1. I'm new to the forum so hello!

Apologies if this question is too simple, if it is perhaps you could point me to another forum?

I'm teaching myself maths because I've never been very good with numbers yet I've always been interested in the concepts - it's bothered me!

Anway I'm starting from the begining and would like to know if there are simple ways to multiply large numbers.

I found a good method online for any pairs of two digit numbers (the cross multiply technique) and wondered if it could easily be extended to three and greater digit numbers. I couldn't get it to work!

Is there a quick way?

(Along the way though I found a quick method for finding the square of any number ending in 50 (not that useful, but there you go). The last four digits are always 2500 so then you simply square the number to the left of the '50' part and add itself to the prduct ie: (n^2)+n

so 450^2 is (4x4)+4 stuck on the left of the 2500 part = 202500

Seems to work for all the numbers I tried. I don't understand why though).

g

2.

3. The proof of number sense techniques often involves decomposing numbers into parts. (Such as 450 = 4*100 + 5*10 + 0*1.)

Your technique above could be written as n*100 + 50 = ((n^2)+n)*10000 + 2500, then you'd just have to use some algebra to show that this is (or isn't) true for all positive integers n.

4. best way to deal with large numbers:
Buy yourself a calculator, cant survive without one

5. Originally Posted by MagiMaster
The proof of number sense techniques often involves decomposing numbers into parts. (Such as 450 = 4*100 + 5*10 + 0*1.)

Your technique above could be written as n*100 + 50 = ((n^2)+n)*10000 + 2500, then you'd just have to use some algebra to show that this is (or isn't) true for all positive integers n.
I'm training myself to trust the abstractions of number manipulation - it's hard which is why I tend to impose (right or wrongly) an understanding of it in laymans terms. That's probalby not the right approach though!

So did you mean that particular technique for working out the square could be written:

(n*100+50)^2 = ((n^2)+n)*10000 + 2500

I get get confused if something is missing that I think should be there (ie I'm not being pedantic).

Hmm that looks like proper maths I haven't got to algebra yet, how would you prove it was true for all values of n?

g

6. Originally Posted by Zelos
best way to deal with large numbers:
Buy yourself a calculator, cant survive without one
A calculator, what's that?

I'm trying to grasp and undersatanding of numbers, I'm not interested in the actual result, more on how and why the result comes out.

7. haha funny joke, you know very well what a calculator is.

if you have trouble understanding why you get what you get in multiplication i must ask at what grade youre going in

8. Originally Posted by Zelos
haha funny joke, you know very well what a calculator is.

if you have trouble understanding why you get what you get in multiplication i must ask at what grade youre going in
Dear Zelos

I've avoided maths all my life because no one could explain it to me in a way I could understand - so I thought I was stupid in school. Now I'm older it turns out that I'm not as stupid as I thought, but I do learn at my own pace.

If you (or anyone else) feel my post to be too simple then please point me to another forum, as I asked. Until then I'll continue to post here, regardless of my 'grade level' if that's OK with you?

Thanks so much.

9. Well, here's my 'stupid person's' approach to multiplication:

I've tried to break the problem down and come up with a way that lets me use my limited maths skills to tackle it, so I can multiply two large numbers (by my standards), though it gets a bit unweildly the more digits you add! If anyone can show me how to reduce it further, or a simpler technique I'd appreciate it:

123*456 :

Sorry but I don't know how to write things correctly, hope this makes sense. I represented each digit as a letter:
(123=abc, 456=xyz):

123*456 = (100a+10b+c) * (100x+10y+z)
100x(100a+10b+c) + 10y(100a+10b+c) + z(100a+10b+c)

So I end up with a 3x3 grid of much simpler numbers to multiply, even I can do it quite quickly! Then you just add up the columns and the three remaining sums to get the total:

123*456 :

400*100, 50*100, 6*100
400*20, 50*20, 6*20
400*3, 50*3, 6*3

Well you get the idea.

I notice that in the following...
100x(100a+10b+c) + 10y(100a+10b+c) + z(100a+10b+c)

...there are three duplicate parts, the (100a+10b+c) bit . Can these be combined at all, like 3(100a+10b+c)? I've gone number-blind again.

Sorry that this is totally elementary, but at least it's helped me to visualize what's going on when you multiply, and to see structure inside the numbers which I couldn't before.

best
g

10. Yeah, it's supposed to be (n*100+50)^2 = ((n^2)+n)*10000 + 2500, and you'd just have to use algebra to show that the two sides are equal.

