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Multiplication Staff Tutorial

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In this tutorial we run through some of the basic ideas and tricks in multiplying whole numbers. We do a bit of stuttering on the decimal system and the fact that 6 times 15 is the same as 15 times 6 before we get under way.

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It’s perhaps a little surprising that although many civilisations counted using a decimal system it took a long while for everyone to realize that the decimal system was a good one to use for doing arithmetic. In the Western world it wasn’t till 1202 that Fibonacci wrote his famous book on arithmetic and introduced Europe to a great way to do calculations. It’s actually amazing that it took so long.

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After all the Greeks, who were no mathematical slouches, had done a great deal – Euclid, for instance, had written a book on geometry that sold for over 2000 years. But it took non- Europeans to see what a good idea it was to write three hundred and ninety-four as 394, with the right-most position standing for 4 ones (or units), the 9 just to the left standing for 90, and the 3 out the front standing for 300. That concise notation revolutionized the world.

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Note too the way that the zero has come into action to keep the numbers apart and to distinguish 345 from 3045, 3405, and You might just like to rub over the next couple of numbers and see what the numbers in the different positions stand for. 2367; ;

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That neat notation means that you never again have to multiply a number by ten using a calculator. You see 3 times 10 is three lots of 10 and so we only have to put a 3 in the tens column to give 3 x 10 = 30. But the same trick works at higher levels. What is 53 x 10? Well 53 x 10 is the same as 50 x 10 plus 3 x 10. Now we already know that 3 x 10 = 30.

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So what is 50 x 10? Surely we want to put 50 in the tens column? But it’s too big and the 5 spills over into the hundreds column. That’s not a surprise. Think of it this way. Now 50 is 5 lots of 10. So 50 x 10 is 5 lots of 10 times 10. Since ten tens is a hundred, 50 times 10 is surely 5 lots of 100. So 50 x 10 = 500. Then 53 x 10 = 50 x x 10 = = 530. And all that we’ve done is add a zero on the end. And that always works. Check it out for 28 x 10; 123 x 10; x 10.

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I promised you that we’d see that 6 times 15 is the same as 15 times 6. How can we do that? Which would you rather have? Six bags of sweets with 15 in each or fifteen bags of sweets with 6 in each?

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Let’s take one bag of 15 sweets. We could divide the sweets into Put the sixes into separate bags. That’d give us two bags of 6 sweets and 3 over. But we’d be able to do that for all of the initial six bags. So now we have twelve bags of 6 sweets and six lots of 3 sweets. But 2 lots of 3 sweets gives another bag of 6. What’s more we get 3 of these 2 lots of 6. So now we have the twelve bags of 6 from above plus the 3 bags of 6 from below. That’s 15 bags of 6 if ever I saw them. Why don’t you show that 8 bags of 17 sweets contains as many sweets as 17 bags of 8 sweets?

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But that’s a bit tedious. It’s not so easy to check out 392 bags of 563 sweets against 563 bags of 392 sweets. But that can be done. What you need to do is to is to get 563 different colour stickers (I jest of course). Then stick a different coloured on each of the 563 sweets in the first of the 392 bags. Then stick a different coloured sticker on each of the 563 sweets in the second of the 392 bags. Keep doing that until all the sweets have had a sticker put on them.

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Now take all the sweets with a red sticker on them and put them into a new bag. That should give you 392 red- stickered sweets in the first bag. (I hope that you are doing this without touching the sweets. Someone’s going to want to eat them after you’re finished.). Now take all the sweets with a blue sticker on them and put them into another new bag. That should give you 392 blue-stickered sweets in the second bag. Now take all the sweets with a green sticker on them and put them into another new bag. That should give you 392 green-stickered sweets in the third bag.

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Can you see that you can keep going like this and eventually you’ll have 563 bags each with 392 sweets in them? Now show that 25 bags with $179 in, is worth as much as 179 bags with $25 in. Can you think of a way to do this without using coloured stickers because the President of the Reserve Bank is getting a bit annoyed with people putting stickers all over his nice money.

