MathChat

One of my subscribers asked why the trick for divisibility for 3 actually works. [If you missed the post on that trick, go here:]

But the gist of the trick is this:  3 divides evenly into a number if 3 divides evenly into the sum of the digits of the number.

I’ll prove this trick for a three-digit number, but you’ll see why the proof applies to numbers with as many digits as you’d like.

Let’s call our three-digit number cde, where c is the digit in the 100s place, d is the digit in the 10s place, and e is the digit in the 1s place.

We can state the value of our cde number like this:

cde =  (100 x c)  +  (10 x d)  +  e

This shows that the c-part of the number is made up of 100 groups of c. If you think about it, there’s no reason we can’t re-write the value of this digit as 99 groups of c plus 1c, or just:  (99c + c).

In the same way, the d-part of cde is made up of 10 groups of d, which we can re-write as (9d + d). And of course the e-part of the number is just e.

So now we have this:

cde =  (99c +  c)  +  (9d + d)  +  e

Using the rules of jolly old algebra, we can shuffle the terms around a bit to get this:

cde =   99c + 9d + (c + d + e)

Then, adding a set of parentheses for clarity, we get this:

cde =   (99c + 9d) + (c + d + e)

So far, so good. But what’s the point? Well, we’re just getting to that.

Let’s think a bit more about the 99c term. Factoring out a 3, we see that 99c =  3 x 33c. Since 3 is a factor of this expression, 99c must be divisible by 3. Aha, progress, right?

Factoring the d-term, we see that 9d = 3 x 3d. Again, since 3 is a factor of this expression, 9d is  also divisible by 3.

So we now know that both 99c and 9d are divisible by 3.

We can never forget the Divisibility Principle of Sums (DPS), which says:

If a number, x, divides evenly into both a and b, then it divides evenly into their sum, (a + b).

What does that mean here? It means that since 3 divides evenly into both 99c and 9d, it must divide evenly into their sum:   (99c + 9d).

And remember that our number, cde, equals nothing more than:  (99c + 9d) + (c + d + e)

So, we’ve just found out that 3 divides evenly into the quantity in the first parentheses:  99c + 9d

So to find out if 3 divides evenly into the whole number cde, all that’s left is to find out whether or not 3 divides into what remains, the quantity in the second parentheses:  c + d + e

But guess what? c + d + e is the sum of the digits for our number, cde. So this idea right here is the trick for divisibility for 3:  To find out if 3 divides evenly into a number, just add up the number’s digits and see if 3 goes into that sum.  If it does, then 3 does go in; if not, 3 does not go in.

So this proves the divisibility trick for a three-digit number like cde.

To see why the same trick works for numbers with four or more digits, keep in mind that the larger digits can similarly all be broken up as we broke up the digits of c and d. For example, if we have a four-digit number, bcde, then the value of the leading digit b can be viewed, first, as (1,000 x b). And then this can be split apart once more, into 999b + b. That way, this four-digit number fits into the pattern of the trick. And this same kind of split up can be done for any digit whatsoever.

So if you follow the logic of this proof, you now see that the divisibility trick rests on solid logical/mathematical ground.