Calculator Number Trick: rectangle patterns
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- Опубліковано 30 тра 2019
- Here is the original Fermat's Library tweet.
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We make Think Maths teaching resources to go with Stand-up Maths videos. For this one we have a cheat sheet to an algebraic proof.
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Originally a homage to the Tim and Eric gif. From this video: • Calculating π by hand:...
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CORRECTIONS
- I did not properly think through the hexadecimal example. I meant that any such digits on a base-10 calculator would give multiples of 11. On a true hexadecimal calculator you would get multiples of 17 (or any n+1 for base-n). First spotted by MasterHigure.
- Let me know if you spot anything else!
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"Never too early to generalize" is one of the most beautiful self-referencing random phrases I've met.
How is that self-referential? I'm having trouble seeing it.
@@OrangeC7 It's "never" too early to generalize -> In the general case, it is not too early to generalize
@@OrangeC7 Because the statement itself is a generalization (about when you should generalize)
@@TotalWarbringer I see, thank you!
@@OrangeC7 I missed it too. Language is not my strong suit, hence why I'm here. :P
Well, for a hex calculator, you won't get multiples of eleven but multiples of hexadecimal one-one, which is better known as seventeen, aka a Parker eleven.
Seventeen is Base Plus 1. Eleven is (base10) + 1. So can we generalise that for base 8 calculator you have it with square ones = nine (base + 1 ) again? (i have not tested this.
@@highpath4776 well if its always 11 in each base, as in ( 1 * b^1) + ( 1 * b^0) where b is the base number, then it simplifies to b + 1 and tgete you go.
It works for normal 11 is you consider each key-press as a decending power of ten. For instace 13131313 is not a multpile of 11, but 14443 is.
@@highpath4776 Yes, it is always one more than the base, for the reason jetison333 stated.
At this rate someone will end up naming Parker X to something important in math just to make the ultimate meme
"My goal was not to do it algebraically"
*Proceeds to do an algebraic proof*
@@samgraf7496 Is that not what you got?
Yeah it's about as simply explained there as it could be and you can actually easily figure out the bowtie question if you look at that for a second as well. Because changing the order results in the +A's or +B's to not be in adjacent slots you end up with adding a number that's a multiple of 101 or 1001 instead at some point and that isn't a multiple of 11 so it breaks it. Trying to work it out with numbers is probably a bit more tricky but from the picture it's much clearer.
Why didn't you unbox that calculator?!
He definitely did
Suppose he didn't unbox it ,then it would still had been in a lonely box
@@palmomki BADA BOOM BADA BANG!! 😂
(3b1b style of QED)
@@Peter_1986 and he means hes joking
@@hamiltonianpathondodecahed5236 Ah, but thats assuming there was a box when it was created in the first place.
More information is needed
@@Peter_1986 r/woosh
I saw, that the 1st and 2nd number have the same difference as the 3rd and 4th number - but in the other direction. So we have
a*1000 + (a+n)*100 + b*10 + (b-n)
= 1100a + 100n + 11b - n
= 11 * (100a + b + 9n)
Same thinking i see
Yes. And this shows that it works for parallelograms too, which Matt’s method did (I think) not.
DorFuchs just casually commenting under a Matt Parker Video without anyone knowing who he is and how many German students he has entertained and saved
This proof is sooo clean and clever. I really love it!
@@teeesen pretty sure it does. A paralellogram would just mean that an additional integer is added to A or B, which wouldn't change the final result since it's generalised anyways. But you're right that he didn't touch upon this.
6:50 "even though it is a prime number"
Hahahahahahah of course
If you include 0, the keys form a Parker square.
0:57 "Go ____clockwise"
Parker script flub + audio editing?
You can sort of hear him breathe three times at that moment.
@@meithecatte8492 You can read his lips saying anticlockwise.
loool i didn't even notice. thanks for that one xD
Great catch, haha.
Parker clockwise
"This calculator will still work. Well not as a calculator obviously"
*theorists be like*
It has an even and an odd number, you can construct any number with them
Matt, you specifically said I could use any calculator that I choose and this would work. I picked my Curta Type I!!!!
