Mental Math: A New Divisibility Rule for Three-Digit Numbers (and more!)
DISCLAIMER: I am not aware of this proof being published by anyone else.
What is fascinating about mathematics is how a simple rule can open the door to a plethora of others. I find great joy in playing with numbers on my calculator to try to uncover yet another secret. Sometimes, my little conjectures (statements based on a pattern and not solid proof) are ripped apart by the most obvious counterexamples. Other times, I feel so shocked that the pattern I noticed works for more than three numbers (yes, my standards are quite high) that I know for fact there must be something wrong, which there inadvertently is.
Then, once in a blue moon, I find something that withstands test after test. Tentatively, I work out a proof. Let me just say that working out a proof for a conjecture, no matter how silly (as this one is), is perhaps my greatest source of joy in math.
We’ve all learned the divisibility rule of three at school: if the sum of the digits in a number is divisible by three, then so is that number. But what if there is another rule for three digit numbers? I love patterns, so I worked with three-digit numbers whose digits are related in some sort of way.
So, I worked with three-digit numbers formed by consecutive digits. This means that the three digits in a number come right after another, like 123 (or 321, 213, 132, 231, 312 — the order doesn’t matter so long as the digits are right after another).
Punching numbers into my calculator, I found the following pattern:
Any three digit number formed of consecutive digits is divisible by three.
This was a surprise. Unsure whether it was pure luck, I decided to attempt to write a proof. Unbelievably, it worked. Here it is below:
- Assume the three digit number has the digits XYZ, where X, Y, and Z are consecutive.
- For the sake of convention, let’s represent X by n.
- Since the integers are consecutive, then Y = n + 1 AND Z = n + 2
- Notice that it does not matter whether the number is XYZ or YZX: it will not change the value.
- Add X, Y, and Z. You will get n + (n+1) + (n+2) = 3n+3
- Divide (3n + 3) by 3 and you will get n + 1, without any remainder. Since the sum is divisible by three, then the number is divisible by three. It is also shown that the order of the consecutive integers does not matter.
Q.E.D
This proof can be extended to six-digit numbers and nine-digit ones.
What are the uses of this trick? Well, not much, if only to avoid tedious addition. But take a moment and think how many other hidden tricks lie out there, waiting to be discovered.
Perhaps the next time you find yourself bored, you can take out the calculator, punch in some numbers, and find your next best pattern.
