Multiplication Tables

Toast · 2007-02-14 23:47:14

If you have a multiplication table, pick any 4 by 4 square and add the numbers on each diagonal. For e.g

24 32 40 48
27 36 45 54
30 40 50 60
33 44 55 66

The numbers in diagonal 1 would add up to (24 + 36 + 50 + 66) = 176
and in diagonal 2, (33 + 40 + 45 + 48) = 166

I've noticed that D_1 = D_2 + 10

1.How would I prove that this statement will always be true?

2.How can I predict what will happen if I pick a 5 by 5 square, a 6 by 6 square... an n by n square, and add all the numbers on the diagonal?

I think I've pretty much solved q.1, using algebra, although I'm not sure if it's correct (it should be uploaded as an image) but i'm having most trouble with question 2.

So far for it I've come up with:
1 by 1 square, 0 difference
2 by 2 square, 1 difference
3 by 3 square, 4 difference
4 by 4 square, 10 difference
5 by 5 square, 20 difference
6 by 6 square, 35 difference...
n by n square, ? difference?

mathsyperson · 2007-02-15 02:20:35

You're nearly right with the proof for question 1. Your method is perfect, but strictly speaking you should just manipulate the expression without first assuming that it's equal to 10. You can't assume what you're trying to prove.

It doesn't matter in this case, because you haven't actually done anything to the 10 throughout the proof, but it shouldn't technically be there.

Multiplying out all the brackets and then doing lots of cancellations is a perfectly good method, but it's a bit inefficient and would cause trouble when you do it for the general case. It might be easier to do some cancelling before you expanded.

So, you have [xy + (x+1)(y+1) + (x+2)(y+2) + (x+3)(y+3)] - [x(y+3) + (x+1)(y+2) + (x+2)(y+1) + (x+3)y].

By pairing up terms, you can simplify this without having to expand:
[xy - x(y+3)] + [(x+1)(y+1) - (x+1)(y+2)] + [(x+2)(y+2) - (x+2)(y+1)] + [(x+3)(y+3) - (x+3)y]
= x(-3) + (x+1)(-1) + (x+2)(1) + (x+3)(3).

This way might be a bit easier when you're doing it with bigger squares, and especially with the nxn ones.

Toast · 2007-02-15 22:26:13

Can you give me some more clues on how to work out the difference in an n by n square? I'm still quite stumped . Thanks.

mathsyperson · 2007-02-16 02:13:57

They're not really clues, more like ideas. I haven't done this investigation before, so the things I'm saying might not work.

Anyway, for the n x n square, you would just do it the same way as you did before but with another variable considered:

[xy + (x+1)(y+1) + (x+2)(y+2) + ... + (x+n)(y+n)] - [x(y+n) + (x+1)(y+n-1) + (x+2)(y+n-2) + ... + (x+n)y]

Pairing similar terms:
[xy - x(y+n)] + [(x+1)(y+1) - (x+1)(y+n-1)] + [(x+2)(y+2) - (x+2)(y+n-2)] + ... + [(x+n)(y+n) - (x+n)y]

And then you can factorise those and make some cancellations and get a nice formula.

Alternatively, an easier but less rigourous method would be to use the sequence you already have and find the nth term of that by finding differences. I'd recommend the first of those, because the second doesn't actually prove that it will always work, it just finds a pattern.

John E. Franklin · 2007-02-17 09:50:09

× 3 4 5 6

8 24 xx xx 48
9 xx 36 45 xx
10 xx 40 50 xx
11 33 xx xx 66

From mathsyp: x(-3) + (x+1)(-1) + (x+2)(1) + (x+3)(3).
Pick apart the terms as the columns above in grid.
Set x = 3 for the below exercise:
x(-3) column 3 #'s in column differ by 3, times 3 moves to get from 24 to 33.
+ (x+1)(-1) column 4 #'s in column differ by 4, times 1 move to get to adjacent square below 36 is 40.
+ (x+2)(1) column 5 45 to 50, one square apart, five apart because in the five tables column.
+ (x+3)(3). column 6 three moves from 48 square down to 66 square times six apart each square.

And I love the distributed property demonstrated by mathsyperson!! Nice way to make it easier!!

Last edited by John E. Franklin (2007-02-17 09:50:56)

Toast · 2007-02-17 14:10:49

K... well so far I've been able to turn it into a sum, with the first term being Diagonal 1 and the second being Diagonal 2. However, I don't know how to simplify it...

When I plug it into my calculator it gives the correct answer:

Can you please tell me... this is the first time I've used sums in my homework and although I know how to simplify arithmetic and geometric sums I don't know how to do this one. Thanks.

Last edited by Toast (2007-02-17 14:11:33)

Toast · 2007-02-17 17:07:59

So... is there actually a method, or do I have to use a calculator?

mathsyperson · 2007-02-17 23:42:17

You can factorise the (x+k) out of both terms to give:
(x+k)(y+k-(y+n-1+k))
= (x+k)(2k+1-n).

I think the best step after that would be to split it into two summations, so that you can find the x term and the constant term. The x term should always be 0, and the constant should fit the formula you just posted.

Incidentally, that formula can be factorised further, into

.
It's up to you whether you use that though, they're both equally neat.

Toast · 2007-02-18 00:03:38

So like:

Then what do I do?

Last edited by Toast (2007-02-18 00:08:21)

Toast · 2007-02-18 19:07:20

I tried using arithmetic progression on both. For the first summation it came out as 0. For the second I got

Geeze... Soooo close. Was I on the right track?

EDIT: I think it was supposed to be

well... maybe not that close, but maybe getting there...

Last edited by Toast (2007-02-18 19:09:21)

Toast · 2007-02-24 00:01:14

I think I may have solved it...

You split up the summation:

Then apply the following summation identities:

for

, and normal arithmetic summations for the other two:

Here we go:

Simply adding together will get

The part I'm confused with is that for the arithmetic simplifications, I'm saying that there is 'n' terms, not 'n-1' terms? Why do I have 'n' terms for the arithmetic ones but 'n-1' for the 2k^2 one?
Because that's the only way it will equal the formula.

Last edited by Toast (2007-02-24 01:02:25)

Toast · 2007-02-24 00:01:14

Editing my previous post seems to have created this new post... which I can't delete ?

Last edited by Toast (2007-02-24 00:04:26)

Math Is Fun Forum

#1 2007-02-14 23:47:14

Multiplication Tables

#2 2007-02-15 02:20:35

Re: Multiplication Tables

#3 2007-02-15 22:26:13

Re: Multiplication Tables

#4 2007-02-16 02:13:57

Re: Multiplication Tables

#5 2007-02-17 09:50:09

Re: Multiplication Tables

#6 2007-02-17 14:10:49

Re: Multiplication Tables

#7 2007-02-17 17:07:59

Re: Multiplication Tables

#8 2007-02-17 23:42:17

Re: Multiplication Tables

#9 2007-02-18 00:03:38

Re: Multiplication Tables

#10 2007-02-18 19:07:20

Re: Multiplication Tables

#11 2007-02-24 00:01:14

Re: Multiplication Tables

#12 2007-02-24 00:01:14

Re: Multiplication Tables

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