This video was much better than I expected. This is a much more general thing than just a quirk of particular calculators!
1:07 imagine if every youtuber was so nice to their video editor :]
What a co-incidence.
I was just thinking you might release a new video!
When I see some interesting number patterns I think "thats interesting" then carry on with my life.
Mathematicians are great because they stop and ask why.
Feels really nice to work out the problem by yourself and then watch Matt go through very similar steps in a different way.
Amazing pattern, great fun to look at the algebra behind it.
Matt, I'm always amazed at how intuitively you explain maths. You're an awesome guy :)
How come these videos always cheer me up?
Best UA-cam personality
Also works with parallelograms such as [1,5,6,2] or [8,9,2,1] and also with two diagonal numbers like 2 and 6 [2,6,6,2]
That's cool! And the 2662 works because it's a parallelogram with side length zero.
OK, I was amused and intrigued with the puzzle and found the logic of the proof for why this works fun to work out.
Then Matt threw in the parallelogram bit and my mind suddenly blew. Like, after a second I could see that sure, it would extend that way as well, but that was the bit that just suddenly hit me out of nowhere. Nicely presented.
"This will still work. Well, not as a calculator, that's ridiculous" xD lol
In fact, it doesn't work at all with two-digit numbers. [insert Parker square joke here].
This stuff is just great! Love your videos, excellent work. One question: Are you like a genius or something?
A square, you say?
A PARKER square?
That'd be if it worked in every area, but one.
And a parker cube in the background
It became a parker square when he generalised it to parallelagrams.
type 58008 and turn your calculator upside down....hehe
...auto rotation
Congratulations on 800,000 subs!
So Simple, SO Brilliant. Love Your Videos Sir.
Any parallelogram works too.
Also two squares, or a rectangle and a square, or any combo as long as you complete the pattern.
Eg: 31,795,478÷11
Also like Matt's 5555, you can do 1199 or 1166 or even 22554477 or a combo 118,822,775,632 crazy
Numerals gone wild! But in a surprisingly orderly way.
I recognise that gif and am very happy that you used it
Love you Matt!!
I love that smug little half-smile after the joke at the end Matt! :D
Benjamin Bradley just like herbert gross at mit
3:35 "It's never too early to generalize." This might be one of the funniest thing I hear for weeks. I have no idea why I found that sop funny.
That's pretty cool! Love your videos!
brilliant, love your proof!
It works for any almost regular polygon, besides some triangles, as long as the polygon has an even amount of button presses to create, and after you've finished a shape you can start a new one adding on and it'll still be divisible by 11.
Just from some quick testing it seems like it works for parallelograms as well... and the 2,4,6,8 square
Edit: oh he said that at the end welp...I tried
good job tho
But isn't a rectangle just a special case of parallelogram?
this really does work with any rectangle, I tried some simple examples like the rectangle [8,9] making 8998/11=818 and [8,6,2,4] being a rotated square making 8624/11=784
However I was surprised when this worked with a rotated square with the base as [1,6] if you extend beyond the limits of just the 3x3 square you can get the number above 8 to be 11 because when you go up a layer it's +3 and then the number to the left of 7 would be 6 because going left is -1 so you get a square of [1,6,11,6] and when you add them together like this:
1000
+600
+110
+6
you get 1716/11=156 and the cool thing is, [1,7,1,6] is not a possible rectangle on the grid, yet it still works, because it's possible to make if you break the boundaries.
