Solution to

Frame 1-15

addition. For example:

4 9 9 13

9 x 4 = 9 + 9 + 9 + 9 = 36.

Also note that 9 x 4 = 4 x 9 = 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 = 36.

5003

-1 0 0 9

NOTE: When a multiplication problem is set up, the number with the

3994

most digits is usually chosen to be the multiplicand (the number on top).

4000+900+90+13

The problem 367 x 97 can also be solved by writing 367 down

-1000 -000 -00 -9

times and adding the numbers together or by writing 97 down

3000+900+90 + 4

times and adding the numbers together.

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Solution to

Like addition, you need to carry when the product is more than nine.

Frame 1-16

example: 24 x 3.

97

1

24

4 x 3 = 12.

Write the "2" and carry the 1 (really 10), then add

367

x3

2x3= 6

the number carried (1) to the product (6)

72

[really six 10's]. 6 + 1 = 7. Write down the "7".

(You can see that

multiplication takes less

Another way of stating the problem is:

time and results in fewer

math errors.)

24 = 20 +

4

x3

x3

x3

60 + 12 = 72

Now you work this problem: 3,415 x 4

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Solution to

When the multiplier has more than one digit, the multiplicand is

Frame 1-17

multiplied by each digit in the multiplier beginning with the right digit of

the multiplier. The products are added together to obtain the final

1

2

answer. Note that the right digit of the product is in the same column as

3415

the digit of the multiplier being used. You may wish to fill in the empty

x4

digit places in the product with zeros to help keep the columns straight.

13660

For example: 621 x 27.

5x4 =

20

1

1

621

621

621

621

621

10 x 4 =

40

x 27 or x 27

x27 =

x7 +

x 20

400 x 4 = 1,600

4347

4347

4347 + 12420 = 16,767

3000 x 4 = 12,000

1242

12420

13,660

16767

16767

You work this problem: 367 x 97

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MD0900

1-7

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