Expert: Sherry Wallin - 4/16/2011

Question
QUESTION: I read in a book that a number is divisible by 3 if sum of the digits is divisible by 3 . I dont understand what is the logic behind this rule .

ANSWER: Abhinav~

First, what you read is true. Without knowing your mathematical background it is hard to decide how to 'show' this to you. Can you please tell me what courses you have taken? Do you know about proofs by induction? Do you know what congruency is in modular arithmetic?

There are only 3 possible remainders if you divide by 3? Do you know which ones they are? Right, 0,1, or 2. Any number can be written as a power of 10. For example 12,324 = 1x10^4 + 2x10^3 + 3x10^2 + 2x10^1 + 4x10^0 or simpler yet: 12,324 = 10,000+2,000+300+20+4

Now divide each of the numbers in the last sum by 3: 10,000/3 = 3,333 1/3; 2,000/3 = 666 2/3; 300/3 = 100; 20/3 = 6 2/3; 4/3 = 1 1/3
now add the results: 3,333 1/3 + 666 2/3 + 100 + 6 2/3 + 1 1/3 = (3,333 + 666 + 100 + 6 +1) + (1/3 + 2/3 + 2/3 + 1/3) = 4,106 + 2 = 4108
which just so happens to be 12,324/3 = 4108. Incidentally, 1+2+3+2+4 = 12 which is divisible by 3 so we knew it would work with a zero remainder.

Math Prof

---------- FOLLOW-UP ----------

QUESTION: Sorry I did not exactly get what you say . I understood that every number can be expressed in the powers of ten . But after that when u divided thd number by 3 , added the results , the way u approached the final point of your explanation and what did that exactly mean  that part i did not understand .

Answer
Ok, try this: count by 3's:3,6,9,12,15,18,21,24,27,30. Notice that multiples of three can end in any of the first 10 digits including 0. So you can't tell if a number is divisible by 3 by looking at it's last digit but you can tell by the last two digits if it is a 2 digit number, for example:
12,15,18,21,24,27,30,33,36,39,42,...,99. Now  extend this idea to 3 digit numbers. In order for a 3 digit number to be divisible by 3 then all three digits have to be divisible by 3. We've already established that a number that is divisible by 3 can end in any single digit not necessarily divisible by 3, for example 1,2,4,5,7,8, none are divisible by 3 but if they are the ones digit in a 2 digit number then they can make a 2 digit number divisible by 3, likewise with a 3 digit number,..., n digit number.

Im sorry without knowing the depth of your mathematical background this makes it difficult to answer your question!

Math Prof

I thought of another way to look at this:

Suppose a number is NOT divisible by 3, then it would have a remainder of 1 or 2 (It can't have a 0 remainder because then it would be divisible by 3 and it can't have any higher remainder than 2 because then another 3 could be divided out). So if the number was not divisible by 3 it could be written as 3n + 1 or 3n + 2. Here is a concrete example: 1022 = 3(340) + 2 => n = 340 or  31 = 3(10) + 1 => n = 10 BUT a number could also look like 3(a+b+c+...) + 1 or 3(a+b+c+...) + 2. For example:

12350 = 10,000 + 2000 + 300 + 50
= 3(3333 + 666 + 100 + 16) + 1 + 2 + 0 + 2
=3(4115) + 5
=3(4115) + 3 + 2
=3(4116) + 2 => n = 4116 or that a+b+c+d = 4116

Hopefully this helps you understand better why only numbers divisible by 3 have digits that add up to numbers divisible by 3.

Math Prof