another cool example using the similar numbers, a 12 instead of the first 1, this creates a parallelogram, where 12 is 9+3 (going up a layer)
[12,7,1,6]
12716/11=1156
and in reverse order:
6000
+100
+70
+12
6182/11=562
[6,1,8,2] again is not a valid pattern on the 3x3 grid normally, but can be found through this boundary extension method
Here's a similar approach:
The arrangement of numbers on the keypad tells us that any digit can be represented by 3v+h
v=vertical coordinate
h=horizontal coordinate
(we won't actually have to specify the possible values of v and h, but lets just say v is in the range of 0-2 and h in the range of 1-3)
Any orthogonal rectangle drawn with vertices on the grid of numbers will have length l and (vertical) width w
Assuming length and width can be positive _and_ negative means we can actually define only one way of
tracing a rectangle, just flipped around depending on the sign of l and w
assume 5698 is the correct way of tracing, with length 1 and width 1, starting where the relative l and w both = 0
then the number can be represented as 3v+h,3v+h+l,3v+h+l+w,3v+h+w
in base-10 positional notation, this evaluates as 1000(3v+h)+100(3v+h+l)+10(3v+h+l+w)+3v+h+w
=3000v+1000h+300v+100h+100l+30v+10h+10l+10w+3v+h+w
=3000v+300v+30v+3v+1000h+100h+10h+h+100l+10l+10w+w
=3333v+1111h+110l+11w
All the coefficients are divisible by 11!
11(303v+101h+10l+w)
(And we also get a nice formula for determining the value of a rectangular pattern from just the properties of the rectangle and where it is on the keypad)
Your such quicker smart thinker . I admire you crestivity ingenuity in this field as well as all your passion in what you do .
Just finished Humble Pi and then a notification of an upload! Awesome
The diamond shaped 8426 works. 8426 / 11 = 766
Also, 7238 / 11 = 658. 1562 / 11 = 142
So all parallelograms work, even if the width is 0: 2662 / 11 = 242
Diamond shape was the first thing I checked, too.
I think that's a coincidence. Setting up the same visualization as Matt would look like:
8 + 0 + 0 | 8 - B - A | 8 - 2B + 0 | 8 - B + A
As you can see, the multiples of eleven are not as clearly obvious as they were in the video. Any number has a 1/11 chance of "accidentally" having this property.
@@wildgoosechase4642 Toward the end, Matt says it works for all parallelograms, which includes the diamond shape.
Here's a slightly different expansion that avoids the negatives:
use any bottom left "origin" for the diamond, n: n+A | n+B | n+A+B+B | n+A+A+B
remove the simple nn00 and nn portions: A | B | A+B+B | A+A+B
remove the simple adjacent As and Bs: A | 0 | 0 | A
realize that 1001 is a multiple of 11, so any x00x is as well: Done.
@@wildgoosechase4642 : I don't think this a fluke. Imagine extending the number pad like so:
*6* 7 8 9
*3* 4 5 6 *7*
*0* 1 2 3 *4*
The diamonds 7 5 1 *3* and 9 *7* 3 5 works also. Dividing by 11 = 683 and 885 respectively.
Cool. I discovered something similar way back in 1975, when I would spend hours just playing with numbers on my first calculator (a Rockwell 5-function).
Nice find Matt!
5:50 I think we can stretch even further from parallelograms if we have a large enough panel of numbers. We can draw a polygon with 4n sides and still get a multiple of b+1 when b is the base as long as that polygon follows this rule. There must be 2n pairs of parallel sides with the same lengths, and each pair must be separated by an odd number of sides.
I can do my proof, but not sure if I can put here in the comment box.
Fantastically fun video! Thanks!!
Can I suggest you look into "dividing heads" which machinists use to rotate things precise fractions of a turn. They use plates with circles of different numbers of holes for example three plates with circles of holes. Perhaps you can explain why they use these numbers and if it is an optimal arrangement or not.
i like the power outlet you got in the backround ... it is wireless in so many ways
That unsolved 2x2x2 in the background gives me the hibbi jibbies, but you knew that already, didn't you 😏
Parker Rubik's cube
He can definitely solve it too, which makes it even worse
I bothers me even more when people leave a 1x1x1 Rubik's cube unsolved
i think its a infinity cube thing, doesnt look like a 2x2x2
@@preferablygeneric That's funny. I've never seen a 1x1x1 unsolved.
I find they're harder to scramble than to solve.
I love how this video's comment section is people genuinely interested in this and having a conversation about the maths.
