Multiplication of Two Numbers

Objectives

At the end of this lesson, you should be able to:

• Multiply by powers of 10 or decompose a natural number in powers of 10.
• Multiply two entire numbers utilizing the following methods:
  • Rectangular multiplication using a concept of the sum of the areas.
  • Multiplication in a grid
  • Standard Multiplication

Decompose a natural number in powers of 10

A natural number can be decomposed like the sume of every one of their figures multiplied by a power of 10. Every power of 10 that will be multiplied at the figure will depend on the positioned value of the figure, counting right to left:

Example: Decompose 315 in powers of 10.

Solution:
        $315=(3\times {\mathrm{10}}^2)+(1\times {\mathrm{10}}^1)+(5\times {\mathrm{10}}^0)$
               $=(3\times 100)+(1\times 10)+(5\times 1)$
               $=300+10+5$

Utilize the following applet to decompose the natural numbers in power of 10.

Multiplication of two numbers decomposed in powers of 10

You decompose a number in powers so that you can facilitate the multiplication between numbers. We only need to know the basic tables of multiplication, which will be shown in a continuation:

To multiply two numbers that are multiples of 10:

1. Count the amount of zeros that are in the two factors, from the right to the left, until a figure different than zero presents itself. Then write this quantity of zeros in the solution.
2. Multiply the previous figures a the zeros that we counted and write the product before the zeros that are already written in the solution.

Example: $20\times 300$

Solution: We have two natural numbers that are multiples of 10 and don't have to decompose.

We count the quantity of zeros between the two numbers, from right to left, until we find a figure different than zero and place the total quantity of zeros in the solution, leaving a white space for when we multiply the different figures of zero.

20 × 300 = __000
Now we multiply the previous figures of the zeros, in this case the product of 2 and 3, in the space that we left white, before the zeros of the solution, and like this we complete the exercise.

20 × 300 = 6000

The decomposition of a number in powers can facilitate the multiplication between numbers. First we decompose the numbers and apply the distributive property.

Example: $36\times 45$

Solution: We decompose different numbers in power of 10.

      36 × 45 = [(3 × 10) + 6] × [(4 × 10) + 5] = (30 + 6) × (40 + 5)

Now we apply the distributive property to multiply every one of the formed factors.

      (30 + 6) × (40 + 5) = (30 × 40) + (30 × 5) + (6 × 40) + (6 × 5)

We proceed to find the product of every one of the multiplications, following the instructions in the previous example:

      (30 × 40) + (30 × 5) + (6 × 40) + (6 × 5) = 1200 + 150 + 240 + 30

Finally we sum up the results.

      1200 + 150 + 240 + 30 = 1620

In general, 36 × 45 = 1620.

Rectangular Multiplication

Find the solution of a rectangular multiplication that consists of finding the areas of all the rectangles that are formed, then to sum them up.

In the application below, you can move the buttons to practice the multiplication of numbers that are 1 digit, with many positives as well as negatives.

Observe that the regions where some numbers have the same sign produce a positive number (blue areas); while the multiplication of numbers with distinct signs produce a negative number (area red). This way we can conclude the following information in the table:

Rectangularly Multiply the sums of two numbers

We can visualize the rectangular multiplication of two numbers in the total sum of the areas of the internal rectangles, which is the total of the large rectangle. The product of the quantity of sums for every factor, gives us the quantity of internal rectangles to sum.

• The first method in finding the area of the large rectangle.

Using this technique, the corresponding area is $9\times 5=45$.

• The second method consists in finding the area of all the internal rectangles and summing them up.

In this case, the area is the same as $21+14+6+4=45$. Like the area of a similar rectangle cannot be changed, the two solutions give the same answer. Then in terms of the multiplication of binomial expressions, we can conclude that $\left(2+7\right)\left(3+2\right)=3\times 7+2\times 7+2\times 3+2\times 2=45$.

Rectangularly multiply two decomposed numbers in powers of 101.

Here we can assume that the premise that can be applied to the decomposition of numbers in powers of 10 to create sums and with this multiply them by another sum that is decomposed in the same form.

To be able to visualize it like the area of a rectangle, we should first draw the rectangle with a factor as a base and the other as the height. Then we decompose some factors to know how many internal rectangles we have to sum.

