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Posts tagged ‘Divisibility by 3’

How to See Why the Divisibility Trick for 3 Works

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) + (cde)

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) + (cde)

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.


Divisibility: Find out if 3 divides evenly into an integer

Quick:  What English word has 12 letters, almost half of which are are the letter “i” — well, 5 of the 12, to be exact?

Why it’s the word “D-I-V-I-S-I-B-I-L-I-T-Y” — a great thing to understand if you’re going to spend any amount of time doing math. And guess what:  virtually ALL students do a fair amount of math, so everyone would do well to master the tricks of divisibility.

With the tricks for divisibility in your command, you will have a much easier time:

—  reducing fractions
—  multiplying fractions
—  dividing fractions
—  adding and subtracting fractions
—  finding the GCF and LCM
—  simplifying ratios
—  solving proportions
—  factoring algebraic expressions
—  factoring quadratic trinomials
—  Need I say more?

I’m sure  you get the point — divisibility tricks are handy to know.

Since the tricks of divisibility are fun and interesting, too, I’ll share as many as I can think of. If, after I’m done, you know tricks I have not mentioned, feel free to share them as comments. Or, if you know any additional tricks for the numbers I’m covering, share those! It’s always fun to learn ways to get faster at math.

Today, I’ll share the trick that tells us whether or not a number is divisible by 3. Now many of you probably know  the basic trick. But even if you do, don’t skip this blogpost. For after I show how this trick is usually presented, I’ll share a few extra tricks that most people don’t know, tricks that make the basic trick even easier to use.

Here’s how the trick is usually presented.

Take any whole number and add up its digits. If the digits add up to a multiple of 3 (3, 6, 9, 12, etc.), then 3 divides into the original number. And if the digits add up to a number that is not a multiple of 3 (5, 7, 8, 10, 11, etc.), then 3 does not divide into the original number.

Example A:  Consider 311.

Add the digits:  3 + 1 + 1  = 5

Since 5 is NOT a multiple of 3, 3 does NOT divide into 311 evenly.

Example B:  Consider 411.

Add the digits:  4 + 1 + 1  =  6

Since 6 IS a multiple of 3, 3 DOES divide into 411 evenly.

Check for yourself:

311 ÷ 3 = 103.666 … So 3 does NOT divide in evenly.

But 411 ÷ 3  =  137 exactly. So 3 DOES divide in evenly.

Isn’t it great how reliable math rules are? I mean, they ALWAYS work, if the rule is correct. In what other field do we get that level of certainty?!

Corollary #1:

Now, to make the rule work even faster, consider this trick. If the number in question has any 0s, 3s, 6s, or 9s, you can disregard those digits. For example, let’s say you need to know if 6,203 is divisible by 3. When adding up the digits, you DON’T need to add the 6, 0 or 3. All you need to do is look at the 2. Since 2 is NOT  a multiple of 3, 3 does NOT go into 6,203.

So now try this … what digits do you need to add up in the following numbers? And, based on that, is the number divisible by 3, or not?

a)  5,391
b)  16,037
c)   972,132


a)  5,391: Consider only the 5 & the 1. DIVISIBLE by 3.
b)  16,037: Consider only the 1 & 7. NOT divisible by 3.
c)   972,132: Consider only the 7, 2, 1 & 2. DIVISIBLE by 3.

Corollary #2:

Just as you can disregard any digits that are 0, 3, 6, and 9, we can also disregard pairs of numbers that add up to a sum that’s divisible by 3. For example, if a number has a 5 and a 4, we can disregard those two digits, since they add up to 9. And if a number has an 8 and a 4, we can disregard them, since they add up to 12, a multiple of 3.

Try this. See which digits you need to consider for these numbers. Then tell whether or not the number is divisible by 3.

a)  51,954
b)  62,497
c)  102,386


a)  51,954: Disregard 5 & 1 (since they add up to 6); disregard the 9; disregard the 5 &4 (since they add up to 9). So number is DIVISIBLE by 3. [NOTE:  If you can disregard all digits, then the number IS divisible by 3.]
b)  62,497: Disregard 6; disregard 2 & 4 (Why?); disregard 9. Consider only the 7. Number is NOT divisible by 3.
c)  102,386: Disregard 0, 3, 6. Disregard 1 & 2 (Why?). Consider only the 8. Number is NOT divisible by 3.

See how you can save time using these corollaries?

Using the trick and the corollaries, determine which numbers you need to consider, then decide whether or not 3 divides into these numbers.

a)  47
b)  915
c)  4,316
d)  84,063
e)  25,172
f)  367,492
g)  5,648
h)  12,039
i)  79
j)  617
k)  924


a)  47:  Consider the 4 and 7. Number NOT divisible by 3.
b)  915:  Consider no digits. Number IS divisible by 3.
c)  4,316:  Consider the 4, 1. Number NOT divisible by 3.
d)  84,563:  Consider only the 5. Number NOT divisible by 3.
e)  71,031:  Consider the 7, 1, 1. Number IS divisible by 3.
f)  367,492:  Consider only the 7. Number NOT divisible by 3.
g)  5,648:  Consider only the 5. Number NOT divisible by 3.
h)  12,039:  Consider no digits. Number IS divisible by 3.
i)  79:  Consider only the 7. Number NOT divisible by 3.
j)  617:  Consider the 1, 7. Number NOT divisible by 3.
k)  927:  Consider no digits. Number IS divisible by 3.