Wow. The explanation is so simple.
We're also on Bulb, they're great! 🙌
Me too! Matts getting free electricity now...
My mind is blown more than that Parkerized mind_blown.gif (which I absolutely need in my life)
Elliot Grey That sure is a sick gif
The pattern actually works with any parallelogram you can draw with the buttons! Even if you use 0, although you have to pretend it's underneath the 3 (where it should be)
Pretty sure there is a much simpler way. Rule of divisibility for 11 is that the sum of the digits in even positions minus the sum of digits in odd positions must be a multiple of 11.
Because of the cw/CCW way we generate the number that means that the two groups of digits are the two diagonals in the rectangle.
Now since the difference between the top two digits and the bottom two is the same but their positions are reversed that means that when you subtract their sum on top you get +-A and on the bottom you get -+A which cancels out to get 0 a multiple of 11 so the whole number is a multiple of 11.
You can also think about it in terms of the rule to check for divisibility of 11, add and subtract alternating digits in reverse and see if that's divisible by 11, e.g. 1-3+9-7, will always give zero (divisible by 11) for any rectangle because you're adding two pairs of numbers, one which has a difference of +A and one which has a difference of -A.
When working this out myself, my first thought was "this is a divisibility problem, it wouldn't hurt to put this in the language of modular arithmetic"
(For those not familiar but want to follow along, taking a number mod n means looking at the remainder after dividing by n. The key insight of why this is useful is you can apply the modular operation at any point during the problem as many times as you want and it will work out the same as applying it at the end).
Let a b c and d refer to the points of our parallelogram, such that the route a->b->c->d->a is a valid path.
Part 1: Re-framing our question in modular arithmetic and simplifying to a easier to handle expression
So abcd (referring to concatenation) being divisible by 11 is the same as taking abcd mod 11 = 0.
What is abcd mod 11?
Well abcd is equal to a*10^3+b*10^2+c*10+d. Taking a*10^3+b*10^2+c*10+d mod 11 gives a*(-1)^3+b*(-1)^2+c*(-1)+d, or d-c+b-a.
So now we can test numbers for divisibility by 11 by doing d-c+b-a mod 11. Awesome!
Part 2: Show d-c+b-a mod 11 = 0
Now, in the case of this grid, if we draw 2 parallel lines of equal length, the difference between the end points will remain constant. Translating the line left and right just increments the end points by 1 and a-b=(a+1)-(b+1). Translating up and down just increments the end points by 3, and a-b=(a+3)-(b+3).
What this means is that a+k=d and b+k=c, where k is just some integer.
We can now write our expression as (a+k)-(b+k)+b-a. Everything cancels out to 0, and 0 mod 11 is 0. Therefore, abcd is divisible by 11.
This was an interesting video: thanks for making it!
I think your last example is missing a detail to make it work. For instance, if we use the parallelogram 2, 6, 26, 22 as shown in the video, we cannot simply type the digits into a calculator and divide by 11: 262622 is not a multiple of 11. To make this example (or any rectangle or parallelogram) work, one of two things must be done:
Option A: add 2 * 1000 + 6 *100 + 26 * 10 + 22. The result, 2882, is a multiple of 11.
Option B: put a leading zero in front of each one-digit number, giving us 02062622. Then, instead of 11, divide by 101: 2062622 is a multiple of 101.
I believe that the generalization in Option B will work for any "calculator rectangle" with individual numbers up to N-digits long and dividing by 10^N + 1. Sadly, this UA-cam comment area is too small to contain my proof. (In reality, I haven't done the proof but I'm pretty sure this is true nonetheless!)
I was about to go to my calculator, then Matt is like “Renewable energy...”
Great video, as always.
@standupmaths we can also get multiple of 11 by diagonal rectangles as well like 4862 or 8426 etc
The parallelogram thing also works on a regular calculator
@@donaldasayers no?
@@donaldasayers I tried making parallelograms including the 0, but they are not a multiple of 11.