$23=(20+3)$
$31=(30+1)$
Hence, $23\times 31=(20+3)\times (30+1)$

Then the rectangle will stay segmented like a rectangle 2 × 2:

Now we find the area of every one of the internal rectangles, to later sum them all up.

In this case, the area is the same as $90+3+600+20=713$. We know this should be at every one of the sume is one of the partial products that we obtained to multiply by decomposition.

Then we can canclude that

To be able to visualize how the area of a rectangle, we should first draw the rectangle with one factor as the width and the other as the height. Then we decompose some factors to know how many internal rectangles we have to sum up.

$231=(200+30+1)$
$52=(50+2)$
Hence, $231\times 52=(200+30+1)\times (50+2)$

Then the rectangle that stays segmented like a rectangle 3 × 2:

Now we find the area of every one of the internal rectangles, to later sum them all up.

In this case, the area is the same as $50+2+1500+60+10000+400=12012$. This we know throught that every one of the sms is one of the partial products that we obtained in the multiplication by decomposition.

Then we can conclude that

To be able to visualize this like the area of a rectangle, we should first draw the rectangle with a factor as the width and another as the height. Then we decompose some factors to know how many internal rectangles we have to sum up.

$486=(400+80+6)$
$529=(500+20+9)$
In general, $486\times 529=(400+80+6)\times (500+20+9)$

Then the rectangle stays segmented like a rectangle 3 × 3:

Now we find the area of every one of the rectangles, to later sum them all up.

In this case, the area is the same as $3000+120+54+40000+1600+720+200000+8000+3600=257094$. This we know through that every one of the sum is one of the partial products that we obtained at the multiplication by decomposition.

Then we can conclude that

Multiplication in the grid

Also known as hindu multiplication, reticulated multiplication, multiplicatin of matrix; and in English as lattice multiplication. The multiplication in a grid combines the rectangular multiplication knowing that only the basic multiplications can find the product of two factors.

Elaboration of the grid

Create a grid that requires drawing a rectangular table m × n, with the variables m and n representing the number of columns and lines they should have in the table, as well as the quantity of digits that every one of the factors has. In this case, we will make a 3 × 2 grid.

Later, we will add diagonals to every one of the cells.

We will color the diagonal columns alternately.

Finally, we will add an extra box and color it the color that is at its side. Every one of the digits of the product can be found in their respective box.

You place every digit from the first factor in a column; and every digit of the second factor in a line, like is shown in the image:

Later you will find the product of every column by every line. The tens of every one of those products are going in the left side of the cell, and the units to the right. You should follow this form:

We sum up the diagonals, writing the resulting numbers in the corresponding spaces. To have to regroup them, the tens result in being summed in the following diagonal.

Then we know that 920 × 58 is the same as 53360.

In the following applet you can observe how you multiply using a grid 2 × 2:

Standard Multiplication

The standard multiplication is the traditional method utilized in classrooms for decades to find the product of two numbers. A difference of the multiplication of a grid, in the order that multiplies the numbers is very important, that we regroup and multiply garden. Also the calculations with verticales, should be that the factors are in a column.

Example: Find the product of 764 and 18.
Solution: We mentally decompose the second factor, so that now we now we have to multiply 764 × 10 and 764 × 8; then later find the sum of those partial products.

764 × 8:

4 units × 8 = 32 units = 3 tens, 2 units
We write the 2 below the line and regroup the 3

(6 tens × 8) + 3 tens = 48 tens + 3 tens = 51 tens = 5 hundreds, 1 ten
We write the 1 below the line (on the left side of the 2) and regroup at the 5.

(7 hundreds × 8) + 5 hundreds = 56 hundreds + 5 hundreds = 61 hundreds = 6 millions, 1 hundredth
Like we don't have any more numbers that we ahve to multiply by 8, the 61 hundredths lower. In general, 764 × 8 = 6112

764 × 10:

We repeat the same to multiply by 10, where we begin in the place of the tenths, assuming that we are multiplying by 1. Then we don't write the zero of the units, we know that 764 × 10 = 7640

.

Sum of partial totals:

We find the total of the two partial products and like this we know that the product of 764 and 18 is 13752.

Summary

Now that you have completed this lesson, you should be able to:

• Multiply by powers of 10 or decompose a natural number in powers of 10.
• Multiply two entire numbers utilizing the following methods:
• Multiplication of a rectangle utilizing the concept of the sum of the areas.
• Multiplication in a grid
• Standard multiplication