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So you should be able to see now that 564 x 702 is the same as 702 x 564. What’s more you can justify that the two are the same. But there is an easier way than the sticker route. Let’s do some tiling. Get a whole stack of tiles and make a big rectangle of them. In fact make them so that you have 564 rows of tiles with 702 tiles in each row. 564 rows 702 tiles in each row

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But now turn your big rectangle through 90º. Now you should see that you have 702 rows and 564 tiles in each row. And that should convince you that 564 x 702 = 702 x rows 702 tiles in each row

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This business about the order of multiplication not mattering has a fancy mathematical name. It’s called the commutative law. It just means that it doesn’t matter which way we do it we get the same answer. Actually there are not too many things around that commute. Try interchanging the order of putting on your socks and your shoes. Shoes first give quite a different result. Can you think of anything but multiplication that is commutative?

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OK, so now we’ve got some tools in place that might come in handy. We can multiply by 10 and we know that it doesn’t matter what order we multiply numbers in, the answer is still the same. So what can we do to get 23 x 11? Well I’m not convinced that 11 x 23 is any easier so let’s try something else. We can all do 23 x 10. That’s 230. And 23 x 1 is 23. So 23 x 11 = = 253.

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Actually we can see this using the tiling idea. If we had 23 tiles one way and 11 the other then the rectangle of tiles would be like the picture below

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If we cut the 11 side into a 10 and a 1 we get the picture below

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And we can do the odd calculation. Clearly = 253 – the number of tiles in the rectangle

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But can you find another way to represent 23 x 11? How about using bags of sweets? And then have a crack at finding 47 x 12 and 13 x 88. After that you might like to see what you can do with 19 x 91.

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Actually 19 x 91 can be done in a number of ways. First you might break 19 up into Then all you have to do is to multiply 10 x 91 and add it to 9 x 91. Of course you might prefer to do it the other way and first do 91 x 10 and add it to 91 x 9. But 19 = 20 – 1. So you could do 91 x 20 – 91 x 1. And all of this could be done using tiles. Work out 19 x 91 using tiles in all of these ways. Can you think of another way?

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But what about big numbers? How about 187 x 56? You can do all of that using tiles too. Look here. You can see that the product of 187 and 56 is the sum of the six smaller areas. We’ve talked about multiplication by the tens and the units but not about the hundreds. But that’s easy isn’t it? x x x x 6 7 x 50 7 x 6

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1 x 100 is surely 100 and 5 x 100 and so on work the same way. But what about 10 x 100? Well here comes the commutative law to the rescue. 10 x 100 = 100 x 10 and we know that 100 x 10 = So 10 x 100 = What about 23 x 100 though? Well that’s 20 x x 100. Obviously 3 x 100 = 300. And 20 x 100 = 100 x 20 = 100 x 10 x 2 = 1000 x 2 = So 23 x 100 = 2300.

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So let’s go back to 187 x 56. We’ve seen how to do that the tiling way and it comes out like this: 187 x 56 = 187 x x 6. Now 187 x 6 = 100 x x x 6 = = 1122, And 187 x 50 = (187 x 5) x 10 = (100 x x x 5) x 10 = ( ) x 10 = (935) x 10 = So 187 x 56 = =

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But you have just seen the good old multiplication algorithm done on its side. Let’s do it the usual way up this line is just 100 x x x this line is just 100 x x x Notice that we get lazy in practice. The ‘50’ line we don’t consciously multiply by 50 each time. We simply dump a 0 at the end and multiply 187 by 5. Hopefully it’s now clear that the algorithm is exactly the same as the tiling method.

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If you want to use the standard algorithm to multiply 2345 by 1234, then talk your way through why you put no zeros at the end of the first row, one zero at the end of the second row, two zeros at the end of the third row, and three zeros at the end of the last row. Then do the algorithm method and check it against the tiling method.

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