Also, 100x(100a+10b+c) + 10y(100a+10b+c) + z(100a+10b+c) does work, but 3(100a+10b+c) doesn't help. If you work out parts ahead of time, then what you'd get is 100x(abc)+10y(abc)+z(abc), where abc is just the original number.

You should try googling Number Sense. It's a competition based on doing math problems in your head. Most of what you're discussing falls under number sense techniques. You might also try looking around the math section of Wikipedia, or at Mathworld.

11. Originally Posted by MagiMaster
Yeah, it's supposed to be (n*100+50)^2 = ((n^2)+n)*10000 + 2500, and you'd just have to use algebra to show that the two sides are equal.

Also, 100x(100a+10b+c) + 10y(100a+10b+c) + z(100a+10b+c) does work, but 3(100a+10b+c) doesn't help. If you work out parts ahead of time, then what you'd get is 100x(abc)+10y(abc)+z(abc), where abc is just the original number.

You should try googling Number Sense. It's a competition based on doing math problems in your head. Most of what you're discussing falls under number sense techniques. You might also try looking around the math section of Wikipedia, or at Mathworld.
Thanks for the help - I've been reading some of the wiki entries, but keep on having hop arround articles because I don't understand some of the terms used, end up forgetting what the first question I was looking up was! But it it useful still. Mathworld - don't know that one , thanks will check it out.

ta
g

12. Originally Posted by gruff
...there are three duplicate parts, the (100a+10b+c) bit . Can these be combined at all, like 3(100a+10b+c)?
Sometimes it might turn out that way, but usually not. There is a strict rule in math that says you must solve the expression inside the parentheses first (actually that's what the parentheses are for)

Sorry that this is totally elementary
Hey! If we all had to apologize for asking questions some brain-box thought elementary, we'd be neck high in sorriness. Don't fret, you're doing just fine.

Oh. Zelos: You have been helped here by members more knowledgeable than you, you might try to return the favour, rather than just mocking.

13. Oh. Zelos: You have been helped here by members more knowledgeable than you, you might try to return the favour, rather than just mocking.
This should go for me too. (I'm, jocking, I'm not mocking anyone, just laughing with everybody, nothing bad)

14. The beauty of numbers ahhhhhh

The great thing with our current hindu-arabic numerals is that they work in a very intuitive way. Long numbers like 14905 is our way of saying: 5*1 + 0*10 + 9*100 + 4*1000 + 1*10000. This is very basic I realise, but before the discovery of 0, numbers were represented in far uglier ways. You see, each digit occupies a column and each column represents successive powers of 10, the digit you apply to a column indicates that you multiply the power of 10 by that digit, you leave a column empty by use of the 0.

This is very simple (and I hope I'm not insulting your intellignece), but if you understand this you should be able to multiply larger numbers. The key to succesful multiplication is getting the numbers in the right column. So an easy example might be 652*45. You lay it out pictorally almost like a counting frame or abacus (I can't type it out neatly for you sorry).

You then proceed to multiply each digit in the number 652, starting from the right (i.e. the "units") by 5. 2*5=10, so you put 0 in the unit column and "carry the 1" This just means you've got one 10, and no units, carrying the one just means you put the one in the ten column and add it to whatever you get next (in that column)......... so now you multiply 5*5=25. The five goes into the "tens column", but remembering that you have already got one ten waiting you add them together to get 6, the 2 carries over to the "hundreds column". 5*6=30, so 0+2 in the hundreds, and you put the 3 in the thousands. So now you have effectively multiplied 652 by 5, giving you 3260.
Now you have to repeat the process by multiplying all the digits by 4, you write the new number directly underneath the number you have just obtained taking care to get your digits in the right column. It's exactly the same basic principle. But now you see that the 4 starts in the "tens column", (i.e. this 4 in that column really means 4*10=40). So you start off by filling your unit column with a zero, and then proceed to multiply 4*2=8, you stick that straight in your tens column. 4*5=20, stick the zero in the hundreds column and carry the 2 into the thousands. 4*6=24, so it would be 4+2=6 in the thousands, and the 2 goes into the ten-thousands column. So by effectively multiplying 625 by 40 you get 26080.
To cap it all off you add your two results, so 3260+26080=29340. Hey presto that's the right answer, I just checked it on my calculator.

