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In the primary school, children are taught multiplication using a formal written method that is based on
This algorithm (often called long multiplication) and its teaching is illustrated below in several steps. Teaching the algorithm proceeds in three steps: multiplication by a single digit, then multiplication by a multiple of ten and then to multiplication by numbers with two or more digits. Now calculators are widespread, not all children need the third stage.
Example 1: 23 x 4
23 is 2 tens and 3 ones. 3 ones multiplied by 4 gives 12 ones and 2 tens multiplied by 4 gives 8 tens (that is 80). 80 and 12 are added to give the final product 92.
Children should write multiplication in this form for some time, until the procedure is familiar and the concepts (especially the distributive law) is well understood. Ruling up and labelling columns for tens and ones is recommended in the early stages. Later it can be reduced to a more compact form:
The 3 ones are first multiplied by 4 giving the product 12, which is 1 ten and 2 ones. 2 is written in the ones column and the 1 is recorded in the tens column. Now the 2 tens are multiplied by 4 to give 8 tens. The 1 ten recorded before is added on, so the product has 9 tens. Click here to see how multiplication is explained with Multi-Base Arithmetic Blocks (MAB).
Children must learn how to multiply by multiples of ten. It is very important that they know that to multiply a whole number by ten a zero can be added to the number. It is better to say that the digits move into the next higher place value column. 10 x 2 = 10 x 2 ones = 20 ones = 2 tens = 20 Example 2: 10 x 152 10 x 152 = 10 x (1 hundred + 5 tens + 2 ones) = 10 hundreds + 50 tens + 20 ones = 1 thousand + 5 hundreds + 2 tens = 1520 (see also Multi-Base Arithmetic Blocks) After learning how to multiply by ten, children can see how to multiply by multiples of ten. Example 3: 21 x 30 To multiply by 30, first multiply by ten (by putting down the zero) and then by 3.
Example 4: 21 x 300 To multiply by 300, first multiply by one hundred (by multiplying by ten and then by ten again i.e. putting down two zeros) and then by 3.
These multiplications require understanding of all that has come before. They are less important now that calculators are common so not all children need to master the algorithm. Extensive practice is no longer a high priority. Example 5: 57 x 46
To compute this product, 57 is first multiplied by 6 ones and then by 4 tens. The two results are then added to get the final result. It will be written down as follows.
Next, 57 is multiplied by 40 (this is done by multiplying by 10; putting down the zero - and then by 4).
There are a variety of slightly different ways of setting out the algorithm. The choice is unimportant, except that omitting zeros (as in the final example) is inadvisable. Children are more likely to keep columns aligned if they put in the zeros. Move carry digits somewhere else, move multiplication sign to the other side.
Move carry digits somewhere else and multiplication sign on top of other side.
Not advised.
Lattice multiplication The lattice method of multiplication appears in the first printed arithmetic book, printed in Treviso (Italy) in 1478. It shows this exact method and a variation as well as some variations of the long multiplication algorithm commonly taught today. Lattice multiplication and variations of today's standard long multiplication were introduced into Europe by Fibonacci (whose correct name is Leonardo of Pisa). He was an Italian who learnt the use of Arabic numerals from a Moorish teacher in North Africa. Before the Hindu-Arabic system was used in Europe, multiplication was often done with counters because Roman numerals were ill-suited to calculation and very few people knew how to multiply. The Hindu-Arabic system has made calculation fairly simple. In lattice multiplication, the partial products are laid out in a lattice and adding along the diagonals gives the answer to the multiplication. The lattice method of multiplication is illustrated by the following examples. Example 6: 28 x 57 As 28 and 57 have two digits each a lattice is set out with two columns and two rows. The diagonals are drawn in each cell as shown below. 28 is written above the lattice with 2 above the first column and 8 above the second. 57 is written to the right of the lattice with 5 along the first row and 7 along the second.
The partial products of these digits taken two at a time is set out in the corresponding cells with the tens above the diagonal and ones below. eg. the partial products in this case are 5 x 8 (= 40), 5 x 2 (= 10), 7 x 8 (=56) and 7 x 2 (=14).These are set out as shown below.
