If a, b, c are three digits of a three-digit number, prove that abc + cab + bca is a multiple of 37. Share with your friends Share 0 Arnab Gupta answered this Dear Student, any 3 digit number abc can be written as 100a+10b+cHence we got abc+cab+bca⇒100a+10b+c+100c+10a+b+100b+10c+a⇒111a+111b+111c⇒111×a+b+cnow factors of 111 are 3×37Hence abc+cab+bca=3×37×a+b+ctherefore the number abc+cab+bca is divisible by 3 and 37 Regards, 1 View Full Answer