@@donaldasayers Oh I misunderstood what you mean by 'under the 3' 😅. I thought you meant when the 0 button is right below the 3 button...
Edit: oh I guess that did understood you the first time and the numbers do check out. I don't remember what I tried, but I must have done it wrong.
what do you call a rectangle that went to the body shop?
a fixedangle
I'll show myself out
while playing around with this I also realized that any line between 3 numbers diagonal, horizontal, or vertical all are multiples of three. Same with one number typed 3 times
Any such line going back again is a multiple of 37, like 159951. Same with one number typed 3 times.
5:45 You get into trouble with the double digits (at least as long as you just concatenate the numbers). So it doesn't work on any grid. And the hexadecimal one at 5:25 will make numbers that are divisible by 17 instead (since that's what the hexadecimal number "11" represents).
That’s a super good point. I was still thinking of a base-10 calculator but with keys for 0 to 16. Which was careless of me! Have added you to the corrections.
That's actually amazing
You are a good teacher.
Messing around with the calculator, I also found that this works for "right triangle" patterns with 6 numbers. e.g. 159,874 is divisible by 11; it equals 14,534. It even works if you start from any position, but you must maintain the right triangle pattern. e.g. 896,357 / 11 = 81,487.
The alternating sum of digits will always be = 0 -> divisible by 11
That was the explanation I came up with as well.
very exciting & noticed the mini-football in the background too!
11 divisibility rule: +-+-... lattice since 10 mod 11 = -1. The numbers are right next to each other in the rectangle pattern, so the difference between each of the start/end pair and the middle pair is the same. Then, you simply add the larger value in calculating divisibility in one pair and add the smaller value in the other while subtracting the number not picked. The disparity always cancels out to 0 when you sum up the digits in the +-+-... fashion, so it always is divisible.
If you hit all six digits in succession in that 8923 rectangle Matt chooses, not just the corner ones, you'll get a multiple of 111, wherever you start and in either direction. For example 258963, 896325, 985236. This goes for any such 3 by 2 rectangles on the pad, including "tiny trivial" ones created by mashing a button 3 times.
Bloody hell! My head is spinning... Need to watch again and grab my calculator, pen and paper!! Well done Matt.
Since it works with a single number 4 times - rectangle with 0 length sides and with regular rectangles, it works with lines - only one 0 length side. When you hit the end of the line go back until you get 4 digits. Like 2585 or 5852. Works with the diagonals too.
Interesting is that the algebraic proof is simpel and very basic.
It immediately explains why it works for bigger fields and also for doubling every number and so on.
More interesting is how to come to the idea considering rectangles mod 11 on a numberfield. :)
I thought of it in terms of the old way of determining if something is a multiple of 11 (alternately adding and subtracting the digits), but to understand how that works, we need to understand modulus functions. If you already get how that works, skip to the §. This is probably going to be confusing, but it’s how I figured it out, and it was a lot more intuitive in my head,.
Basically, if a mod b=c, that means that dividing a mod b has a remainder of c; for example, 36 mod 5=1 since 36 divided by 5 is 7 with remainder 1. It can also be expressed as d~e mod f, if d and e have the same remainder mod f (which is equivalent to saying their difference is a multiple of f); for example, 36~11 mod 5, since both are 1 mod 5, and their difference is 25, a multiple of 5.
Now, an important quality of modulus is that the modulus of a product of two numbers is the product of their moduli; for example, to find 35 mod 3, you could factor 35 into 7*5, find their remainders mod 3 (1 and 2), and multiply them for 2*1=2.
This works because if two numbers a and b are respectively equivalent to c and d mod e, they can be expressed as that modulus plus a multiple of e; multiplying them gets you c*d and a bunch of terms with a factor of e that vanish when you take the modulus.
For example, with 35, 7 and 5 become 2*3+1 and 1*3+2; FOILing those two gets you (2*3)*(1*3) + (2*3)*2 + 1*(1*3) + 1*2< all terms besides the last have a 3 factor, and thus vanish when you take mod 3, leaving 2.