That's how I learned it in primary school although I only recently understood why it works. Don't worry about zelos, nobody ever listens to him anyway, good luck with the numbers :wink:

15. [quote=...That's how I learned it in primary school although I only recently understood why it works. Don't worry about zelos, nobody ever listens to him anyway, good luck with the numbers :wink:[/quote]

Thanks, billiards (is it Bill for short? :)

Yes this is the same way I learned it at school too. But, I'm not comfortable with just learning the rules, I'm stubborn and I have to find out for myself how things tick. The way I worked it out is more complicated but the process helped me understand what was going on. My arithmetic is terrible BUT just by practicing a little its getting better, and doing it 'the old way' really isn't that that bad after all:)

16. Gruff,

I had the same problem, I needed to understand it from the bottom up. The problem is maths is not 'real' it's a man made concept. There are many short cuts though. Stick at it mate.

17. Originally Posted by Megabrain
Gruff,

I had the same problem, I needed to understand it from the bottom up. The problem is maths is not 'real' it's a man made concept. There are many short cuts though. Stick at it mate.
Good to know I'm not the only one :) thanks for the encouragement guys.

I used some maths for the first time in my job today. I could have found out the result I needed experimentally, but decided to try to find a formula instead. It wasn't a biggie: given a regualr grid what is the relationship between the total numbers of vertices, polygons and edges. Forgive my use of letters in the following, I don't know what the proper way to notate things is!

I figured first that the total number of vertices is Vt=(Vs1)(Vs2) (hmm, how do you do subsrcipt online? The s1 and s2 is meant to mean side1 and side2). Ie the area of a rectangle. The number of polygons, Pt, is ((Vs1)-1)x((Vs2)-1). The number of edges is the total vertices + total polygons -1:

Et=Vt+Pt-1

This is only when both sides of the grid are the same. If its not a square then the equation is different.

Here's my train of thought for the rest of the problem:

If the grid is only one polygon on one side (ie it's just a single row), then the number of edges is simply:

3P+1
where P is just the number of polygons in the row.

For any subsequent rows though its only:

2P+1

So the total edges is:

Et=(3P+1) + n(2P + 1)
where n is the number of subsequent rows after the first one.

I didn't like the look of that, so I tried again a different way and got something better I think:

Et=2Pt+(Ps1+Ps2)
where s1 and s2 is each side of the rectangle.

And I think you can put it in terms of just edges because each Ps and Es are equivelent so:

Et=2(Es1 x Es2)+(Es1+Es2)

But that's only in terms of polygons or edges. Ideally I want a single formula that relates Vertices, Polygons and Edges for any rectangular grid. Im sure its possible, but how?

Maybe I'm making it more complex than it is, but that's where I've got to so far.

cheers
g
(PS what I'm finding very interesting is that the abstraction of the equations makes it a lot easier to see realtionships, that's cool!)

18. On the 3-d side you might like to know...

This is one of eulers theorums,

Take a cube, Number of faces = 6, corners = 8 edges = 12
Tetrahedron, Number of faces = 4, corners = 4 edges = 6

Faces +corners-2 = edges,

Same for any regular 3-d shape

I think you might have got something wrong there gruff, I think you mean
(on a 2-d drawing) Number of edges = Polygons+intersections-1

I checked this by drawing a 3x3 grid which has 24 'sides' 16 'intersections' and 9 'polygons'

19. Originally Posted by Megabrain
On the 3-d side you might like to know...

This is one of eulers theorums,

Take a cube, Number of faces = 6, corners = 8 edges(vertices) = 12
Tetrahedron, Number of faces = 4, corners = 4 edges(vertices) = 6

Faces + corners -2 = vertices,

Same for any regular 3-d shape

So how do you find the number of edges for any rectangular grid? Or relate edges, faces and vertices in a single formula?

think I'll go and look up Euler, see if there's any info on 2D surfaces.

g

20. OK here's the one for the chess board - UN TESTED

Intersections = (rows+1)times(Columns+1)
polygons = (rows times columns)

edges = intersections plus polygons - 1

So edges = (r+1)times(c+1) + rc -1

= rc + r + c + rc + 1 - 1

Nuber of edges = 2rc +r+c

Where r= rows
c = columns.

So a chess board should be 144

21. Originally Posted by Megabrain
OK here's the one for the chess board - UN TESTED

Intersections = (rows+1)times(Columns+1)
polygons = (rows times columns)

edges = intersections plus polygons - 1

So edges = (r+1)times(c+1) + rc -1

= rc + r + c + rc + 1 - 1

Nuber of edges = 2rc +r+c

Where r= rows
c = columns.

