Answer to Puzzle 4

Take a number. Say ABCD. It can be written as 1000A+100B+10C+D
= 999A + A + 99B + B + 9C +C + D
= 9(111A+11B+C) + A+B+C+D
In the above the first term 9 (111A+11B+C) is divisible by 9 obviously since 9 is a factor.
So if the remaining terms A+B+C+D are divisible by 9 then the entire
number is divisible by 9. So the divisibility rule for nine is if the sum of the digits of a given
number is divisible by 9 then the entire number is divisible by 9. the same rule applies for
3 also and the proof is same as above. For 6 also the same proof is applicable and the rule is the number should be an even number and the sum of the digits shall be divisible by 3.