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Mathematics 18 Online
OpenStudy (anonymous):

if x represents the sum of all the positive 3-digit nos that can be constructed using each of the distinct non zero digits a,b,c exactly once, wat is the largest integer by wic x must be divisible?

OpenStudy (anonymous):

no its 222

OpenStudy (anonymous):

see it is clear for the sum of all 3-digited number is always divisible by 111

OpenStudy (anonymous):

total no. of 3 digited nos as per ur question is 504 do u need the highest integer which divides the sum of all 504 numbers? or something else..

OpenStudy (anonymous):

if i am not wrong the sum of all 3 digited numbers is 279720 with each of the distinct non zero digits a,b,c exactly once it is divisible by integers greater than 222

OpenStudy (anonymous):

?? r u there

OpenStudy (anonymous):

how did u gt tis?? .. bit confused

OpenStudy (anonymous):

see in any order the each digit 1 through 9 will appear as many as(8X7) times in the unit , tenth and hundredth place

OpenStudy (anonymous):

now the sum of 1 through 9=45 hence required sum of all reqd + 3digited number is=45*56*(100+10+1)=279720

OpenStudy (anonymous):

i hope it is clear to u or point where i am making mistake

OpenStudy (anonymous):

see u might be rite but its jst not helping.... no offense but see its 3digit no so it shud be 9X8X7 Possibilities

OpenStudy (anonymous):

yes u r correct that is total 504 numbers possible

OpenStudy (anonymous):

k then

OpenStudy (anonymous):

now each of the digits appear 504 /9 =56 times in each column

OpenStudy (anonymous):

do u agree

OpenStudy (anonymous):

k

OpenStudy (anonymous):

now can u guess the sum total of each of the three columns

OpenStudy (anonymous):

will it not be 56*(1+2+3....+9)=56*45=2520

OpenStudy (anonymous):

am i correct...

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