Calculator 11s

[Phil Reynolds]

Apologies if I'm trampling all over Charlie's new subforum with an OT thread, as this isn't strictly a game or a puzzle. But it's an interesting numerical phenomenon which I discovered by accident many years ago and for which I've never been able to find any rationale or explanation. Given the number of maths gurus on this board, one of you may be able to explain why this works the way it does.

OK, here's a thing: imagine you have in your hands a pocket calculator. (That probably gives you some idea of how many years ago it was that I discovered this.) The digits 0-9 are arranged on the keypad like this:

Code: Select all

7   8   9
4   5   6
1   2   3
        0

(If you try this on an actual calculator (or a calculator program like the one on Windows) that has the 0 positioned under the 1, you'll need to pretend it's under the 3 for this to work.)

Now, tap in any 4-digit number that conforms to the following rule: Pick a starting digit, and then imagine that digit as being at one corner of a regular 4-sided shape on the keypad. By "regular 4-sided shape", I mean a square, rectangle, parallelogram or rhombus. Tap in the 4-digit number formed by tracking the corners of that shape around the keypad, either clockwise or anticlockwise. For example:

• Starting with the 7 and tracing a clockwise square round the outer corners of the main block of 9 digits gives you 7931.
• Starting with the 8, you could trace an anticlockwise parallelogram to give 8129. Also, you can bring the 0 into play by entering 8206.
• Starting with the 5 and working clockwise, you could trace a square to give 5632, or a rhombus to give 5302.
• You can also start with a notional leading 0 to produce a 3-digit number such as (0)341.

OK, here's the weird part. Whatever 3- or 4-digit number you tap in - whichever shape you traced out, wherever you started from on the keypad and whichever way round the shape you went - provided you followed the basic rule, the number you end up with is always exactly divisible by 11.

Why?

[Howard Somerset]

It's a result of the fairly well known rule for divisibilty by 11, which goes something like:

For any interger, regard oddly positioned digits as the units, the hundreds, etc, i.e. an odd number of digits from where the decimal point would be, and evenly positioned digits as the rest, e.g. tens, thousands, etc.
For example, in the number 25491, 2, 4 and 1 are the oddly positioned digits and 5 and 9 are the evenly positioned digits.

Now, add up all the evenly positioned digits, and add up all the oddly positioned digits, to give two totals.

The original number will be divisible by 11 if and only if the two totals are equal, or if they differ by a multiple of 11.

(Carol used to mention this on occasion, when noticing that the first and last digit of Cecil's target added up to the middle digit.)

Now, in the way you describe the shape you're making, you'll find that the first and third digit total, must equal the second and fourth digit total. Hence any number traced out in the way you've described must be a multiple of 11.

[Paul Howe]

Neat trick. The divisibility by 11 test that Howard mentioned works because 99, 9999, 999999, etc are all divisible by 11, and so are 11 (duh), 1001, 100001, 10000001, etc, which can be checked by long division if you're not convinced.

As an example, suppose we want to test 4791237 for divisibility by 11, we can write

4791237 = 4*1000000 + 7*100000 + 9*10000 + 1*1000 + 2*100 + 3*10 +7
= 4*(999999+1) + 7*(100001-1) + 9*(9999+1) + 1*(1001-1) + 2*(99+1) + 3*(11-1) + 7
= (4*999999 + 7*100001 + 9*9999 + 1*1001 + 2*99 + 3*11) + (4+9+2+7) - (7+1+3)

The stuff in the first set of brackets is then divisible by 11, so we need the stuff in the second set of brackets (the odd digits) and the stuff in the third set (the even digits) to be equal or differ by a multiple of 11 to make the whole thing divisible by 11.

[Howard Somerset]

Interesting explanation of why the 11 rule works, Paul.

Just to complete the story. I simply said that the sum of the first and third digits of the shape in this puzzle must equal the sum of the second and fourth digits. Just in case the reason for that is not completely clear, let's suppose that the first corner is at a and the third corner is at b.

Movement from first corner to second corner is equal and opposite to movement from third corner to fourth corner.

So, suppose from first to second corner we move right x columns, and up y rows. x and y can of course be zero or negative.
Going up one row adds 3 to the number, and going right one column adds 1. So an x,y movement as described adds x+3y, thus making the second corner a+x+3y

Going from third to fourth corners, we're now moving left x columns (-x cols right) and down y rows (-y rows up). So this -x,-y movement adds -x-3y, thus making the fourth corner b-x-3y.

It's now easy to see that adding first and third corners gives a+b, and adding second and fourth corners gives a+x+3y+b-x-3y, which equates to a+b.

[Dinos Sfyris]

Tutor me Howard!

[Ian Volante]

Boobless!

Oh, wrong calculator thing.

[Naomi Laddiman]

Boobless!

Oh, wrong calculator thing.