Now in terms of numbers mod 11, the place number system means each digit in a number indicates that digit times a power of 10. Now, in mod 11, 10~ -1, since they differ by 11. Thus, the tens digit is effectively multiplied by -1, and the hundreds? -1*-1=1, so 100 mod 11=1 (makes sense, since 99 is a multiple of 11). Each next digit is multiplied by another 10 (equivalent to -1), so the powers of 10 alternate between 1 and -1 mod 11. Hence why you alternately add and subtract digits: you’re effectively multiplying each digit by what its power of 10 is equal to mod 11, and then adding them to get the overall modulus; if you get 0, or a multiple of 11, the number was a multiple of 11.
§ Now that that is established, I tend to use the rule a bit differently, in a way that will make my idea more understandable. If given a number like abcdefg, instead of adding a, subtracting b, adding c, etc. I tend to separate the added digits from the subtracted ones; I’ll put a, c, e, and g in one group and add them together, b, d, and f in another and add them, and subtract one from the other; if the result of 0 or a multiple of 11, the initial number was a multiple of 11.
Now apply this to the numbers generated from the keyboard. To determine their mod 11, you’d take the first and third digits, add them to create a sum I’ll call A, and subtract the sum of the second and fourth digits, which I’ll call B. Now, think of each number in the sum B in terms of a transformation from the previous digit, which was in sum A. As Parker mentioned, moving once in a direction on the keypad is equivalent to adding or subtracting a specific value from the number.
Since the opposite sides of rectangles are equal lengths, and the way you go around them while creating the number means the side between the first and second digits is traversed in the opposite direction to that between the third and fourth. Since going from the first digit (in A) to the second (in B) require going the same distance in the opposite direction as going from the third (in A) to the fourth (in B), the change between A and B added by one is subtracted by the other. Thus, the two changes cancel out, A and B are equal, and the difference is 0, making the number a multiple of 11.
For example, in the 5698 example, doing from 5 to 6 requires one right (adding 2 to B compared to A), whole going from 9 to 8 involves going left once, which subtracts one. Or with 1793, 1->7 involves adding 6, while 9->3 involves subtracting 6. Thus, the two sums are the same, making the remainder 0 and the number a multiple of 11.
I love the ending here!!
Since you like shapes, who doesn't, I picked up a "Euclidean Cube" and have been baffled with it's many hidden faces even though it's a "fairly simple" joining of a few squashed looking Tetrahedron's linked together by the edges (with some magnets as well). I'd love to see if there are any interesting properties about it that a aren't as obvious to those of us that haven't studied math as much. You can even put a bunch of them together to make even crazier looking objects and at the moment find it as my most interesting cube shaped desk toy. Also if you know of any other cube shaped desk toys I'm all ears, I have a collection growing at this point.
He is the guys from 'you 've been warned' and 'numberphile' love u bro.
Works on the diagonals with 4, 8, 6 and 2 combinations too.
thank you. this really helps
if you were adding the corners and edges(the steps have to be symmetrical!), starting at the number x, then:
sum - x * 11111111 / 12221 is an integer.
this can be generalized for n numbers per edge.
the series of deviders is:
12221
12333321
12344444321
12345555554321
etc.
Mind.blown!
(1:45) Dang it, now you tell us!? I already paused it at 1:25 and examined the proof... Thank you very much Mr. Matt Parker!
This was really fun to do!!
We need to make the "Mindblown" GIF at 1:09 a Parker-Meme.
If you know the 11s multiplication trick, this is obvious. Why the 11s trick works is more difficult.
In the 11s trick, for a number whose digits are abcd, if a-b+c-d is divisible by 11, then abcd is divisible by 11. So in one case, if a-b+c-d=0, then abcd is divisible by 11. So a-b=d-c.
In any rectangle, the difference between the two numbers on top is the same as the difference between two numbers on the bottom. But if you are going either clockwise or counterclockwise, then you will be moving in the opposite direction across the top than you will across the bottom. So if you label the corners clockwise from the top left abcd, you will have a-b= either -1 or -2 and d-c is the same number. If yiu don't start with a or if you go the other way, you're just swapping the minus signs in a way that won't change the end result.
