Without actual calculation write the quotient when the sum of 3-digit number abc and the number obtained by changing the order of digits cyclically i.e. bca and cab is divided by 111
Let the digits of the number be a, b and c taken in order.
i.e 100a+10b+c (original number) .....(1)
1 cyclic rotation will give: 100b + 10c + a .....(2)
2nd cyclic rotation will give: 100c + 10a + b .....(3)
sum of (1), (2) and (3) will give: (100a + 10b + c)+( 100b + 10c + a)+(100c + 10a + b ) =111a+111b+111c
on dividing 111a+111b+111c with 111 we will get: a+b+c
Hence, quotient is a+b+c.
i.e 100a+10b+c (original number) .....(1)
1 cyclic rotation will give: 100b + 10c + a .....(2)
2nd cyclic rotation will give: 100c + 10a + b .....(3)
sum of (1), (2) and (3) will give: (100a + 10b + c)+( 100b + 10c + a)+(100c + 10a + b ) =111a+111b+111c
on dividing 111a+111b+111c with 111 we will get: a+b+c
Hence, quotient is a+b+c.