The sum along each diagonal is then recorded as shown below and these digits 1, 5, 9 and 6 form the answer to the multiplication.
Thus 28 x 57 = 1596 Example 7: 183 x 49 The lattice set out for this multiplication will have 3 columns and two rows as 183 has 3 digits and 49 has 2 digits. As before the numbers are set out as shown below and the partial products are written down in their respective positions. The numbers along the diagonals are added to give the answer.
Note that in this example adding along the third diagonal gives 19 which needs 1 to be carried to the diagonal above it. Therefore the addition should begin with the lowest diagonal ( the product of the ones from the two numbers). The Russian Peasant Algorithm This interesting algorithm of multiplication, which was used long ago, is based on the principle of doubling and halving. Of the two numbers to be multiplied, one is halved and the other is doubled successively until the number in the halves column is reduced to 1. Remainders in halving odd numbers are ignored. The numbers in the doubles column corresponding to the odd numbers in the halves column are chosen and the rest are ignored. The sum of these selected numbers in the doubles column gives the product. The following examples illustrate the use of the algorithm. Example 8: 28 x 57 Two columns are set up as shown below, one for halves and the other for doubles. 28 is placed in the halves column and 57 in the doubles (They can be placed the other way round as well).
In the next step 28 is halved and 57 is doubled and written below the respective numbers.
Next 14 is halved to get 7 and 114 is doubled to give 228. This process is continued as shown below until the number in the halves column becomes 1. Note that remainders are ignored when halving odd numbers.
The rows containing the even numbers in the halves column are ignored. The remaining numbers in the halves column are 7, 3 and 1. The numbers in the doubles column corresponding to these, are added to give the product. 228 + 456 + 912 =1596 which is the product of 28 and 57. Example 9: 183 x 49 To use the Russian peasant algorithm, 49 is placed in the halves column and 183 in the doubles column. The halving and doubling process is represented below.
The odd numbers in the halves column are 49, 3 and 1. The corresponding numbers in the doubles column are added to give the product. 183 + 2928 + 5856 = 8967. 183 x 49 = 8967 Why does it work? The method is simple enough to follow and most people who come across the method have no difficulty in doing it. However what surprises most people is why it works at all. While the principle of halving and doubling is logical and familiar enough, it simply ignores remainders and some of the rows. Here are some observations about the algorithm which help in seeing why it works.
 The ability to use the memory function on a calculator is essential to perform quick and easy calculations involving more than one operation. A calculator with a memory function will contain the following buttons.
Example 10: Using a calculator to solve, 4 x 8 + 15 x 3
Example 11: Using a calculator to solve, 40 + (100 - 77)/2 - (19 x 3)
Example
12: Using a calculator to solve,
For more information on Using a calculator see Negative Numbers, Key Ideas, Using a calculator and Percent Examples, Calculator Short Cuts.
5. Complete the multiplication tables below. Time yourself for each set and see if you can beat your time for each new table.
Further questions: 1. If a car goes 9 km on a litre of petrol how far can it go on 45 litres of petrol? 2. When a number is divided by 15, the quotient is 38. What is the number? 3. If a carton holds 36 cans of coke, how many cans of coke are there in 125 cartons? 4. At a concert there are 26 rows. Each row has 130 chairs. How many people can be seated at the concert? 5. Brenda is 4 times older than Tim and Tony is three times older than Brenda. If Tim is 7, how old is Tony? 6. Light travels at 300000 km/sec. What is the distance from the earth to the sun if light from the sun takes 8 minutes to reach the earth? 7. When a certain number is divided by 8 the answer is 12 with a remainder of 1. What was the number that was divided? 8. In a certain theatre there are 24 seats to a row. At a charity function in this theatre our group occupied 16 rows and there were 8 extra people from our group sitting in another row. How many people were in our group? 9. A car travels at 100 km/hr. How far will it travel in 12 hours? 10. A metronome is set to give 4 beats per second. How many beats will it give in one hour?
If
you would like to do some more questions, click
here to go to the mixed operations quiz at the end of the division
section.
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