm{640}\right]\rightarrow\mathrm{0} \\ $$$$\left.\mathrm{2}\right){when}\:{we}\:{add}\:{all}\:{digit}\:{of}\:{second}\:{place}\:+{carry} \\ $$$${forward}\:\mathrm{64}\:\left[\mathrm{640}+\mathrm{64}=\mathrm{704}\right]\:{so}\:{tenth}\:{digit}\:{of}\:{sum}\rightarrow\mathrm{4} \\ $$$$\left.\mathrm{3}\right){hundred}\:{place}\:\left[\mathrm{640}+\mathrm{70}=\mathrm{710}\right]\rightarrow\mathrm{0} \\ $$$$\left.\mathrm{4}\right){thousand}\:{place}\:\left[\mathrm{640}+\mathrm{71}=\mathrm{711}\right]\rightarrow\mathrm{1} \\ $$$${so}\:{answer}\:{is}\:\mathrm{711040} \\ $$$$ \\ $$

Commented by mr W last updated on 27/Dec/18

answer is correct sir∙ thank you!

$${answer}\:{is}\:{correct}\:{sir}\centerdot\:{thank}\:{you}! \\ $$

Answered by mr W last updated on 27/Dec/18

sum of all possible numbers is equal  to the sum of all values which their  digits represent:  (4^4 /4)×(1000+100+10+1)×(1+2+3+4)  =(4^4 /4)×1111×10=711040

$${sum}\:{of}\:{all}\:{possible}\:{numbers}\:{is}\:{equal} \\ $$$${to}\:{the}\:{sum}\:{of}\:{all}\:{values}\:{which}\:{their} \\ $$$${digits}\:{represent}: \\ $$$$\frac{\mathrm{4}^{\mathrm{4}} }{\mathrm{4}}×\left(\mathrm{1000}+\mathrm{100}+\mathrm{10}+\mathrm{1}\right)×\left(\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}\right) \\ $$$$=\frac{\mathrm{4}^{\mathrm{4}} }{\mathrm{4}}×\mathrm{1111}×\mathrm{10}=\mathrm{711040} \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 27/Dec/18

unique method...excellent...

$${unique}\:{method}...{excellent}... \\ $$

Commented by ajfour last updated on 27/Dec/18

standard old method..

$${standard}\:{old}\:{method}.. \\ $$

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