Speed Math for Kids Achieve Their Full Potential

(Marvins-Underground-K-12) #1

Using factors expressed as division


To multiply 48 × 96, we could use reference numbers of 50 and 100, expressed as (50 × 2) or (100 ÷ 2).
It would be easier to use (100 ÷ 2) because 100 then becomes the main reference number. It is easier to
multiply by 100 than it is by 50.
When writing the multiplication, write the number first which has the main reference number, so
instead of writing 48 × 96 we would write 96 × 48. The completed problem would look like this:


That worked out well, but what would happen if we multiplied 97 by 48? Then we have to halve an
odd number. Let’s see:


In this case we have to divide the 3 below the 97 by the 2 in the brackets. Three divided by 2 is or
1½. Subtracting 1½ from 48 gives an answer of 46½. Then we have to multiply 46½by 100. Forty-six
by 100 is 4,600, plus half of 100 gives us another 50. So, 46½ times 100 is 4,650.
Then we multiply 3 by 2 for an answer of 6; add this to 4,650 for our answer of 4,656.
Let us check this answer by casting out the nines:


We check by multiplying 7 by 3 to get 21, which reduces to 3, the same as our answer. Our answer is
correct.
What if we multiply 97 by 23? We could use 100 and 25 (100 ÷ 4) as reference numbers.


We divide 3 by 4 for an answer of ¾. Subtract ¾ from 23 (subtract 1 and give back ¼). We then
multiply by 100.


23  −   ¾   =   22¼
22¼ × 100 = 2,225 (25 is a quarter of 100)
The completed problem looks like this:
Free download pdf