FRAME 1-16.
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
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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FRAME 1-17.
Solution to
Like addition, you need to carry when the product is more than nine.
Frame 1-16
Remember to add the number you carried after you multiply. For
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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FRAME 1-18.
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