And the same formula holds true for the other grids to which you generalized.
And it even works if your rectangle is askew. 4862 is divisible by 11 as well.
also works for "diamond" patterns (eg 4862), "hexagon" patterns (eg 148962), and "6-digit triangle" patterns (eg 159632) !!
I mean a diamond _is_ a parallelogram
Props for reshooting that mind blown meme yourself.
I feel that the explanation was somewhat lacking and didn't, but since I'm lazy I can't be bothered to investigate further so I'm in the "not having fun investigating it myself" camp :D
Too effective to be Parker squares.
Maybe the gods are telling us something with these things all being rectangles.
5,555.
Jesus, I didn’t expect 60 likes
Big fan of the explanation
The hexadecimal calculators work too, but only if you interpret "11" as if it was in hex - same for other bases (just replace 16 with any other number)
If think about the divisibility rule for 11(difference of the sum of the two alternating series of digits must be divisible by 11), all the proofs are quite easy. The numbers on the diagonals would be in alternate positions(thats why the bowtie shape does not work) and their sums are always equal.
The center will hold as long as matt parker is protected at all costs.
I just bought your humble pi book, I can't wait for it to arrive ^_^
Damn! I knew I should have swaped out my abbacus for a calculator
Well, if you keep your rectangles even hight, on an abacus, they will also be divisible by 11...
parallelograms work on 3x3 calculators too! eg 1265/11=115 or 7425/11=675
8624 works as well
a pretty simple trick with whole numbers that are multiplied by 11. lets say you have 29x11, split the 29 so you have 2 and 9, add these two and you get 11. now the 9 stays in the single digit area while now you add 20+11, this will give you 31. tack the 9 at the end and you get 319, now with most numbers you do this with you will just tack the first digit to the front, the last digit to the end, and the sum is in the center. example 34x11 is 374. So lets do this with 319. Split them, 3/1/9. 3+1 = 4, 1+9 = 10, 40+10 = 50. the answer is 3509. it is a pretty cool math trick that can save you time, takes a bit more time to go backwards though, best to just use a calculator. fun showing my college class mates this since most don't know about it
Hey Matt!
This also works for longer, complex patterns as long as they retain a certain amount of symmetry. Like 7 5 9 5 3 5 1 5. You still get 11.
Or 1 8 3 4 9 2 7 6.
Or even 1 4 7 8 9 6 3 2.
Very cool trick!
My answer: ( before watching what you came up with )
I know n|11 if sum of all k's * 10^2m+1 = to the sum of all t's * 10^2m - a little formally
When we make a square we move clockwise or anticlockwise - it's easy to see, that 1st + 3rd number has to be equal 2nd + 4th. Let's call them A, B, C, D in right order.
If our assumption is right 10^3*A+10^2*B+10*C+D ( where we can swap pair AC with BD and make permutations inside those pairs ) is divisible by eleven.
Then we pick some number p to begin with as our A. B = p + x ( x represents horizontal movement )
D = p + y ( y represents vertical movement )
C = p + x + y
We check, that A+C indeed equals B + D ; = 2p + x + y
Therefore it is proved whenever we make a square in our calculator, the number we create is divisible by eleven. Literally just because of how it was designed. For example:
7 8 9
6 5 4
1 2 3
Wouldn't work; but:
1 4 7
2 5 8
3 6 9
Would've been exactly same to the actual one, just rotated.
I did not to really deep in this, but I hope effort I put finding the answer will pay off. :)
If you look at the criteria of multiple of 11 it also becomes pretty easy
Clockwise:
First Digit: X
Second digit: X + A
Third Digit (or fifth): Y
Forth Digit: Y - A
Adding the odd position digits is X+Y. Adding the Even position digits is also X+Y. Therefore, it is a multiple of 11.
Anti-Clockwise is the same idea:
First Digit: X
Second digit: X - A
Third Digit (or fifth): Y
Forth Digit: Y + A