Question

let abc,bca and cab be three digit number. prove that abc+bca+cab is divisible by 3​

Answer 1

Answer:

THE ANSWER IS AS FOLLOWS

Step-by-step explanation:

TO PROVE ABC+BCA+CAB

in general form it can be generalized

100a+10b+c + 100b+10c+a+100c+10a+b

= 111a+111b+111c

= 111(a+b+c)

= 3*37 (a+b+c)

= 37(a+b+c)

hence it is divisible 3

hence proved