nice on to try :)

the no. itself n shud be a multiple of 57

now the digit 7 in 57 is important
just for a minute suppose dat the no we want is abcde
so condition is 57 x bcde = abcde
now it is clear that last term of bcde = last term of abcde

so we need to find such numbers , such that multiplying it with 7 retains the last digit
now such a no shud hav 5 or 10 at d end bcz
7 x 5 = 35
and 7 x 10 = 10
and no oder no frm 1 to 10 retains the last digit

if it is the case wid 7 , same will be it with 57

so all numbers ending with 5 or 0 will retain the last digits viz 5,0 wen multiplied wid 7 or 57

so all numbers ending with 5 or 0 will retain the last digits viz 5,0 wen multiplied wid 7 or 57

so we check nos ending wid 5 , that if dey r multiplied wid 7 retain that no itself , bcz we want 57 x bcde = abcde

now we will find 25 x 7 = 125 but it doesnt satisfy the given condition . but notice that after 5 , 25 retained the entire digit. i.e 5 and 5² retained the entire digit

so there is something here relating to powers of 5 so b4 checking 30 x 7 , chck 5³ = 125

and yes 125 x 57 = 7125

so we get it !!!!

so sum of nos = 15

obviously its not d an ideal ans , but its one way of doing it [3]and dont lol at it