So a chess board should be 144
LOL, look how much simpler your working out is to mine! :)

We got the same result though, but yours is a hell of a lot neater looking!

So, is it possible to add in the vertices to that so that given any one you have a formula to work out the other two?

g

22. You need two to work out the third.

If there is 1 unknown you can solve it with 1 equation.
if there are two unknowns you need two equations
3 for three etc.

One unknown;
4 apples cost 24P
well one unknown (the price of an apple), one equation, soulution = 6P

TWO unknowns;

Suppose I say 2 apples and 2 pears cost 40p altogether, the individual prices could be almost anything like 1P per apple 19p a pear or vice versa
but if I say

2 apples + 2 pears cost 40P
2 apples + 3 pears cost 50P

then you can see 1 pear costs 10P. Take the cost of three peras away leaves two apples for 20P and you find they are also 10P each.

23. Originally Posted by Megabrain
You need two to work out the third.

If there is 1 unknown you can solve it with 1 equation.
if there are two unknowns you need two equations
3 for three etc.

One unknown;
4 apples cost 24P
well one unknown (the price of an apple), one equation, soulution = 6P

TWO unknowns;

Suppose I say 2 apples and 2 pears cost 40p altogether, the individual prices could be almost anything like 1P per apple 19p a pear or vice versa
but if I say

2 apples + 2 pears cost 40P
2 apples + 3 pears cost 50P

then you can see 1 pear costs 10P. Take the cost of three peras away leaves two apples for 20P and you find they are also 10P each.
Yes of course that makes perfect sense - and a good rule to learn :)

But the cost of pears in not dependant on the cost of apples. the values for each could be anything at all so I can see the sense of the above. My instinct is still niggling at me that in the case of a grid, Vertices Polygons and Edges should be relatable. You can't add a polygon to a grid without knowing exactly how many vertices and edges you are also adding.

Is my mind (well, my gut) playing tricks on me?

24. OK number of polygons = 24
this could be either a 2x12,3x8,4x6,6x4,8x3,12x2 (rows by columns)

Now remember 2rc+r+c

so working out for the above example then you would get

48+14, 48+11,48+10,48+10,48+11,48+14.

In order to sort out which one is correct you must also know 'r' or 'c'

If R=c then it is different since you 'lose' one of the variables then you are down to one equation as you now only have one unknown.
say n=r=c

now edges = 2(n squared)+2n

25. These examples assume the grids are all a single rectangle in overall shape. For non-rectangular shapes any maths will depend on how many you add, and where to.

26. Originally Posted by Megabrain
These examples assume the grids are all a single rectangle in overall shape. For non-rectangular shapes any maths will depend on how many you add, and where to.
Yes, that's the case I'm considering. The next step was to work out what happens with holes or different shaped grids ! I think I'll save that for another day :)

I get what you're saying now about the variables r and c, thanks for clearing that up.

cheers
g

27. The beauty in math....

This is the way I count 123 * 456

456 x 1000/8 - 456(2)

so quickly, divide 456 by 8 and multiply it by 1000, 57000

then, 57000 - 912 = 56088

LOL

28. Is anyone familiar with the techniques applied on an abacus?

I used to know how to use the Chinese one, but lack of practice made me forget... As I understand, people who are really good at it routinely outperform even the fastest calculator users on large multiplications. I am wondering if the relatively simple rules to slide the beads could teach you an equivalent approach for doing the math on paper or in your head. Any ideas?

29. I've heard of people that got good enough with an abacus, they can basically use one in their head. The one I saw on TV went through the motions with his fingers, but without an actual abacus.

30. Originally Posted by M
Is anyone familiar with the techniques applied on an abacus?

I used to know how to use the Chinese one, but lack of practice made me forget... As I understand, people who are really good at it routinely outperform even the fastest calculator users on large multiplications. I am wondering if the relatively simple rules to slide the beads could teach you an equivalent approach for doing the math on paper or in your head. Any ideas?
Yes, I read an article about them recently online - one I saw used intermediate groups of 5 between the units and tens rows (and tens and hundreds etc), it looked very simple and certainly adaptable to in-head use. I guess it would just be practice to be able to do it quickly in your head. I suspect it would mean juggling fewer numbers than the normal carrying system, though haven't tested it.

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